TS Inter 2nd Year

Here students can locate TS Inter 2nd Year Physics Notes 3rd Lesson Wave Optics to prepare for their exam.

TS Inter 2nd Year Physics Notes 3rd Lesson Wave Optics

→ Huygens principle: Each point on the wave-front is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wave lets emanating from the wavefront are usually referred to as secondary wavelets.
From Huygens principle every wave is a secondary wave to the preceding wave.

→ Wavefront: The locus of points which oscillate in phase is called “wavefront”.
(OR)
A wavefront is defined as a surface of constant phase.

→ Plane wave: Generally wavefront is spherical in nature when radius of sphere is very large. A small portion of spherical wave can be treated as plane wave.

→ Geometrical optics: It is a branch of optics in which we are completely neglects the finiteness of the wavelength is called geometrical optics. A ray of light is defined as the path of energy propagation. In this concept wavelength of light is tending to zero.

→ Snell’s law of refraction: Let nj and n2 are the refractive indices of the two media and i’ and ‘r’ are angle of incidence and angle of refraction then
n₁ sin i = n₂ sin r. This relation is called Snell’s law.
From Snell’s law for a given pair of media sin i / sin r = n₂ / n₁ is constant also called refractive index of the medium. Where nj is air or vacuum.

→ Critical angle (I_c): It is define as the angle of incidence in denser medium i for which angle of refraction in rarer medium r = 90°.
μ = \(\frac{1}{\sin c}\)
i. e, when r = 90° then i = C in denser medium.

→ Doppler’s effect in light: When there is relative motion between source and obser¬ver then there is a change in frequency of light received by the observer.

→ Red shift: If the source moves away from the, observer then frequency measured by observer is less (i.e., wavelength increases) as a result wavelength of received light moves towards red colour. This is known as “red shift”.

→ Blue shift: When source of light is approach¬ing the observer frequency of light received increases, (i.e., wavelength of light decreases.) As a result wavelength of received light will move towards blue colour. This is known as “blue shift”.

→ Superposition principle: According to super¬position principle at a particular point in the medium the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.

→ Coherent soures: Two sources are said to be coherent when the phase difference pro¬duced by each of two waves does not change with time.
Note: For a non – coherent waves the phase difference between them changes with time.

→ Interference: Interference is based on the superposition principle. According to which at a particular point in the medium the resultant displacement produced by a number of waves is the vecotor sum of displacements produced by each of the wave.
Note: In light when two coherent waves are superposed we will get dark and bright bands.

→ Constructive interference Or bright band:
When two coherent waves of path difference λ /2 or its integral multiples or a phase difference of or integral multiples of 2π are superposed on one another then the displacements of the two waves are in phase and intensity of light is 4I₀ where I₀ is intensity of each wave.
Condition, for constructive interference
is path difference = nλ or Φ = 0, 2π, 4π ………….. etc i.e., even multiples of π.

→ Destructive interference or dark band:
When two coherent waves of path difference \(\frac{\lambda}{2}\) or (n + \(\frac{1}{2}\)) λ or phase difference of \(\frac{\pi}{2}\) or odd multiples of \(\frac{\pi}{2}\) superposed at a given point then their displacements are out of phase and resultant intensity its is zero. This is called dark band.
For dark band to from path difference
= (n + \(\frac{1}{2}\)) λ (OR)
phase difference Φ = π, 3π, 5π …. odd multiples of n.

→ Diffraction: Bending of light rays at sharp edges (say edge of blade) is called”diffraction”.
As a result of diffraction we can see dark and bright bands close to the region of geometrical shadow.

→ Resolving power of Telescope:
Resolving power of telescope Δθ = \(\frac{0.61 \lambda}{\mathrm{a}}\)
Where 2a is aperture or vertical height (dia-meter) of lens used:
Resolving power of telescopes is its ability to show two distant object clearly when angular separation between them is Δθ.
When Δθ is less then resolving power of that telescope is high.
From above equation to increase resolving power of telescope its aperture or diameter of lens used must be high.

→ Resolving power of microscope:
The resolving power of the microscope is given by the reciprocal of the minimum separation of two points seen as distinct.
Minimum distance d_min = \(\frac{1.22 \lambda}{2 n \sin \beta}\), The term n sin β is called numerical aperture.
Resolving power of microscope = \(\frac{2 \mathrm{n} \sin \beta}{1.22 \lambda}\)
Note: The resolving power of microscope increases if refractive index n is high. In oil immersion objectives the lenses are placed in a transparent oil with refractive index close to that of objective lens to increase magnification.

→ Fresnel distance: The term z = a²/ λ is called fresnel distance.
In explaining the spreading of beam due to diffraction we will use the equation z = a²/ λ,
Where after travelling a distance z\(\frac{\lambda}{\mathrm{a}}\) size of beam is comparable to size of slit or hole ‘a’.
Note: When we are travelling from aperture to screen a distance z then width of diffrated beam z\(\frac{\lambda}{\mathrm{a}}\) is equals to aperture a’.
Beyond this distance ‘z’ divergence of the beam of width ‘a’ becomes significant. When distances are smaller than z spreading due to diffraction is small when compared with size of beam.

→ Polarisation: If is a process in which vib-rations of electric vectors of light are made ot oscillate, in a single direction.

→ Polaroids: A polaroid consists of a long chain of molecules aligned in a particular direction.

→ Malus’ Law: Let two polariods say P₁ and P₂ are arranged with some angle ‘θ’ between their axes. Then intensity of light coming
I = I₀ cos²θ
where I₀ is intensity of polarised light after passing through 1st polaroid P₁. This is known as Malus Law.

→ Uses of Polaroids: Polaroids are used

• to control intensity of light.
• They are used in photography,
• polaroids are used in sunglasses and in window panes.

→ Unpolarised light: In an unpolarised light electric vectors can vibrate in 360° direc¬tion perpendicular to direction of propaga¬tion. All these electric vectors can be grouped into two groups.

• Dot components they are vibrating perpendicular to the plane of the paper.
• Arrow components ‘↔’ i.e., their plane of vibration is along the plane of the paper.
  Thus an unploarised light can be shown as a group of dot components and arrow components.

→ Polorisation by scattering: The sky appears blue due to scattering of light. The light coming from clear blue portion of sky is made to pass through a polariser when it is rotated the intensity of light coming from polariser is found to be changing. Which shows that the scattered light consists of polarised light.

→ Polarisation by reflection:

• When unpolarised light falls on the boundary layer sepa-rating two transparent media the reflected light is found to be partially polarised. The amount of polarisation depends on angle of incidence i.
• It is found that when reflected ray and refracted ray are perpendicular the reflected ray is found to be totally plane polarised. The angle of incidence at this stage is known as Brewster angle.

→ Brewster’s angle: When reflected ray and refracted ray are mutually perpendicular then reflected ray is plane polarised. This particular angle of incidence i_B for which the reflected ray is plane polarised is called Brewster angle.

Explanation:
At Brewster angle i_B + r = \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \left(\pi / 2-\mathrm{i}_{\mathrm{B}}\right)}\)
n (OR) μ = \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \left(\pi / 2-\mathrm{i}_{\mathrm{B}}\right)}\)
= \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\cos \mathrm{i}_{\mathrm{B}}}\) = tani_B (OR) μ = tan i_B
∴ The tangent of Brewster’s angle tan (i_g) is equals to refractive index, i.e., μ = tan i_B.
Note: Refractive index can be shown by the symbol μ or n.

→ From the super position principle the resultant displacement is y = y₁ + y₂.
For constructive interference (bright band) y = y₁ + y₂) ; Intensity I = (y₁ + y₂)²
For destructive interference (dark band) y = y₁ ~ y₂; Intensity I = (y₁ – y₂)²

→ In interference, the resultant intensity
I = 4I₀ cos²\(\frac{\phi}{2}\) (Where I₀ is maximum intensity)
Resultant phase θ = \(\frac{\phi}{2}\) (Where Φ is initial phase difference)

→ Condition for formation of bright band is
(a) Path difference x = mλ, where m = 0,1, 2 ………….. etc.
(b) Phase difference Φ = 0, 2π ………… even multiples of π.

→ Condition for formation of dark band
(a) Path difference x = \(\frac{λ}{2}\) and odd multiples
(b) Phase difference Φ = π, 3π, 5π, ………….. odd multiples of n.

→ Relation between path difference (x) and
phase difference (Φ) is λ = \(\frac{\lambda}{2 \pi}\). Φ

→ Fringe with β = \(\frac{\lambda L}{\mathrm{~d}}\); Angular fringe width \(\frac{\beta}{L}=\frac{\lambda}{d}\)

→ Distance of mth bright band from central bright band is x₂ = \(\frac{\mathrm{m} \lambda \mathrm{L}}{\mathrm{d}}\)

→ Distance of m dark band from central dark band x₂ = \(\left(m+\frac{1}{2}\right) \lambda \frac{L}{d}\)

→ For two waves with intensities I and 12 with phase 4 resultant intensity
I = I₁ + I₂ + 2\(\sqrt{I_1 I_2}\)cos Φ

→ When a glass plate of thickness (t) is introduced in the path of one light wave then interference fringes will shift. Thickness glass plate t = \(\frac{m \lambda}{(\mu-1)}\)
m = number of fringes shifted;
λ = wavelength of light used.

→ When unpolarized light of intensity I, passes through a polarizer Intensity of emergent light I = \(\frac{\mathrm{I}_0}{2}\)

→ When polarized light falls on a polarizer with an angle θ to the axis then Intensity of refracted light I = I₀cos²θ

→ If polarized light falls on 1st polarIzer with an angle θ, and angle between the axes of given two polarizers is θ then intensity of light coming out of 2nd polarlzer I = I₀cos²θ₁cos²θ₂

→ For polarizatIon through reflection wIth Brewster angle i_B then μ or n = tan i_B

→ In diffraction radIus of central bright region
r₀ = \(\frac{1.22 \lambda \mathrm{f}}{2 \mathrm{a}}=\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)

→ ResolvIng power of telescope Δθ = \(\frac{0.61 \lambda}{\mathrm{a}}\)
Where Δθ Is the minimum angular separation between two distant objects.

→ Resolving power of microscope
d_min = \(=\frac{1.22 \mathrm{f} \lambda}{\mathrm{D}}=\frac{1.22 \lambda}{2 \tan \beta}=\frac{1.22 \lambda}{2 \sin \beta}=\frac{1.22 \lambda}{2 \mathrm{n} \sin \beta}=\frac{1.22 \lambda}{2 \mathrm{~N} \cdot \mathrm{A}}\)
Where d_min is the minimum separation required between two points in object.
N.A is numerical aperture (n sin β)
β is the angle subtended by the object at object lens.

→ Fresnel distance z = \(\frac{\mathrm{a}^2}{\lambda}\) Where ‘a’ is size of hole or slit.

Here students can locate TS Inter 2nd Year Physics Notes 2nd Lesson Ray Optics and Optical Instruments to prepare for their exam.

TS Inter 2nd Year Physics Notes 2nd Lesson Ray Optics and Optical Instruments

→ In ray optics, light will travel from one point to another point along a straight line. The path is called a “ray of light”. A bundle of such rays is called “a beam of light”.

→ Laws of reflection:

• Angle of reflection is equals to angle of incidence (r = i).
• The incident ray, the reflected ray and normal to the reflecting surface lie in the same plane.

→ Spherical mirrors:

• Pole (P): The geometrical centre of spherical mirror is called ‘pole ‘P’.”
• Principal axis: The line joining the pole (P) and centre of curvature ‘C’ of a spherical mirror is known as “principal axis”. Principal focus: For mirrors, after reflection a parallel beam of light w.r.to principal axis will converge or appears to diverge from a point on principal axis. This point is called “principal focus” ‘F’.
• Focal length (f): Distance between principal focus and pole of mirror or centre of lens is called “focal length”.

In spherical mirrors:

• Focal length of concave mirror is positive.
• Focal length of convex mirror is ve1.
• Relation between radius of curvature of mirror R and focal length f is R = 2f

→ Cartesian sign convention:

• All distances must be measured from the pole of the mirror or the optical centre of lens.
• The distances measured along the direction of the incident ray are taken as positive and the distances measured against the direction of incident ray are taken as negative.
• Distances measured above the principal axis of mirror (or) lens are taken as positive. The distances measured below are taken as ve’.

→ Mirror equation = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

→ Linear Magnification (m): The ratio of height of image (ht) to height of object (hj is called “linear magnification”. Linear magnification (m) = \(\frac{\mathbf{h}_{\mathrm{i}}}{\mathrm{h}_{\mathrm{o}}}=\frac{-\mathrm{v}}{\mathbf{u}}\)

→ Refraction: The bending nature of light rays at refracting surface while travelling from one medium to another medium.

→ Laws of refraction:

• The incident ray, the refracted ray and normal to the interface at the point of incidence all lie in the same plane.
• Snell’s law: The ratio of sine of angle of incidence to sine of angle of refraction is constant for a given pair of media. sin i / sin r = n₂₁.
  where n₂₁ = refractive index of 2nd medium w.r.to 1st medium.
• When light rays are travelling from rarer medium to denser medium they will bend towards the normal.
• When light rays are travelling from denser medium to rarer medium they will bend away from normal.

→ Total internal reflection: When light rays are travelling from denser medium to rarer medium for angle of incidence i > i_c light rays are notable to penetrate the boundary layer and come back into the same medium. This phenomena is known as “total internal reflection”.

→ Critical angle (ic): The angle of incidence in the denser medium for which angle of refraction in rarer medium is 90° is called “critical angle” of denser medium.

→ Applications: Due to total internal reflection:

• Formation of mirages on hot summer days on tar roads and in deserts.
• Sparkling of well cut diamonds.
• Prisms designed to bend light rays by 90° or by 180° make use of total internal reflection.
• Optical fibre provides loss less trans-mission overlong distances with total internal reflection.

→ Refraction through lenses:
Lens formula I = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Lens makers formula \(\frac{1}{f}\) = (n₂₁ – 1) \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
For any curved spherical surface \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}}\)

→ Power of a lens (P): Power of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre.
i.e, tan δ = \(\frac{h}{f}\) when δ is small tan δ = δ
δ = \(\frac{h}{f}\) (or) P = \(\frac{1}{f}\)
Power P = j unit: dioptre. Here, f is in meters.

→ Lens combination: When lenses are in contact then their combined focal length
\(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}\)+…………
Combined power P = P₁ + P₂ + ……………

→ Magnification of combined lens is
m = m₁ × m₂ × m₃ i.e., total magnification is the product of individual magnifications of lenses on that combination.

→ Refraction through prism: In prism path of light ray inside prism is parallel to prism base.
Angle of prism, A = r₁ + r₂.
Angle of deviation δ = i + e – A
where e is angle of emergence.
At minimum deviation position
r₁ = r₂ ⇒ i₁ = r₂
r = \(\frac{A}{2}\) and I = \(\frac{\left(\mathrm{A}+\mathrm{d}_{\mathrm{m}}\right)}{2}\)
Refractive index n₂₁ = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\)
= \(\frac{\sin \left(\mathrm{A}+\mathrm{d}_{\mathrm{m}}\right) / 2}{\sin \mathrm{A} / 2}\)
For small angled prism dm = (n₂₁ – 1) A.

→ Dispersion: The phenomenon of splitting of light into its constitute colours is known as “dispersion”.
Dispersion takes place due to change in refractive index of medium for different wave lengths.

Note:

• The bending of red component of white light is least due to its longest wavelength.
• Bending of violet component of white light is maximum due to its short wave length.
• In vacuum all colours will travel with same velocity and velocity in vacuum is independent of wavelength of light.

→ Rainbow: Rainbow is due to dispersion of white light through water drops.

→ Scattering: Change in the direction of light rays in an irregular manner is called”scatte-ring”.
Ex: Sun light gets scattered by atmospheric particles.
Note: Light of shorter wavelengths will scatter much more than light of longer wave-length. Due to this, reason blue light will scatter more than red light.

→ Rayleigh scattering: The amount of scattering is inversely proportional to fourth power of the wavelength.
Ex: Blue has shorter wavelength than red. Hence blue will scatter much more strongly. Note: Wavelength of violet is shorter than blue. So violet will scatter more than blue. But our eye is not sensitive to violet. So we will see the blue colour in sky.

→ Human eye: Human eye contains rods and cones. Rods will respond to intensity of light.

→ Accommodation of eye: Our eye consists of a crystalline lens. The curvature and focal length of this lens is controlled by ciliary muscles. When ciliary muscles are relaxed focal length of eye lens is high. So light rays from distant object are focused onto the retina while viewing the near by objects eye lens will become thick and focal length will become less.
The property of eye lens to change its focal length depending on the objects to be viewed is called “accommodation”.

→ Hypermyopia: It is one type of eye defect. Where light rays coming from distant object are converged at a point infront of retina. This defect is called “short sighted-ness or Hypermyopia”. This defect can be compensated by using a concave lens.

→ Hypermetropia: This is one type of eye defect. In this defect eye lens focusses the incoming light at a point behind the retina. This is called “far sightedness or Hyper-metropia”. This defect can be compensated by using a convex lens.

→ Astigmatism: In this type of eye defect cornea will have large curvature in vertical direction and less curvature in horizon¬tal direction. A person with this type of defect can not clearly see objects because carnia has non-uniform radius.
This defect can be reduced by using cylindrical lenses of desired radius of curvature.

→ Simple microscope: A simple convex lens can be used as a simple microscope. Given object is placed between centre of lens and its focus. A virtual image is made to form at near point.
Near point magnification m = (1 + D/f).
If final image is made to form at infinity (far point).
Far point magnification m = D / f.

→ Compound microscope: Magnification of microscope
M = m₀ . m_e = \(\frac{v}{u}\left(1+\frac{D}{f_e}\right)\)
Or
m ≈ \(\frac{L}{f_e} \frac{D}{f_e}\), m₀ = \(\frac{h^{\prime}}{h}=\frac{v}{u}\)

For near point me = (1 + D/f_e)
For far point m = D/f_e.

→ Telescopes: It consists of two convex len¬ses mounted coaxially. Telescope is used to see large objects which are very far away.
Object lens will form a point size real image of object in between the lenses. This first image will act as an object to eye lens. Generally eye lens will form final image at infinity. This is called “Normal adjustment”.

Length of telescope L = f
Manification m = \(\frac{\beta}{\alpha}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
Reflecting telescopes are also called “cassegrain telescopes”.
In Telescopes if final image is inverted it is called Astronomical telescope. If final image is erect with respect to object it is called terrestrial telescope.

→ Velocity of light ¡n vacuum c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
= 3 × 10⁸ m/s
where μ₀ = permeability and E₀= permittivity of vacuum or air (free space).
In a medium the velocity of light (c) = \(\frac{v_1}{v_2}=\frac{v \lambda_1}{v \lambda_2}=\frac{\lambda_1}{\lambda_2}\)
also c = υλ

→ Refractive index of a medium μ = c / V;
where y = velocity of light in medium
From Sneils law μ = sin i / sin r.

→ In refraction or reflection the frequency of light remains constant
₁μ₂ or n₂₁ = \(\frac{v_1}{v_2}=\frac{v \lambda_1}{v \lambda_2}=\frac{\lambda_1}{\lambda_2}\)
Wavelength In medium λ_med = \(\frac{\lambda_{\text {vacuum }}}{\mu(\text { or }) n}\)

→ In refraction n₁λ₁ = n₂λ₂ (In refraction frequency of light υ is constant)

→ Relative refractive index n_r = \(\frac{\mathrm{n}_1}{\mathrm{n}_2}\)
Ex: _wn_g = \(\frac{\mathbf{n}_{\text {glass }}}{\mathbf{n}_{\text {water }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {glass }}}\)

→ In prism

• r₁ + r₂ = A (angle of prism);
• i₁ + i₂ = A + D (angle of deviation) (or) D = (i₁ – i₂) – A
• n = sin \(\frac{\mathrm{A}+\delta}{2}\)/sin(A/2)where δ = angle of minimum deviation.
• At minimum deviation position i₁ = i₂ and r₁ = r₂.
  So r = A/2 and i = \(\frac{\mathrm{A}+\delta}{2}\)

→ For small angled prisms 6 = (n – 1) A.

→ Relation between critical angle ‘c’ and refractive index n is
n = \(\frac{1}{\sin c}\) or c = sin^-1\(\left[\frac{1}{n}\right]\)

→ Lens makers formula \(\frac{1}{f}\) = (n – 1) \(\left[\frac{1}{n}\right]\)

→ Relation between object distance ‘u’, image distance y and focal length f is = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

→ In lens combination \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\) when in contact \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}\) when lenses are separated by a distance ‘d.

→ Condition to eliminate spherical aberration by lens combination is separation
d = f₁ – f₂.

→ Conditions to eliminate chromatic aberration:

• In lens combination with separation
  d = \(\frac{\mathrm{f}_1+\mathrm{f}_2}{2}\)
• In achromatic doublet \(\frac{\omega_1}{\mathrm{f}_1}+\frac{\omega_2}{\mathrm{f}_2}\) = 0

→ In simple microscope

• Near point magnification m = \(\left(1+\frac{D}{f}\right)\)
• When image is at Infinity m = \(\frac{D}{f}\)
  where D is the least distance of distinct vision.

→ In compound microscope

• Total magnification m = m₀ × m_e or m = \(\frac{v_0}{u}\left(1+\frac{D}{f_e}\right)\)
• When final Image is at infinity m = \(\frac{L D}{f_0 f_e}\)
• Object lens will form a real image, so m₀ = \(\frac{\mathrm{v}_0}{\mathrm{u}_0}\)
• Eye lens will form a virtual image at near point; so m₀ = (1 + \([latex]\)[/latex])

→ In Telescopes
(a) magnification m = \(\frac{\alpha}{\beta}=\frac{\tan \alpha}{\tan \beta}\) or m = \(\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
At normal adjustment, magnification
m = \(-\frac{\mathrm{f}_0}{\mathrm{f}_e}=\frac{\text { focal length of object lens }}{\text { focal length of eye lens }}\)

(b) When final image is at near point
m = \(-\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{D}}\right)\)

(c) Length of telescope L = f₀ + f_e

(d) In terrestrial telescope length of telescope L = f₀ + f_e + 4f.

→ Dispersive power of prism
w = \(\frac{d_v-d_R}{\frac{d_v+d_g}{2}}=\frac{d_v-d_g}{d}=\frac{\mu_v-\mu_g}{(n-1)}\)
where d_v = Deviation of violet colour;
d_g = Deviation of red colour
\(\frac{\mathrm{d}_{\mathrm{v}}+\mathrm{d}_{\mathrm{g}}}{2}\) = Average deviation

→ Angular dispersive power to = (n_v – n_R)A

→ In prisms the condition for no dispersion is A₁ (n_g – n_R)₁ + A₂ (n_g – n_R)₂ = 0

→ In prisms the condition for no deviation is A₁ (n₁ – 1) + A₂ (n₂ – 1) = 0

→ In lens combination power of combined lens is P = P₁ + P₂ + ……………………

Here students can locate TS Inter 2nd Year Physics Notes 1st Lesson Waves to prepare for their exam.

TS Inter 2nd Year Physics Notes 1st Lesson Waves

→ Wave: A wave is a physical manifestation of disturbance that propagates in space.

→ Transverse waves: In these waves, the con-stituents of the medium will oscillate perpen-dicular to the direction of propagation of the wave.

→ Longitudinal waves: In these waves the constituents of the medium will oscillate paral¬lel to the direction of propagation of the wave.

→ Wave motion can be represented as a func¬tion of both position ‘x’ and time ‘t’.

→ Generally for x – ‘+ve’ direction equation of a wave is y = a sin (kx – ωt – Φ)

→ Crest: It is a point of a maximum positive displacement.

→ Trough: It is a point of maximum negative displacement.

→ Amplitude (a): The maximum displacement of constituents of the medium from means position is called “amplitude” ‘a’.
a = y_max

→ Phase (Φ) Phase gives the displacement of the wave at any position and at any instant.

→ Initial Phase: At initial condition (when x = 0 and t = 0) phase of the wave is called initial phase.

→ Wavelength (λ): The minimum distance between any two successive points of same phase on wave is called “wavelength” A.

→ Propagation constant (OR) angular wave number (k):
\(\frac{2 \pi}{\lambda}\) or A = \(\frac{2 \pi}{\mathrm{k}}\) where ‘A’ is A k

→ Time period (T): Time taken to produce one complete wave (or) time taken to complete one oscillation is known as “Time period T”.

→ Frequency ‘υ’: Number of waves produced per second. (OR) Number of oscillations com¬pleted per second is known as “frequency υ.”
Frequency υ = \(\frac{1}{\mathrm{~T}}\) (Or) Time period T = \(\frac{1}{v}\)

→ Angular frequency (or) velocity (ω):
Angular frequency (w) = 2πυ (or) ω = \(\frac{2 \pi}{\mathrm{T}}\)

→ Relation between velocity v, wavelength A and frequency o is v = υλ (OR) v = λ/T.

→ Speed of a wave in stretched strings:
Speed of transverse wave in stretched wires v =\(\sqrt{\mathrm{T} / \mu}\)
where µ = linear density = mass / length. S.I. unit = kg/metre.

→ Speed of longitudinal waves in different media
In liquids:
v = \(\sqrt{\frac{B}{\rho}}\)
where B = Bulk modulus
ρ = Density.

→ In solids:
v = \(\sqrt{\frac{Y}{\rho}}\)
where Y = Young’s modulus
ρ = Density.

In gases:
According to Newton’s formula
v = \(\sqrt{\frac{P}{\rho}}\)
where P = pressure of the gas and
ρ = density of the gas.
According to Newton – Laplace’s formula
v = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\) where γ = the ratio of specific heats of the gas.

→ Principle of superposition: If two or more waves moving in the medium superposes then the resultant wave form is the sum of wave functions of individual waves.
i.e y = y₁ + y₂ + y₃ + ………

→ Stationary waves (or) standing waves :
When a progressive wave and reflected wave superpose with suitable phase a steady wave pattern is set up on the string or in the medium.
A standing wave is represented by y (x, t) = 2 a sin kx cos ωt.

→ Fundamental mode : The lowest possible natural frequency of a system is called “fundamental mode.”

→ Frequency of fundamental mode is called fundamental frequency (or) first harmonic.

→ Harmonics: Sounds with frequencies equal to integral multiple of a fundamental frequency (n) are called “harmonics.”

→ Resonance : It is a special condition of a system where frequency of external periodic force is equal to or almost equal to natural frequency of a vibrating body.

→ Beats: When two sounds of nearly equal frequencies are produced together they will pro-duce a waxing and waning intensity of sound at observer. This effect is called Heats.”
Beat Frequency Δυ = υ₁ ~ υ₂
Beat Period T = \(\frac{1}{v_1 \sim v_2}\)

→ Doppler’s effect: The apparent change in frequency of sound heard due to relative motion of source and observer is called” Doppler’s effect.”

→ Doppler’s effect is applicable to mechanical waves and also to electromagnetic waves. In sound it is “asymmetric” whereas in light it is “symmetric”.

→ Velocity of sound in a medium v = υλ
where υ = \(\frac{1}{T}\)

→ Propagation constant of wave (k) = \(\frac{2 \pi}{\lambda}\) (Also k is known as angular wave number) Angular velocity of wave (©) = \(\frac{2 \pi}{\lambda}\) = 2πυ; frequency υ = \(\frac{1}{T}=\frac{2 \pi}{\omega}\)

→ Equation of progressive wave in x-positive direction is
y’= a sin (ωt – kx) (or) y = a cos (ωt – kx)
Along – ve direction on X-axis y = a sin (ωt + kx) (or) y = a cos (ωt + kx)

→ From the superposition principle, the dis-placement of the resultant wave is given by y = y₁ + y₂

→ Equation of stationary wave is
y = 2 A sin kx cos ωt or Y = 2A kx sin ωt Here kx and ωt are in separate trigonometric functions.

→ In stretched wires of string
(i) Velocity of transverse vibrations
v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
where T = tension applied
and ρ = linear density,

(ii) Fundamental frequency of vibration
υ₀ = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)

→ The laws of transverse vibrations in stretched strings

• 1st law, υ ∝ \(\) (OR) \(\frac{v_1}{v_2}=\frac{l_2}{l_1}\)
• 2nd law υ ∝ √T (OR) \(\frac{v_1}{v_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}\)
• 3rd law υ ∝ \(\frac{1}{\sqrt{\mu}}\) (OR) \(\frac{v_1}{v_2}=\sqrt{\frac{\mu_2}{\mu_1}}\)

→ Newton’s equations for velocity of sound in different media.
1. In solids υ_s = \(\sqrt{\frac{Y}{\rho}}\)
Y = Young’s modulus of wire

2. In liquids υ₁ = \(\sqrt{\frac{B}{\rho}}\)
B = Bulk modulus of liquid

3. In gases υ_g = \(\sqrt{\frac{\mathrm{P}}{\rho}}\)
P = Pressure of the gas Laplace corrected the formula for velocity of sound in gases as υ_g = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)
where γ = \(\frac{C_P}{C_V}\)
γ = Ratio of specific heats of a gas.
Where Y = Young’s modulus of solid,
K = Bulk modulus of the liquid and P is pressure of the gas.

→ In case of closed pipes

• Length of pipe at the fundamental note is l = \(\frac{\lambda}{4}\) ⇒ λ = 4l
• Fundamental frequency of vibration
  υ = \(\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{4 l}\), υ’ = 3υ, υ” = 5u.
  υ’ and υ” are second harmonic and third harmonics.
• Closes pipes will support only odd har-monics.
  Ratio of frequencies or harmonics is 1: 3: 5: 7 etc.

→ In case of open pipes .
1. Length of pipe at the fundamental note is l,
l = \(\frac{\lambda}{2}\) ⇒ λ = 2l

2. Fundamental frequency of vibration
υ = \(\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}\)
υ’ = 2υ, υ” = 3υ.
υ’ and υ” are second and third harmonics.

3. Open pipe will support all harmonics of a fundamental frequency.
Ratio of frequencies = 1: 2: 3: 4

→ Beat frequency Δυ = υ₁ ~ υ₂

→ When a tuning fork is loaded, its frequency of vibration decreases.
Due to loading, beat frequency decreases ⇒ frequency of that fork υ₁, < υ₂ (2nd fork).
When a tuning fork is field then its frequency of vibrating increases.
Due to filing beat frequency increases ⇒ frequency of that fork υ₁ > υ₂ (2nd fork).

→ General equation for Doppler’s effect is
υ’ = \(\left[\frac{v \pm v_0}{v \mp v_s}\right]\)υ
When velocity of medium (v^ is also taken into account apparent frequency
υ’ = \(\left[\frac{v+v_0 \pm v_m}{v \mp v_s \mp v m}\right]\)υ Sign convention is to be applied.
Note: In sign convension direction from observer to source is taken as + ve direction of velocity.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 1st Lesson Waves Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 1st Lesson Waves

Very Short Answer Type Questions

Question 1.
What does a wave represent?
Answer:
A wave is a disturbance which moves through a medium.

Waves transmit energy without trans-mitting matter. So waves will carry energy from one place to another place.

Question 2.
Distinguish between longitudinal and transverse waves.
Answer:
Longitudinal wave :
If the particles of the medium vibrate parallel to the direction of propagation of the wave then that wave is called “longitudinal wave”.
Ex : Sound waves in air.

These waves can be produced in solids, liquids and gases.

Transverse wave :
If the particles of the medium vibrate perpendicular to the direction of propagation of the wave, then that wave is called “transverse wave”.
Ex : Waves on a stretched string.

These waves can be produced only in solids.

Question 3.
What are the parameters used to describe a progressive harmonic wave?
Answer:
For a progressive wave (y) = a sin (ωt – kx)
The parameters in the above equation
1) a = Amplitude
2) ω = Angular velocity
3) υ = Frequency
4) T = Time period
5) λ = Wavelength
6) v = Velocity (v)
7) Φ = Phase
8) K = Propagation constant.

Question 4.
Obtain an expression for the wave velocity in terms of these parameters.
Answer:
Wave velocity :
It is the distance travelled by the disturbance (energy) along the wave in one second. It is represented by V.
Wave Velocity ‘v’

Question 5.
Using dimensional analysis obtain an expression for the speed of transverse waves in a stretched string.
Answer:
Speed of transverse wave in a stretched string depends on tension (T) and linear density (µ).
Let v ∝ T^a µ^b ;
Dimension of velocity (v) = LT^-1
Dimension of tension (T) = MLT^-2 ;
Linear density (p) = ML^-1
∴ v = K T^a µ^b
Where k is a dimensionless constant.
LT^-1 = K (MLT^-2)^a (ML^-1)^b
= M^a L^a T^-2a M^b L^-b
M°L¹T^-1 = M^{a + b} L^a-b T^-2a
Equating the powers of mass, length and time,
a + b = 0, a – b = 1 ⇒ – 2a = – 1

Question 6.
Using dimensional analysis obtain an ex-pression for the speed of sound waves in a medium.
Answer:
Speed of sound depends on wavelength and time period. v ∝ λ^a T^b
Dimensions of Velocity (v) = LT^-1;
Dimensions of Wavelength (λ) = L;
Dimensions of Time period (T) = T;
L¹T^-1 = L^aT^b
a = 1, b = – 1 ; v = λ¹ T^-1
v = \(\frac{\lambda}{\mathrm{T}}\)

Question 7.
What is the principle of superposition of waves?
Answer:
When two waves are pulses overlap at a point the resultant displacement is the al¬gebraic sum of displacements due to each wave. Their resultant is also a wave.
y = (y₁+ y₂)

Question 8.
Under what conditions will a wave be reflected?
nswer:
Wave well be reflected if it falls on a rigid surface. Because at rigid surface the particles of medium does not vibrate.

If a wave falls on the interface of two different elastic media, than a Part of two different elastic media, than a part of wave is reflected and a part of incident wave will be refracted. During refraction they obey Snell’s Law.

Question 9.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary?
Answer:
At rigid boundary phase difference between the incident and reflected wave = 180° (or) (π). Because at rigid boundary a node is formed.

Question 10.
What is a stationary (or) standing wave?
Answer:
When two progressive waves having same wavelength, amplitude and frequency travelling in the medium in opposite directions superposed stationary waves are formed.

Question 11.
What do you understand by the terms ‘node’ and ‘antinode’?
Answer:
Node :
The point where the displacement is minimum (zero) of a wave is called Node.

Antinode :
The point where the displacement is maximum of a wave is called Antinode.

Question 12.
What is the distance between a node and an antinode in a stationary wave?
Answer:
The distance between a node and an antinode = \(\frac{\lambda}{\mathrm{4}}\)

Question 13.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’?
Answer:
Natural frequency :
Vibrations produced by a body with elastic properties due to application of a constant force are known as natural frequency.
Ex : For a tuning fork natural frequency depends on elastic nature of the material, the mass distribution and the dimensions of the prongs of the fork.

Question 14.
What are harmonics?
Answer:
A harmonic is defined as a ‘tone’ of sound having a frequency which is an integral multiple of the fundamental frequency.

Question 15.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string?
Answer:
The fundamental frequency of vibration and their harmonics are possible if a string is stretched between two rigid supports. If T is the natural frequency of vibration of the string, then possible their harmonics are 2f, 3f, 4f so on.

Question 16.
The air column in a long tube, closed at one end, is set in vibration. What harmonics are possible in the vibrating air column?
Answer:
If the fundamental frequency of the air column is denoted by f, then the frequencies at which the second, third, fourth and later modes occur are 3f, 5f, 7f …………… (2n – 1) f. A closed pipe will support only odd Har-monics.

Question 17.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible?
Answer:
If the tube is open at both the ends is set in vibration, the frequencies of the harmonics present in an open pipe are integral multiples of fundamental frequency of the air column. Let f is fundamental frequency then possible harmonics are f, 2f, 3f…. etc.

Question 18.
What are ‘beats’?
Answer:
Beats :
When two sounds of nearly equal frequency are superposed, they will create a waxing and warning intensity of sounds. This affect is called “beats”. Beats are produced due to interference of sound waves.
Beat frequency υ_beat = υ₁ – υ₂

Question 19.
Write down an expression for beat frequency and explain the terms therein.
Answer:
Beat frequency (υ_beat) = υ₁ – υ₂
Where υ₁ and υ₂ are the frequencies is of the two sound waves.

Question 20.
What is ‘Doppler effect’? Give an example.
Answer:
The apparent change in the frequency of source of sound due to relative motion between the source and observer is known as doppler’s effect.
Ex : The whistle of an approaching train appears to have high pitch. When the train is moving away pitch of its whistle decreases.

Question 21.
Write down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.
Answer:
When source and observer are moving in the same direction equation for observed
(or) apparent frequency υ = υ₀ (\(\frac{v+v_{0}}{v+v_{s}}\))

Short Answer Questions

Question 1.
What are transverse waves? Give illustrative examples of such waves.
Answer:
Transverse Waves :
In these waves, the particles of the medium vibrate perpendicular to the direction of propagation of the wave.

These waves can propagate through solids and liquids.

Let a rope or string fixed at one end. At the other end continuous periodic up and down jerks are given by a external agency then transverse waves are produced.
Example:
Vibrations in strings, ripples on water surface and electromagnetic waves.

Question 2.
What are longitudinal waves? Give illustrative examples of such waves.
Answer:
Longitudinal waves :
In these waves particles of the medium vibrate parallel to the direction of propagation of the wave.

• These waves can propagate through solids and liquids and gases.
• A longitudinal wave travels in the form of compression and rarefaction.
• These waves are also known as “Compression waves”.
• In air sound waves are longitudinal waves.

Example :
Let a long pipe filled with air has a piston at one end. If the piston is pushed and pulled continuously then a series of compressions and rare fractions are produced. It represents a longitudinal wave.

Question 3.
Write an expression for a progressive harmonic wave and explain the various parameters used in the expression.
Answer:
Equation for progressive harmonic wave towards positive x-direction.

where y = displacement at any given time
A = amplitude of wave;
ω = angular velocity
along negative x-direction
y = Asin(ωt + kx)
General expression for wave motion y = A sin (ωt ± kx)

Question 4.
Explain the modes of vibration of a stretched string with examples.
Answer:
Equation of fundamental frequency :
If the wire of length ‘l’ is stretched between points ‘A’ and ‘B’ with tension T vibrates as a single loop then the frequency of the vibrations is known as fundamental frequency and is denoted by ‘υ’.

Than length of the wire (l) = \(\frac{\lambda}{\mathrm{2}}\) ⇒ λ = 2l.
But velocity of a wave (v) = υλ ……………. (1)
In case of stretched strings
v = \(\sqrt{\frac{T}{\mu}}\) …………. (2)
where v is the velocity of transverse vibrations in the string.
T = the tension.
µ = the linear density of the string,
υλ = \(\sqrt{\frac{T}{\mu}}\) (or) υ = \(\frac{1}{\lambda}\) \(\sqrt{\frac{T}{\mu}}\) From (1)
When one loop is formed
∴ υ = \(\frac{1}{2l}\)\(\sqrt{\frac{T}{\mu}}\) (∵ λ = 2l)
If the wire vibrates for p loops, then the frequency υ_p = \(\frac{p}{2l}\)\(\sqrt{\frac{T}{\mu}}\)

Laws of Transverse vibrations in a stretched string:

I. First Law (or) Law of length :
The fundamental frequency of a vibrating string is inversely proportional to the length of the string, when the tension (T) in the string and linear density p are constant.
υ ∝ \(\frac{1}{l}\) ⇒ υl = constant
when T and ‘µ’ are constant.

II. Second law (or) Law of Tension :
The fundamental frequency of a vibrating string is directly proportional to the square root of stretching force T (Tension), when the length of the string and linear density ‘µ’ are constant.
υ ∝ √T ⇒ \(\frac{υ}{\sqrt{T}}\) = constant
when T and ‘µ’ are constant.

III. Third law (or) Law of linear density (or) Law of mass :
The fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (µ) when the length (l) and tension (T) in the string are constant.

when ‘l’ and ‘T’ are constant

Examples :
An exciting tuning fork, the plucked wire of a stringed instrument, a bell struck with a hammer, a vibrating air column in a trumpet are some of examples of stretched string.

Question 5.
Explain the modes of vibration of an air column in an open pipe.
Answer:
Harmonics in open pipe: In the fundamental mode of vibration, an antinode is formed at each end with a node formed between them.
If ‘l’ is the vibrating length and λ₁ is the corresponding wavelength.
l = \(\frac{\lambda_{1}}{2}\),
The frequency of fundamental mode
(or) 1st harmonic υ₁ = \(\frac{v}{\lambda_{1}}\) ∴ υ₁ = \(\frac{v}{2l}\)

Second Harmonic (or) first overtone :
It will have three antinodes and two nodes.
If λ₂ is the corresponding wavelength

Third harmonic (or) Second overtone :
It will have four antinodes and three nodes. If λ₃ is the corresponding wavelength,

In open pipe harmonics are in the ratio of 1 : 2 : 3 …………..

Question 6.
What do you understand by ‘resonance’? How would you use resonance to deter-mine the velocity of sound in air?
Answer:
Resonance :
If the natural frequency of the body coincides with frequency periodic force impressed on it, then the body is said to be at resonance.

At resonance, the body vibrates with increasing amplitude.

Determination of velocity of sound using Resonance :
Consider a closed tube where air column length can be changed.

For the First resonance :
Length of air column is equal to = l₁ + e = \(\frac{\lambda}{4}\) → (1)

where e is endcorrection

For the Second resonance :
Length of air column is equal to = l₂ + e = \(\frac{3\lambda}{4}\) → (2)

Length of the second resonating air column is approximately equal to three times the length of first resonating air column.
(2) – (1)
l₂ – l₁ = \(\frac{3\lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2}\)
⇒ λ = 2(l₂ – l₁)

Velocity of sound in air
v = υλ
v = 2υ(l₂ – l₁)
where υ = the frequency of tuning fork, l₁, l₂ = first and second resonating lengths.

Quetion 7.
What are standing waves? Explain how standing waves may be formed in a stretched string.
Answer:
Stationary waves (or) standing waves :
When a progressive wave and reflected wave superpose with suitable phase a steady wave pattern is set up on the string or in the medium.
A standing wave is represented by y (x, t) = 2a sin kx cos ωt.

Formation of standing waves in stretched strings :
Let a wire of length V and mass’m’ is fixed between two fixed supports with some tension T.
Let the wire is plucked at the mid point then it will vibrate with maximum amplitude. So antinode (A.N) is formed at centre of wire. At the fixed ends molecules of the wire are not free to vibrate. So nodes will be formed at fixed ends as shown in figure.
Now length of wire l = \(\frac{\lambda}{2}\).
If two loops are formed,

Quetion 8.
Describe a procedure for measuring the velocity of sound in a stretched string.
Answer:
Let a wire of length ‘l’ and mass M is fixed between two rigid supports.

Practical method :
To measure velocity of sound in stretched strings we will adjust length of wire until stationary waves are formed in the given string by means of the tuning fork in the sonometer expt.. When
length of string (l) is equals to \(\frac{\lambda}{2}\)
l = \(\frac{\lambda}{2}\)
But υ = nλ or υ = n × 2l

Question 9.
Explain, using suitable diagrams, the formation of standing waves in a closed pipe. How may of this can be used to determine the frequency of a source of sound?
Answer:
Stationary waves formed in a closed pipe :
If one end is closed and the other end is open, then it is called a ‘Closed pipe’.

Let a longitudinal wave be sent through a closed pipe. It gets reflected at the closed end. These incident and reflected waves which are of same frequency, travelling in opposite directions gets superposed along the length of the pipe. As a result of it, longitudinal stationary waves are formed. In a closed pipe
υ_n = \(\frac{(2n+1)υ}{4l}\)
For 1st harmonic, when n = 0, υ₁ = \(\frac{υ}{4l}\)
For 2nd harmonic, when n = 1, υ₂ = 3\(\frac{υ}{4l}\)
For 3rd harmonic, when n = 2, υ₂ = 5\(\frac{υ}{4l}\)
From resonating air column expt. the frequency of sound can be found by using the formula

Where l₁ ansd l₂ 1st and 2nd resonating lengths and n is the frequency of the tuning fork.

Question 10.
What are ‘beats’? When do they occur? Explain their use, if any.
Answer:
Beats :
When two sounds of nearly equal frequency are superposed, they will create a waxing and warning intensity of sounds. This effect is called “beats”.
Beats are produced due to interference of sound waves.
Beat frequency υ_beat = υ₁ – υ₂

Uses of Beats:

1. To know the frequency of an unknown tuning fork.
2. Harmful gases in a mine can be detected by using beats.

Question 11.
What is ‘Doppler effect’? Give illustrative examples.
Answer:
Doppler effect :
The apparent change in frequency heard by the observer due to the relative motion between source and the observer is called “Doppler’s effect”.

Examples of Doppler effect:

1. The frequency of sound increases as the source moves closer to the observer. Ex: Pitch of the whistle of an approaching train appears to be increased and that of a train going away decreases.
2. In astronomy, the Doppler effect was originally studied in the visible part of the electromagnetic spectrum.

Because of the inverse relationship between frequency and wavelength, we can describe the Doppler shift in terms of wavelengths. Radiation is redshifted when its wavelength increases. It indicates that the stars are moving away and universe is expanding.

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the equations for first, second and third harmonics and also deduce the laws of transverse waves is stretched strings. [TS Mar. 19; AP & TS May 18, 16]
Answer:
Stationary wave :
When two progressive waves of same wavelength, amplitude and frequency travelling in opposite directions and superimpose over each other stationary waves (or) standing waves are formed.

Formation of stationary wave in a stretched string:

• Let us consider a string of length ‘l’ stretched at the two fixed ends ‘A’ and ‘B’.
• Now pluck the string perpendicular to its length.
• The transverse wave travel along the length of the string and get reflected at fixed ends.
• Due to superimposition of these reflected waves, stationary waves are formed in the string.

Equation of Stationary Wave :
Let two transverse progressive waves having same amplitude ‘A’, wavelength λ and frequency ‘n’; travelling in opposite direction along a stretched string be given by
y₁ = A sin (kx – ωt) and y₂ = A sin (kx + ωt)
where ω = 2πn and k = \(\frac{2 \pi}{\lambda}\)
Applying the principle of superposition of waves, the resultant wave is given by
y = y₁ + y₂
y = A sin (kx – ωt) + A sin (kx + ωt)
y = 2A sin kx cos ωt

Amplitude of resultant wave (2A sin kx) is no more constant. It depends on the
value of kx. When x = 0, \(\frac{\lambda}{2},\frac{2\lambda}{2}.\frac{3\lambda}{2}\), …………… etc.

The amplitude becomes zero.
These positions of zero amplitude are
known as “Nodes”, when x = \(\frac{\lambda}{4},\frac{3\lambda}{4}.\frac{5\lambda}{4}\), ……………. etc. The amplitude becomes 2A (Maximum).
These positions of maximum amplitude are known as “Antinodes”.

Equation of fundamental frequency :
If the wire of length ‘l’ is stretched between points ‘A’ and ‘B’ with tension T vibrates as a single loop then the frequency of the vibrations is known as fundamental frequency and is denoted by ‘υ’.
l = \(\frac{\lambda}{2}\)
But velocity of a wave v = υλ – (1)
In case of stretched strings v = \(\sqrt{\frac{T}{\mu}}\) – (2)
From (1) and (2)
υ = \(\frac{1}{\lambda}\sqrt{\frac{T}{\mu}}\)
∴ υ = \(\frac{1}{2l}\sqrt{\frac{T}{\mu}}\) (∵ λ= 2l)
If the wire vibrates for ‘p’ loops _P If
The frequency = υ_p = \(\frac{p{2l}\sqrt{\frac{T}{\mu}}\)

Formation of harmonics in stretched strings :
From above equation if the wire vibrates with one loop then p =1
Frequency of 1st harmonic υ₁ = \(\frac{1}{2l}\sqrt{\frac{T}{\mu}}\)

Laws of Transverse vibrations in a stretched string :
I. First Law (or) Law of length :
The fundamental frequency of a vibrating string is inversely proportional to the length of the string, when the tension (T) and its
linear density µ are constant.
υ ∝ \(\frac{1}{l}\) ⇒ υl = constant ⇒ υ₁l₁ = υ₂l₂ (∵ T, µ are constant)

II. Second law (or) Law of Tension :
The fundamental frequency of a vibrating string is directly proportional to the square root of stretching force T (Tension), when the length of the string l and linear density ‘µ’ are constant.

III. Third law (or) Law of linear density :
The fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m) when the length (l) and tension (T) in the string are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equation for the frequencies of harmonics produced. [(TS May 17, Mar. 16; AP Mar. 18, 17, 16, May 17, 14)]
Answer:
Formation of stationary wave in open pipe :
An open pipe is a cylindrical tube having air inside with both ends open.

Let a longitudinal wave pass through the organ pipe. It gets reflected at the end of the pipe. These incident and reflected waves which are of same frequency, travelling in opposite directions, get superposed along the length of the pipe.

As a result of it, longitudinal stationary waves are formed.

At the open end, the particles of the medium are free to vibrate and the incident and reflected waves will be in phase, so particles have maximum displacement, forming antinode. Thus the air column in- side a open pipe is set into vibration due to stationary waves formed in it with anti- node at each open end.

Harmonics in open pipe :
Fundamental :
In the fundamental mode of vibration, an antinode is formed at each end and a node is formed between them. The note produced is called first harmonic.

If ‘l’ is the vibrating length and λ₁ is the corresponding wavelength.

Second Harmonic (or) first overtone :
2nd harmonic consists of three antinodes and two nodes.

If λ₂ is the corresponding wavelength

Third harmonic or Second overtone :
3rd harmonic consists of four antinodes and three nodes. If λ₃ is the corresponding wavelength,

The ratio of harmonics in open pipes υ₁ : υ₂ : υ₂ : = 1 : 2 : 3 …………

Question 3.
How are stationary waves formed in closed pipes? Explain the various modes of vibrations and obtain relations for their frequencies. [T.S. Mar. ’19, ’15, May, June ’15; AP Mar. June ’15, May ’17, ’16]
Answer:
A closed pipe is a cylindrical tube having air inside, one end closed and the other is open.

Let a longitudinal wave be sent through a closed pipe. It gets reflected at the closed end of the pipe. These incident and reflected waves, which are of same frequency, trav-elling in opposite directions gets superposed as a result longitudinal stationary waves are formed.

At the closed end reflection is on a rigid surface a node is formed at closed end, at the open end antinode is formed.

I. Harmonics in a closed pipe :
Fundamental mode (or) First Harmonic :
In this one node and one antinode is formed. If ‘l’ is the vibrating length and λ₁ is the corresponding wavelength.

II. First overtone or Third Harmonic :

1st overtone consists of two nodes and two antinodes. If ‘l’ is the vibrating length and λ₃ is the corresponding wavelength,

as third Harmonic or first overtone.

III. Fifth Harmonic or Second overtone :
2nd overtone consists of three nodes and three antinodes.

If ‘l’ is the vibrating length, λ₅ is the corresponding wavelength,

υ₅ is called fifth Harmonic.

The ratio of harmonics in closed pipe is υ₁ : υ₃ : υ₅ : = 1 : 3 : 5 : ………….

From the above, in closed pipe only odd Harmonics are formed.

Question 4.
What are beats? Obtain an expression for the beat frequency. Where and how are beats made use of?
Answer:
When two sounds of nearly (or) slightly equal frequency are superposed they will create a waxing and warning intensity of sounds. This effect is called “beats”.

Beats are produced due to interference of sound waves.

Beat frequency ubeat υ_beat = υ₁ ~ υ₂

Expression for beat frequency:

Let us consider two sound waves y₁ and y₂ of nearly equal frequency ‘υ₁‘ and ‘υ₂‘ each of amplitude a’ superpose each other then the resultant wave is given by
y = y₁ + y₂ = a sin ω₁t + a sin ω₂t.
where ω₁ = 2πυ₁ and ω₂ = 2πυ₂;
∴ y = a sin 2πυ₁t + a sin 2πυ₂t

The frequency of the resultant wave is \(\frac{υ_1+υ_2}{2}\)
The frequency of the amplitude is \(\frac{υ_1-υ_2}{2}\)

The intensity of the sound will be maximum when 2a cos 2π(\(\frac{υ_1-υ_2}{2}\)) is maximum.

Where k = 0, 1, 2, …………. maximum sound will be heard at interval
0, \(\frac{1}{υ_1-υ_2},\frac{2}{υ_1-υ_2},\frac{3}{υ_1-υ_2}\)
The time interval between two consecutive maxima = \(\frac{1}{υ_1-υ_2}\)
or, Beat frequency = υ₁ – υ₂
The intensity of sound will be minimum when cos 2π(\(\frac{υ_1-υ_2}{2}\))t is minimum i.e., zero.
The time interval between two consecutive minima = \(\frac{1}{υ_1-υ_2}\)
The number of minima heard per second = υ₁ ~ υ₂

Importance of Beats: Beats can be used

1. in tuning musical instruments.
2. to detect dangerous gases in mines.
3. to produce special effects in cinematography.
4. to determine unknown frequency of a tuning fork.
5. in heterodyne receivers range.

Question 5.
What is Doppler effect? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [(AP Mar. 16, 14; TS Mar. 18, 17]
Answer:
Doppler effect :
The apparent change in the frequency heard by the observer due to relative motion between the observer and the source of sound is called “Doppler effect”.

Expression for apparent frequency when source is in motion and observer at rest:
Let ‘s’ be a source of sound moving with a velocity v_s away from a stationary observer. Let the source produces a sound of frequency υ₀ and its time period is T₀.
At time t = 0 the source produces a crest.

Let the distance between source and observer is ‘L’ and velocity of sound is ‘v’.
Then time taken by observer to detect crest = t₁ = \(\frac{L}{v}\) ……… (1)
The second crest is produced after a time interval T₀.

Distance travelled by source during this time = v_sT₀. So total distance from observer = L + v_sT₀
Time taken to detect 2nd crest

Let the source produced (n + 1)th crest at time nT₀.
Now time taken to detect that crest =

So apparent frequency υ < υ₀.
If source is moving towards observer, then v_s is -ve (as per sign convension).
∴ apparent frequency υ = υ₀ [1 – \(\frac{(-v_s)}{v}\)]
= υ₀ [1 + \(\frac{v_s}{v}\)]
When source is approaching the observer frequency heard υ = υ₀(\(\frac{v+v_s}{v}\)) OR
υ = υ₀ [1 + \(\frac{v_s}{v}\)]. In this case apparent frequency heard υ > υ₀.

Question 6.
What is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer is in motion with respect to a source at rest. [AP June ’15]
Answer:
Doppler shift :
The difference between apparent frequency heard by observer and actual frequency produced by the source is called as “Doppler’s shift”.

Derivation of apparent frequency when source at rest and observer in motion :
Let s’ is a source at rest produces a sound of constant frequency ‘υ₀‘ Let Time Period of wave is ‘T₀‘. Let an observer is moving away from source with a velocity ‘v₀‘.

Carries a device that can count the num-ber of crests / compressions produced by source.

At time t = 0 source produces a crest. Let the distance between source and observer is L’ and velocity of sound is ‘v’.

Now time taken by the observer to detect the crest t₁ = \(\frac{L}{v}\) → 1

2nd crest is produced after a time period T₀.
During this time observer moves a distance v₀T₀
time taken by observer to detect t₂ = T₀ + (\(\frac{L+T_{0}v_{0}}{v}\)) → 2
Let source produces (n + 1) th crest after a time nT₀
Time taken to detect this crest

Solved Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg/m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [AP May 18; TS May 16]
Answer:
Vibrating length (l) = 0.6 m
For fundamental mode l = \(\frac{\lambda}{2}\)
⇒ λ = 2l = 2 × 0.6 ⇒ λ = 1.2 m
Fundamental frequency n = 30 Hz
a) Velocity of transverse wave v = nλ ;
v = 30 × 1.2 ⇒ v = 36 ms^-1

b) Linear density of the string
µ = 0.05 kg m^-1
But, v = \(\sqrt{\frac{T}{\mu}}\) (T is the tension)
T = v²µ = 36 × 36 × 0.05 = 64.8 N.

Question 2.
A steel cable of diameter 3 cm is kept under a tension of 10 kN. The density of steel is 7.8g/cm³. With what speed would transverse waves propagate along the cable?
Answer:
The density of steel is
ρ = 7.8 gm cm^-3 ⇒ 7.8 × 10³ kgm^-3
Diameter of cable D = 3 cm
⇒ r = 1.5 × 10^-2 m
Tension (T) = 10 × 10³ N.

Question 3.
Two progressive transverse waves given by y₁ = 0.07 sin π (12x – 500t) and y₂ = 0.07 sin π (12x + 500t) travelling along a stretched string form nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes? (c) What is the wavelength of the standing wave?
Answer:
Equation of progressive wave is
y₁ = 0.07 sin (12πx – 500 πt) ;
y₂ = 0.07 sin (12πx + 500 πt)
a) Displacement at node = 0
b) Displacement at Antinode = 2A = 2 × 0.07 = 0.14 m.
c) Propagation constant k = 12π

Question 4.
A string has a length of 0.4 m and a mass of 0.16 g. If the tension in the string is 70 N, what are the three lowest frequencies it produces when plucked?
Answer:
Length of string (l) = 0.4 m
Mass of the string (m) = 0.16 × 10^-3kg ;
Linear density (µ) = \(\frac{0.16}{0.4}\) = 4 × 10^-4 kg m^-1
Tension (T) = 70 N.
Fundamental frequency υ₁ = \(\frac{1}{2l}\sqrt{\frac{T}{\mu}}\) ;

Next frequency is υ₂ = 2υ₁
= 2 × 523 = 1046 Hz

Next frequency is υ₃ = 3υ₁
= 3 × 523 = 1569 Hz

Question 5.
A metal bar when clamped at its centre, resonates in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end, what will be its fundamental resonance frequency?
Answer:
Fundamental frequency (υ) = 4 × 10³ Hz
a) If metal bar is clamped at its centre

b) If metal is clamped at one end

Question 6.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column? [TS Mar. ’19; AP Mar. 18, I 7, May 1 7, 14]
Answer:
Length of closed organ pipe l = 70 cm
Velocity of sound (v) = 331 ms^-1
Fundamental frequency of closed pipe

Question 7.
A vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top the tube. If standing waves are produced at two successive water lev-els of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube?
Answer:
Frequency of sound wave (n) = 320 Hz
1st resonating length l₁ = 20 cm
2nd resonating length l₂ = 73 m
Speed of sound wave (v) = 2n (l₂ – l₁)
= 2 × 320 (73 – 20)
∴ v = 640 × 53 × 10^-2
= 33920 cm s^-1 = 339 ms^-1

Question 8.
Two organ pipes of lengths 65 cm and 70 cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes? (Velocity of sound = 330 m/s).
Answer:
First open organ pipe of length (l₁) = 65 cm
Second open organ pipe of length (l₂) = 70 cm

Question 9.
A train sounds its whistle as it approaches and crosses a level-crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, in the speed of the train and the frequency of its whistle.
Answer:
a) Speed of sound (v) = 340 ms^-1 ;
Apparent frequency heard by observer when train approaches υ’ = 219 Hz

When train leaves the observer ap-parent frequency n” = 184 Hz

Question 10.
Two trucks heading in opposite directions with speeds of 60 kmph and 70 kmph respectively, approach each other. The ckiver of the first truck sounds his horn of frequency 400 Hz. What frequency does the driver of the second truck hear? (Velocity of sound = 330 m/s). After the two trucks have passed each other, what frequency does of the second truck hear?
Answer:
Speed of first truck = 60 × \(\frac{5}{18}\) = v_s
Speed of second truck = 70 × \(\frac{5}{18}\) = v₀
Frequency of first truck n = 400 Hz ;
Velocity of sound = 330 ms^-1

a) When they approach each other, frequency heard by second truck driver is

∴ Apparent frequency (n’) = 600 Hz b)

b) When they move away from each other, the frequency heard by second truck driver

∴ Apparent frequency υ” = 270 Hz

Question 11.
A rocket is moving at a speed of 200 ms^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Calculate the frequency of the sound as detected by the target. (Velocity of sound in air is 330 ms^-1) [AP Mar. ’16]
Answer:
Velocity of source (Rocket) (V_s) = 200 m/s
Velocity of sound (V) = 330 m/s.
Frequency of sound emitted (υ₀) = 1000 Hz

Frequency of sound heard by observer υ = 2540 Hz.

Question 12.
A open organ pipe 85 cm long is sounded. If the velocity of sound is 340 m/s, what is the fundamental frequency of vibration of the air column? [TS Mar. ’16]
Answer:
Length of pipe (l) = 85 cm,
Velocity of sound (V) = 340 m/s.
Fundamental frequency of open pipe = n
= \(\frac{V}{2l}\)
n = \(\frac{340 \times 100}{2 \times 85}\) = 200Hz
∴ Fundamental frequency (n) = 200 Hz.

Question 13.
A pipe 30 cm long is open at both ends. Find the fundamental frequency. Velocity of sound in air is 330 ms-1.
Answer:
Length of open pipe (l) = 30 cm;
Velocity of sound (V) = 330 ms^-1
At fundamental frequency λ = 2l.
⇒ λ = 30 × 2 = 60 cm = 0.6 m
Fundamental frequency
v = \(\frac{V}{\lambda}=\frac{330}{0.6}\) = 550 Hz.

Question 14.
If the fundamental frequency in a closed pipe is 300Hz, find the value of third har¬monic in it. [TS June ’15]
Answer:
Fundamental frequencies (v₁) = 300Hz
Frequencies of 3rd Harmonic (v₃) = 3v₁
= 300 × 3 = 900Hz
Note: In closed pipes, the harmonics are in the ratio 1:3:5 even harmonic does not exist.

Intext Question and Answer

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. Then length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Mass of the string, (M) = 2.5 kg
Tension in the string, (T) = 200 N
Length of the string (l) = 20 m
Linear density (µ) = mass per unit length = \(\frac{M}{l}=\frac{2.5}{20}\) = 0.125 kgm^-1
Velocity of transverse wave in the string is

∴ Time taken by the disturbance to reach the other end, t = \(\frac{l}{v}=\frac{20}{40}\) = 0.5 s.

Question 2.
A stone dropped from the top of a tower of height 300 m splashes into the a pond of water near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 ms^-1? (g = 9.8 ms^-2)
Answer:
Height of the tower (n) = 300 m
Initial velocity (u) = 0
Acceleration due to gravity, g = 9.8 m/s²
Speed of sound = 340 ms^-1
The time (t₁) taken by the stone to strike the water in the pond can be calculated using

Time taken (t₂) by the sound to reach the top of the tower, t₂ = \(\frac{h}{u}=\frac{300}{340}\) = 0.88 s
∴ Total time after the splash is heard,
t = t₁ + t₂ = 7.82 + 0.88 = 8.7 s.

Question 3.
A steel wire has a length of 12 in and a mass of 2.1 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms^-1.
Answer:
Length of the wire (l) = 12 m ;
Mass of the wire (M) = 2.1 kg
Linear density (µ) = mass per unit length
= \(\frac{M}{l}=\frac{2.1kg}{12m}\) = 0.175 kgm^-1
We know that velocity of a transverse wave in a stretched string is given by
v = \(\sqrt{\frac{T}{\mu}}\) ⇒ v²µ = (343)² × 0.175
= 20588 × 575 N ≅ 2.06 × 10⁴ N.

Question 4.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 ms-1 and in water 1486 ms^-1.
Answer
a) Frequency of ultrasonic sound,
n = 1000 kHz = 10⁶ Hz
Speed of sound (v_a) = 340 ms^-1
The wavelength (λ_r) of the reflected sound is given by
λ_r = \(\frac{υ_{a}}{n}=\frac{340}{10^{6}}\) = 3.4 × 10^-4 m

b) Frequency of the ultrasonic sound,
n = 1000 kHz = 10⁶ Hz
Speed of sound in water, υ_w = 1486 ms^-1.
The wavelength (λ_r) of the reflected sound is given by
λ_r = \(\frac{υ_{w}}{n}=\frac{1486}{10^{6}}\) = 1.49 × 10^-3 m

Question 5.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound is 1.7 kms^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
Speed of sound in the tissue,
v = 1.7 km s^-1 = 1.7 × 10³ ms^-1
Operating frequency of the scanner,
n = 4.2 MHz = 4.2 × 10⁶ Hz
The wavelength of sound in the tissue is
λ = \(\frac{\mathrm{v}}{\mathrm{n}}=\frac{1.7 \times 10^3}{4.2 \times 10^6}\) = 4.1 × 10^-4 m

Question 6.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2 kg m^-1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Answer:
a) Mass of the wire, M = 3.5 × 10^-2 kg m^-1

The wavelength (λ) of the stationary wave is related to the length of the wire by the relation
λ = \(\frac{2l}{n}\)
where n → no. of nodes in the wire
For fundamental node, n = 1
λ = 2l = 2 × 0.875 = 1650 m.
The speed of the transverse wave is given by
v = nλ = 4.5 × 1.75
= 78.75 ms^-1

b) The tension produced in the string is given by the formula
T = v²m = (78.75)² × 4 × 10^-2
= 248.06 N.

Question 7.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Length of the steel rod, l = 100 cm = 1 m.
Fundamental frequency of vibration,
n = 2.53 kHz = 2.53 × 10³ Hz

When the rod is plucked at its middle, an antinode is formed at its centre and nodes (N) are formed at its two ends as shown in the given figure.
The distance between two successive nodes is \(\frac{\lambda}{2}\).
∴ l = \(\frac{\lambda}{2}\)
⇒ λ = 2l = 2 × l = 2m
The speed of sound in steel is given by the relation:
v = nλ= 2.53 × 10³ × 2
= 5.06 × 10³ ms^-1
= 5.06 kms^-1

Question 8.
A train, standing at the outer signal of a railway station blows a whistle of freq-uency 400 Hz in still air.
i) What is the frequency of the whistle for a platform observer when the train
a) approaches the platform with a speed of 10 ms^-1,
b) recedes from the platform with a speed of 10 ms^-1?
ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms^-1.
Answer:
i) a) Frequency of the whistle (n) = 400 Hz
Speed of the train (v_t) = 10 ms^-1
Speed of sound (v) = 340 ms^-1
The apparent frequency (n’) of the whistle as the train approaches the platform is given by the relation

b) The apparent frequency (n’) of the whistle as the train recedes from the platform is given by the relation :

ii) The apparent change in frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same i.e., 340 ms^-1.

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Circles Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Circles Important Questions Very Short Answer Type

Question 1.
Find the centre and radius of the circle x² + y² + 6x + 8y – 96 = 0. (Mar. ’94)
Solution:
Given the equation of the circle is
x² + y² + 6x + 8y – 96 = 0 …….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
We get 2g = 6 ⇒ g = 3
2f = 8 ⇒ f = 4
c = -96
Centre c = (-g, -f) = (-3, -4)
radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)
= \(\sqrt{(3)^2+(4)^2+96}\)
= √121
= 11

Question 2.
Find the centre and radius of the circle x² + y² + 2x – 4y – 4 = 0
Solution:
Given the equation of the circle is
x² + y² + 2x – 4y – 4 = 0 …….(1)
Comparing (1) with
x² + y² + 2gx + 2fy + c = 0
We get 2g = 2 ⇒ g = 1
2f = -4 ⇒ f = -2
c = -4
Centre c = (-g, -f) = (-1, 2)
radius r = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(1)^2+(-2)^2+4}\)
= √9
= 3

Question 3.
Find the centre and radius of the circle x² + y² + 2ax – 2by + b² = 0
Solution:
Given the equation of the circle is
x² + y² + 2ax – 2by + b2 = 0 …….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
We get 2g = 2a ⇒ g = a
2f = -2b ⇒ f = -b
c = b²
Centre c = (-g, -f) = (-a, b)
radius r = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(a)^2+(-b)^2-b^2}\)
= √a²
= a

Question 4.
Find the centre and radius of the circle 3x² + 3y² – 6x + 4y – 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 6x + 4y – 4 = 0
⇒ x² + y² – 2x + \(\frac{4}{3}\)y – \(\frac{4}{3}\) = 0 ………(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0

Question 5.
Find the centre and radius of the circle \(\sqrt{1+\mathrm{m}^2}\) (x² + y²) – 2cx – 2mcy = 0. (May ’10)
Solution:

Question 6.
If the circle x² + y² + ax + by – 12 = 0 has the centre at (2, 3), then find a, b, and the radius of the circle. (May ’09, ’07, Mar. ’08)
Solution:
Given the equation of the circle is
x² + y² + ax + by – 12 = 0 ……..(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
we get 2g = a ⇒ g = \(\frac{a}{2}\)
2f = b ⇒ f = \(\frac{b}{2}\)
c = -12

Question 7.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3). Find g, f, and its radius. [(AP) May ’19]
Solution:
Given the equation of the circle is
x² + y² + 2gx + 2fy – 12 = 0 ……..(1)
Centre of (1) is C = (-g, -f)
Given that centre C = (2, 3)
∴ (-g, -f) = (2, 3)
⇒ g = -2, f = -3
Now Radius r = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(-2)^2+(-3)^2-(-12)}\)
= √25
= 5

Question 8.
If x² + y² + 2gx + 2fy = 0 represents a circle with centre (-4, -3) then find g, f, and the radius of the circle. [(AP) May ’17]
Solution:
Given the equation of the circle is
x² + y² + 2gx + 2fy = 0 …….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
we get g = g, f = f, c = 0
Centre of (1) is C = (-g, -f)
Given that centre, C = (-4, -3)
∴ (-g, -f) = (-4, -3)
we get g = 4, f = 3
Radius r = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(4)^2+(3)^2-0}\)
= √25
= 5

Question 9.
Find the value of ‘a’, if 2x² + ay² – 3x + 2y – 1 = 0 represents a circle, and also find its radius. [(AP) Mar. ’15]
Solution:
Given equation is
2x² + ay² – 3x + 2y – 1 = 0 …….(1)
Comparing (1) with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 2, h = 0, b = a, g = \(\frac{-3}{2}\), f = 1, c = -1
Now, equation (1) represents a circle
then a = b
⇒ 2 = a
∴ a = 2
The equation of the circle is
2x² + 2y² – 3x + 2y – 1 = 0
\(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\)
Comparing the above equation with x² + y² + 2gx + 2fy + c = 0

Question 10.
If the circle x² + y² – 4x + 6y + a = 0 has a radius of 4 then find ‘a’. [(AP) Mar. ’16]
Solution:
Given the equation of the circle is
x² + y² – 4x + 6y + a = 0 …….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
We get 2g = -4 ⇒ g = -2
2f = 6 ⇒ f = 3
c = a
Given that Radius, r = 4
⇒ \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=4\)
⇒ \(\sqrt{(-2)^2+(3)^2-a}=4\)
⇒ \(\sqrt{4+9-a}\) = 4
⇒ \(\sqrt{13-\mathrm{a}}\) = 4
Squaring on both sides
13 – a = 16
⇒ 13 – 16 = a
⇒ a = -3
∴ a = -3

Question 11.
If x² + y² – 4x + 6y + c = 0 represents a circle with a radius of 6 then find the value of ‘c’.
Solution:
Given equation of the circle is x² + y² – 4x + 6y + c = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0,
we get g = -2, f = 3, c = c
Given that radius r = 6
⇒ \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=6\)
⇒ \(\sqrt{(-2)^2+(3)^2-c}=6\)
⇒ \(\sqrt{4+9-\mathrm{c}}\)
⇒ \(\sqrt{13-c}=6\)
Squaring on both sides
13 – c = 36
⇒ 13 – 36 = c
⇒ c = -23

Question 12.
Find the equation of the circle passing through (-2, 3) having the centre at (0, 0).
Solution:

Given centre C(h, k) = (0, 0)
Let the given point A (-2, 3)
Since A(-2, 3) is a point on the circle.
Radius r = CA = \(\sqrt{(-2-0)^2+(3-0)^2}\) = √13
∴ The equation of the required circle is (x – h)² + (y – k)² = r²
⇒ (x – 0)² + (y – 0)² = (√13)²
⇒ x² + y² = 13
⇒ x² + y² – 13 = 0

Question 13.
Find the equation of the circle passing through the origin and having the centre at (-4, -3). (Mar. ’04; May ’02)
Solution:
Given centre, C(h, k) = (-4, -3)
Let, the given point A(0, 0).
Since, A(0, 0) is a point on the circle.
∴ Radius, r = CA = \(\sqrt{(-4-0)^2+(-3-0)^2}\)
= √25
= 5
∴ The equation of the required circle is (x – h)² + (y – k)² = r²
⇒ (x + 4)²+ (y + 3)² = (5)²
⇒ x² + 16 + 8x + y² + 9 + 6y = 25
⇒ x² + y² + 8x + 6y + 25 = 25
⇒ x² + y² + 8x + 6y = 0

Question 14.
Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4).
Solution:
Given centre C(h, k) = (-3, 4)
Let the given point A = (3, 4)

Since ‘A’ is the point on the circle.
∴ Radius, r = CA = \(\sqrt{(-3-3)^2+(4-4)^2}\)
= √36
= 6
∴ The equation of the required circle is (x – h)² + (y – k)² = r²
⇒ (x + 3)² + (y – 4)² = 6²
⇒ x² + 9 + 6x + y² + 16 – 8y = 36
⇒ x² + y² + 6x – 8y – 11 = 0

Question 15.
Find the equation of the circle which is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (-2, 14). [(TS) May ’17; Mar & May ’14]
Solution:

Given equation of the circle is x² + y² – 6x – 4y – 12 = 0
Equation of the circle concentric with the circle
x² + y² – 6x – 4y – 12 = 0 is x² + y² – 6x – 4y + k = 0 …….(1)
Equation (1) passes through (-2, 14) then
(-2)² + (14)² – 6(-2) – 4(14) + k = 0
⇒ 4 + 196 + 12 – 56 + k = 0
⇒ 156 + k = 0
⇒ k = -156
The required circle from (1) is x² + y² – 6x – 4y – 156 = 0.

Question 16.
Find the equation of the circle with (-4, 3), (3, -4) as ends of a diameter.
Solution:
Let A = (x₁, y₁) = (-4, 3), B = (x₂, y₂) = (3, -4) are the given points.
∴ The required equation of the circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x + 4) (x – 3) + (y – 3) (y + 4) = 0
⇒ x² – 3x + 4x – 12 + y² + 4y – 3y – 12 = 0
⇒ x² + y² + x + y – 24 = 0

Question 17.
Find the other end of the diameter of the circle x² + y² – 8x – 8y + 27 = 0 if one end of it is (2, 3). [(AP) Mar. ’20; May ’12]
Solution:
Given the equation of the circle is x² + y² – 8x – 8y + 27 = 0.
Let A(2, 3) be the given point.
Comparing (1) with x² + y² + 2gx + 2fy + c = 0, we get
g = -4, f = -4, c = 27
Centre of the circle, C(h, k) = (-g, -f) = (4, 4)
Let, B(x, y) be the other end of the diameter

∵ ‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then (4, 4) = \(\left(\frac{2+x}{2}, \frac{3+y}{2}\right)\)
\(\frac{2+x}{2}\) = 4
⇒ 2 + x = 8
⇒ x = 8 – 2
⇒ x = 6
\(\frac{3+y}{2}\) = 4
⇒ 3 + y = 8
⇒ y = 8 – 3
⇒ y = 5
∴ The other end of the diameter is B = (6, 5).

Question 18.
Obtain the parametric equation of the circle represented by x² + y² = 4. (Mar. ’14)
Solution:
Given the equation of the circle is x² + y² = 4
This is of the form x² + y² = r²
Centre C = (h, k) = (0, 0); Radius r = 2
∴ The parametric equations of the circle are
x = h + r cos θ = 0 + 2 cos θ = 2 cos θ
y = k + r sin θ = 0 + 2 sin θ = 2 sin θ
0 ≤ θ ≤ 2π

Question 19.
Find the parametric equations of the circle (x – 3)² + (y – 4)² = 8². [Mar. ’18, ’16 (AP)]
Solution:
Given equation of the circle is (x – 3)² + (y – 4)² = 8²
This is of the form (x – h)² + (y – k)² = r²
we get h = 3, k = 4, r = 8
∴ Centre (h, k) = (3, 4)
Radius r = 8
∴ The parametric equations are
x = h + r cos θ = 3 + 8 cos θ
y = k + r sin θ = 4 + 8 sin θ
0 ≤ θ ≤ 2π

Question 20.
Obtain the parametric equations of the circle represented by x² + y² + 6x + 8y – 96 = 0.
Solution:
Given the equation of the circle is
x² + y² + 6x + 8y – 96 = 0 …….(1)
Comparing (1) with
x² + y² + 2gx + 2fy + c = 0, we get
2g = 6 ⇒ g = 3
2f = 8 ⇒ f = 4
c = -96
Centre, C(h, k) = (-g, -f) = (-3, -4)
Radius, r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)
= \(\sqrt{3^2+4^2+96}\)
= √121
= 11
The parametric equations of the circle are
x = h + r cos θ = -3 + 11 cos θ
y = k + r sin θ = -4 + 11 sin θ
0 ≤ θ ≤ 2π

Question 21.
Find the parametric equations of the circle x² + y² – 6x + 4y – 12 = 0. [(AP) May ’15; Mar. ’10, ’06]
Solution:
Given the equation of the circle is
x² + y² – 6x + 4y – 12 = 0 ……….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6 ⇒ g = -3
2f = 4 ⇒ f = 2
f = -12
Centre, C(h, k) = (-g, -f) = (3, -2)
Radius, r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)
= \(\sqrt{(-3)^2+(2)^2+12}\)
= √25
= 5
The parametric equations of the circle are
x = h + r cos θ = 3 + 5 cos θ
y = k + r sin θ = -2 + 5 sin θ
0 ≤ θ ≤ 2π

Question 22.
Find the length of the tangent from (3, 3) to the circle x² + y² + 6x + 18y + 26 = 0. (Mar. ’05)
Solution:
Given equation of the circle is x² + y² + 6x + 18y + 26 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0, we get
g = 3, f = 9, c = 26
Given point P(x₁, y₁) = (3, 3)
Length of the tangent = \(\sqrt{S_{11}}\)
= \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{(3)^2+(3)^2+2(3)(3)+2(9)(3)+26}\)
= \(\sqrt{9+9+18+54+26}\)
= √116

Question 23.
If the length of the tangent from (2, 5) to the circle x² + y² – 5x + 4y + k = 0 is √37, then find k. [(TS) May, Mar. ’18; (AP) May ’17]
Solution:
Given equation of the circle is x² + y² – 5x + 4y + k = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0, we get
g = \(\frac{-5}{2}\), f = 2, c = k
Let the given point P(x₁, y₁) = (2, 5)
Given that the length of the tangent = √37
∴ \(\sqrt{S_{11}}\) = √37

Question 24.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1, then find ‘k’. [(TS) Mar. ’20, ’15, May 15; (AP) May ’18, Mar. ’15; May ‘16, ‘15]
Solution:
The given equation of the circle is x² + y² + 2ky = 0.
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0 we get
g = 0, f = k, c = 0
Let, the given point P(x₁, y₁) = (5, 4)
Given the length of the tangent = 1
\(\sqrt{S_{11}}\) = 1
⇒ S₁₁ = 1
⇒ \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}=1\)
⇒ (5)² + (4)² + 2(0)(5) + 2k(4) + 0 = 1
⇒ 25 + 16 + 8k = 1
⇒ 8k = -40
⇒ k = \(\frac{-40}{8}\) = -5

Question 25.
Show that the line lx + my + n = 0 is normal to the circle S = 0 if and only if lg + mf = n.
Solution:
The straight line lx + my + n = 0 is normal to the circle.
S = x² + y² + 2gx + 2fy + c = 0
If the centre, C(-g, -f) lies on lx + my +n = 0 then
l(-g) + m(-f) + n = 0
⇒ -lg – mf + n = 0
⇒ lg + mf = n

Question 26.
Find the equation of the polar of (2, 3) with respect to the circle x² + y² + 6x + 8y – 96 = 0. [May ’96]
Solution:
Given equation of the circle is x² + y² + 6x + 8y – 96 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
we get g = 3, f = 4, c = -96
Let the given point P(x₁, y₁) = (2, 3)
The equation of polar P(2, 3) w.r.t the given circle is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(2) + y(3) + 3(x + 2) + 4(y + 3) – 96 = 0
⇒ 2x + 3y + 3x + 6 + 4y + 12 – 96 = 0
⇒ 5x + 7y – 78 = 0

Question 27.
Find the polar of (1, -2) with respect to x² + y² – 10x – 10y + 25 = 0. [(TS) Mar. ’15]
Solution:
Given equation of the circle is x² + y² – 10x – 10y + 25 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
We get g = -5, f = -5, c = 25
Let the given point P(x₁, y₁) = (1, -2)
The equation of polar P(1, -2) w.r.t the given circle is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(1) + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0
⇒ x – 2y – 5x – 5 – 5y + 10 + 25 = 0
⇒ -4x – 7y + 30 = 0
⇒ 4x + 7y – 30 = 0

Question 28.
Find the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x² + y² = 35. [(AP) May ’19, Mar. ’17; (TS) Mar. ’19, ’16, May ’17]
Solution:
Given the equation of the circle is x² + y² = 35
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
we get g = 0, f = 0, c = -35
Let, the given points are A(x₁, y₁) = (1, 3), B(x₂, y₂) = (2, k)
Since the given points are conjugate then S₁₂ = 0
⇒ x₁x₂+ y₁y₂+ g(x₁ + x₂) + f(y₁ + y₂) + c = 0
⇒ 1(2) + 300 + 0(1 + 2) + 0(3 + k) – 35 = 0
⇒ 2 + 3k – 35 = 0
⇒ 3k = 33
⇒ k = 11

Question 29.
If(4, k) and (2, 3) are conjugate points with respect to the circle x² + y² = 17 then find k.
Solution:
Given equation of the circle is x² + y² – 17 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
g = 0, f = 0, c = -17
Let, the given points be
A(x₁, y₁) = (4, k), B(x₂, y₂) = (2, 3)
Since the given points are conjugate then
S₁₂ = 0
⇒ x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c = o
⇒ 4(2) + k(3) + 0(4 + 2) + 0(k + 3) – 17 = 0
⇒ 8 + 3k + 0 + 0 – 17 = 0
⇒ 3k – 9 = 0
⇒ 3k = 9
⇒ k = 3

Question 30.
Find the value of ‘k’ if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x² + y² – 5x + 8y + 6 = 0. [(AP) Mar. ’19; (TS) ’17; May ’14]
Solution:
Given equation of the circle is x² + y2 – 5x + 8y + 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
g = \(\frac{-5}{2}\), f = 4, c = 6
Let the given points are
A(x₁, y₁) = (4, 2) and B(x₂, y₂) = (k, -3)
Since the given points are conjugate then S₁₂ = 0
⇒ x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c = 0
⇒ 4(k) + 2(-3) + \(\frac{-5}{2}\) (4 + k) + 4(2 – 3) + 6 = 0
⇒ 4k – 6 – 10 – \(\frac{5k}{2}\) – 4 + 6 = 0
⇒ \(\frac{-5k}{2}\) + 4k – 14 = 0
⇒ \(\frac{-5 k+8 k-28}{2}\) = 0
⇒ 3k – 28 = 0
⇒ 3k = 28
⇒ k = \(\frac{28}{3}\)

Question 31.
Find the die equation of the circle with centre (2, -3) and radius 4.
Solution:
Given centre c(h, k) = (2, -3)
radius r = 4
The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y + 3)² = (4)²
⇒ x² + 4 – 4x + y² + 9 + 6y = 16
⇒ x² + y² – 4x + 6y – 3 = 0

Question 32.
Find the equation of the circle with centre (-a, -b) and radius \(\sqrt{\mathbf{a}^2-\mathbf{b}^2}\).
Solution:
Given centre c(h, k) = (-a, -b)
radius r = \(\sqrt{\mathbf{a}^2-\mathbf{b}^2}\)
The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + a)² + (y + b)² = \(\left(\sqrt{a^2-b^2}\right)^2\)
⇒ x² + a² + 2ax + y² + b² + 2by = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 33.
Find the equation of the circle with centre c(a, -b) and radius a + b.
Solution:
Given centre c(h, k) = (a, -b)
radius r = a + b
The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – a)² + (y + b)² = (a + b)²
⇒ x² + a² – 2ax + y² + b² + 2by = a² + b² + 2ab
⇒ x² + y² – 2ax + 2by – 2ab = 0

Question 34.
Find the centre and radius of the circle 3x² + 3y² – 5x – 6y + 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 5x – 6y + 4 = 0

Question 35.
Find the equation of the circle whose extremities of a diameter are (1, 2) and (4, 5).
Solution:
Let A(x₁, y₁) = (1, 2) and B(x₂, y₂) = (4, 5) be the two given points.
∴ The equation of the required circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 1)(x – 4) + (y – 2)(y – 5) = 0
⇒ x² – 4x – x + 4 + y² – 2y – 5y + 10 = 0
⇒ x² + y² – 5x – 7y + 14 = 0

Question 36.
Find the position of the point P(3, 4) w.r.t the circle x² + y² – 4x – 6y – 12 = 0.
Solution:
Given equation of the circle is x² + y² – 4x – 6y – 12 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
We get g = -2, f = -3, c = -12
Given point P(x₁, y₁) = (3, 4)
Now, S11 = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (3)² + (4)² + 2(-2)(3) + 2(-3)(4) – 12
= 9 + 16 – 12 – 24 – 12
= 25 – 48
= -23
Since S₁₁ < 0, then the point, P(3, 4) is inside the given circle.

Question 37.
Find the position of the point (1, 5) with respect to the circle x² + y² – 2x – 4y + 3 = 0.
Solution:
Given equation of the circle is x² + y² – 2x – 4y + 3 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
We get g = -1, f = -2, c = 3
Given point P(x₁, y₁) = (1, 5)
Now S11 = \(\mathrm{x}_1{ }^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (1)² + (5)² + 2(-1)(1) + 2(-2)(5) + 3
= 1 + 25 – 2 – 20 + 3
= 29 – 22
= 7
Since S₁₁ > 0, then the point, P(1, 5) is outside the given circle.

Question 38.
Find the power of (1, 2) with respect to the circle x² + y² + 6x + 8y – 96 = 0.
Solution:
The given equation of the circle is x² + y² + 6x + 8y – 96 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
we get g = 3, f = 4, c = -96
Point P(x₁, y₁) = (1, 2)
∴ The power of P(1, 2) w.r.t the given circle is S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (1)² + (2)² + 2(3)(1) + 2(4)(2) – 96
= 1 + 4 + 6 + 16 – 96
= -69

Question 39.
Find the power of the die point (-1, 1) with respect to the circle x² + y² – 6x + 4y – 12 = 0. [(TS) Mar. ’16]
Solution:
The given equation of the circle is x² + y² – 6x + 4y – 12 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
we get g = -3, f = 2, c = -12
Point P(x₁, y₁) = (-1, 1)
∴ The power of P(-1, 1) w.r.t the given circle is
S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (-1)² + (1)² + 2(-3)(-1) + 2(2)(1) – 12
= 1 + 1 + 6 + 4 – 12
= 12 – 12
= 0

Question 40.
Find the length of the tangent from (2, 5) to the circle x² + y² – 5x + 4y – 5 = 0.
Solution:
The given equation of the circle is x² + y² – 5x + 4y – 5 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0,
we get g = \(\frac{-5}{2}\), f = 2, c = -5
Point, P(x₁, y₁) = (2, 5)
∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 41.
Find the length of the tangent from (-2, 5) to the circle x² + y² – 25 = 0. [(TS) May ’16]
Solution:
The given equation of the circle is x² + y² – 25 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
we get g = 0, f = 0, c = -25
Point P(x₁, y₁) = (-2, 5)
∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 42.
Find the length of the tangent from (12, 17) to the circle x² + y² – 6x – 8y – 25 = 0.
Solution:
The given equation of the circle is x² + y² – 6x – 8y – 25 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0,
We get g = -3, f = -4, c = -25
Point P(x₁, y₁) = (12, 17)
∴ The length of the tangent = \(\sqrt{S_{11}}\)

Question 43.
Find the equation of the tangent at P(-1, 1) of the circle x² + y² – 6x + 4y – 12 = 0. [(TS) Mar. ’16]
Solution:
Given equation of the circle is x² + y² – 6x + 4y – 12 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
we get g = -3, f = 2, c = -12
The given point P(x₁, y₁) = (-1, 1)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(-1) + y(1) + (-3) (x – 1) + 2(y + 1) + c = 0
⇒ -x + y – 3x + 3 + 2y + 2 – 12 = 0
⇒ -4x + 3y – 7 = 0
⇒ 4x – 3y + 7 = 0

Question 44.
Find the area of the triangle formed by the tangent at P(x1, y1) to the circle x² + y² = a² with the coordinate axes. [(AP) Mar. ’20; Apr. ’00, ’98, ’93]
Solution:
Given the equation of the circle is x² + y² = a²
The equation of the tangent at P(x₁, y₁) to the given circle is S₁ = 0.

Question 45.
Find the chord of contact of (1, 1) w.r.t the circle x² + y² = 9. [(AP) Mar. ’20]
Solution:
Given the equation of the circle is x² + y² = 9
Comparing this equation with x² + y² + 2gx + 2fy + c = 0,
We get g = 0, f = 0, c = -9
Given point P(x₁, y₁) = (1, 1)
∴ The equation of chord of contact is S1 = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(1) + y(1) + 0(x + 1) + 0(y + 1) – 9 = 0
⇒ x + y – 9 = 0

Question 46.
Find the pole of ax + by + c = 0 with respect to the circle x² + y² = r². [(AP) May ’16]
Solution:
Given the equation of the circle is x² + y² = r²
Let P(x₁, y₁) be the pole of ax + by + c = 0 …….(1)
The polar of P w.r. t the circle x² + y² = r² is S₁ = 0
xx₁ + yy₁ – r² = 0 ……(2)
Now (1) and (2) represent the same line

Question 47.
Discuss the relative position of the pair of circles (x – 2)² + (y + 1)² = 9, (x + 1)² + (y – 3)² = 4.
Solution:
Given equations of the circles are
(x – 2)² + (y + 1)² = 9 ……..(1)
(x + 1)² + (y – 3)² = 4 ………(2)
For circle (1),
Centre c₁ = (2, -1); radius r₁ = 3
For circle (2),
Centre c₂ = (-1, 3); radius r₂ = 2
c₁c₂ = \(\sqrt{(2+1)^2+(-1-3)^2}\)
= \(\sqrt{9+16}\)
= √25
= 5
Now r₁ + r₂ = 3 + 2 = 5
∴ c₁c₂ = r₁ + r₂
∴ The given circles touch each other externally.

Question 48.
Find the number of common tangents that exist for the pair of circles x² + y² = 4, x² + y² – 6x – 8y + 16 = 0. (May ’93)
Solution:
Given equations of the circles are
x² + y² = 4 ……(1)
x² + y² – 6x – 8y + 16 = 0 ………(2)
For circle (1),
centre, C₁ = (0, 0); Radius, r₁ = 2
For circle (2),
centre, C₂ = (3, 4)
Radius, r₂ = \(\sqrt{(-3)^2+(-4)^2-16}\) = √9 = 3
C₁C₂ = \(\sqrt{(0-3)^2+(0-4)^2}\) = √25 = 5
Now, r₁ + r₂ = 3 + 2 = 5
∴ C₁C₂ = r₁ + r₂
∴ The given circles touch each other externally.
∴ No. of common tangents = 3

Question 49.
Find the equation of the circle with centre (-7, -3) and radius ‘4’. (May ’93)
Solution:
Given centre, C(h, k) = (-7, -3)
radius, r = 4
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + 7)² + (y + 3)² = 4²
⇒ x² + 49 + 14x + y² + 9 + 6y = 16
⇒ x² + y² + 14x + 6y + 42 = 0

Question 50.
Find the equation of the circle passing through (2, -1) having the centre at (2, 3).
Solution:
Given centre, C(h, k) = (2, 3)
Let, the given point A(2, -1)
Since, A(2, -1) is a point on the circle
∴ Radius, r = CA = \(\sqrt{(2-2)^2+(-1-3)^2}\)
= \(\sqrt{0+(-4)^2}\)
= √16
= 4
∴ The equation of the required circle is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y – 3)² = 4²
⇒ x² + 4 – 4x + y² + 9 – 6y = 16
⇒ x² + y² – 4x – 6y – 3 = 0

Question 51.
Find the values of a, b if ax² + bxy + 3y² – 5x + 2y – 3 = 0 represents a circle. Also, find the radius and centre of the circle.
Solution:
Given equation is ax² + bxy + 3y² – 5x + 2y – 3 = 0 …….(1)
Comparing the equation (1) with ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = a,
2h = b ⇒ h = \(\frac{b}{2}\)
b = 3
Now, equation (1) represents a circle, then
(i) a = b ⇒ a = 3
(ii) h = 0 ⇒ \(\frac{b}{2}\) = 0 ⇒ b = 0
The equation of the circle is 3x² + 3y² – 5x + 2y – 3 = 0
⇒ \(x^2+y^2-\frac{5}{3} x+\frac{2}{3} y-1=0\)
Comparing the above equation with x² + y² + 2gx + 2fy + c = 0, we get

Question 52.
Find the equation of the circle with (4, 2), (1, 5) as ends of a diameter. [(AP) Mar. ’19]
Solution:
Let A(x₁, y₁) = (4, 2) and B(x₂, y₂) = (1, 5) are the two given points.
∴ The equation of the required circle is (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 4) (x – 1) + (y – 2) (y – 5) = 0
⇒ x² – 4x – x + 4 + y² – 5y – 2y + 10 = 0
⇒ x² + y² – 5x – 7y + 14 = 0

Question 53.
Show that A(3, -1) lies on the circle x² + y² – 2x + 4y = 0. Also, find the other end of the diameter through A. (May ’99, ’95, ’91)
Solution:
Given equation of the circle is x² + y² – 2x + 4y = 0 ……(1)
Given point A = (3, -1)
Now substituting the point A (3, -1) in equation (1)
(3)² + (-1)² – 2(3) + 4 (-1) = 0
9 + 1 – 6 – 4 = 0
10 – 10 = 0
0 = 0
∴ The point A(3, -1) lies on the circle x² + y² – 2x + 4y = 0.

Comparing the equation with x² + y² + 2gx + 2fy + c = 0
We get g = -1, f = 2, c = 0
Centre C = (-g, -f) = (1, -2)
Let B = (x, y) be the other end of the diameter.
Since ‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then
C = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right]\)
(1, -2) = \(\left(\frac{3+\mathrm{x}}{2}, \frac{-1+\mathrm{y}}{2}\right)\)
\(\frac{3+x}{2}\) = 1
⇒ 3 + x = 2
⇒ x = 2 – 3
⇒ x = -1
\(\frac{-1+\mathrm{y}}{2}\) = -2
⇒ -1 + y = -4
⇒ y = -4 + 1
⇒ y = -3
∴ Other end of the diameter B(x, y) = (-1, -3)

Question 54.
Find the equation of the circle passing through (3, -4) and concentric with x² + y² + 4x – 2y + 1 = 0.
Solution:
Given equation of the circle is x² + y² + 4x – 2y + 1 = 0
The equation of the circle concentric with the given circle is
x² + y² + 4x – 2y + k = 0 ……..(1)
The circle (1) passes through (3, -4) then
(3)² + (-4)² + 4(3) – 2(-4) + k = 0
⇒ 9 + 16 + 12 + 8 + k = 0
⇒ 45 + k = 0
⇒ k = -45
∴ The equation of the required circle is x² + y² + 4x – 2y – 45 = 0

Question 55.
Find the lengths of the intercepts made by the circle x² + y² + 8x – 12y – 9 = 0 on coordinate axes.
Solution:
Given equation of the circle is x² + y² + 8x – 12y – 9 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0,
we get g = 4, f = -6, c = -9
Length of the intercept made on X-axis = \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)
= \(2 \sqrt{(4)^2+9}\)
= 2√25
= 2(5)
= 10
Length of the intercept made on Y-axis = \(2 \sqrt{\mathrm{f}^2-\mathrm{c}}\)
= \(2 \sqrt{(-6)^2+9}\)
= 2√45
= 6√5

Question 56.
Find the power of the point (3, 4) with respect to x² + y² – 4x – 6y – 12 = 0.
Solution:
Given equation of the circle is x² + y² – 4x – 6y- 12 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0, we get
g = -2, f = -3, c = -12
Given point P(x₁, y₁) = (3, 4)
Power of (3, 4) is S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)
= (3)² + (4)² + 2(-2) (3) + 2(-3) (4) – 12
= 9 + 16 – 12 – 24 – 12
= -23

Question 57.
Find the polar of (1, 2) with respect to x² + y² = 7. (Mar. ’95)
Solution:
Given equation of the circle is x² + y² – 7 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0, we get
g = 0, f = 0, c = -7
Let the given point P(x₁, y₁) = (1, 2)
The equation of polar of P(1, 2) w.r.t. the given circle is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(1) + y(2) + 0(x + 1) + 0(y + 2) – 7 = 0
⇒ x + 2y – 7 = 0

Question 58.
Find the pole of the line 3x + 4y – 12 = 0 with respect to x² + y² = 24.
Solution:
Comparing 3x + 4y – 12 = 0 with the equation lx + my + n = 0, we get
l = 3, m = 4, n = -12
Comparing x² + y² = 24 with x² + y² = a²
we get a² = 24
∴ Pole = \(\left(\frac{-l \mathrm{a}^2}{\mathrm{n}}, \frac{-m \mathrm{a}^2}{\mathrm{n}}\right)\)
= \(\left(\frac{-3(24)}{-12}, \frac{-4(24)}{-12}\right)\)
= (6, 8)

Question 59.
Show that the points (-6, 1) and (2, 3) are conjugate points with respect to the circle x² + y² – 2x + 2y + 1 = 0. (Mar. ’96)
Solution:
Given equation of the circle is x² + y² – 2x + 2y + 1 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0,
we get g = -1, f = 1, c = 1
Let the given points are P(x₁, y₁) = (-6, 1), Q(x₂, y₂) = (2, 3)
Now S₁₂ = x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c
= (-6)(2) + 1(3) – 1(-6 + 2) + 1(1 + 3) + 1
= 12 + 3 – (-4) + 1(4) + 1
= -12 + 3 + 4 + 4 + 1
= -12 + 12
= 0
Since S₁₂ = 0 then, the given points are conjugate w.r.t. the given circle.

Question 60.
Find the pair of tangents drawn from (0, 0) to x² + y² + 10x + 10y + 40 = 0. (Mar. ’94)
Solution:
Given equation of the circle is x² + y² + 10x + 10y + 40 = 0
Here g = 5, f = 5, c = 40
Let the given point P(x₁, y₁) = (0, 0)
The equation of the pair of tangents drawn from (0, 0) to x² + y² + 10x + 10y + 40 = 0 is \(\mathrm{S}_1^2\) = SS₁₁
⇒ (xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c)² = (x² + y² + 2gx + 2fy + c) \(\left(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\right)\)
⇒ [x(0) + y(0) + 5(x + 0) + 5(y + 0) + 40]² = (x² + y² + 10x + 10y + 40) [0 + 0 + 2(5)(0) + 2(5)(0) + 40]
⇒ (5x + 5y + 40)² = (x² + y² + 10x + 10y + 40)(40)
⇒ 25(x + y + 8)² = (x² + y² + 10x + 10y + 40)(40)
⇒ 5(x² + y² + 64 + 2xy + 16y + 16x) = (x² + y² + 10x + 10y + 40) (8)
⇒ 5x² + 5y² + 320 + 10xy + 80y + 80x = 8x² + 8y² + 80x + 80y + 320
⇒ 3x² – 10xy + 3y² = 0

Question 61.
Find the equation to the pair of tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.
Solution:
Given equation of the circle is x² + y² – 6x + 4y – 2 = 0
Here, g = -3, f = 2, c = -2
Let, the given point P(x₁, y₁) = (3, 2)
∴ The equation of the pair of tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0 is SS₁₁ = \(\mathrm{S}_1^2\)
⇒ (x² + y² + 2gx + 2fy + c) \(\left(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g}_1+2 \mathrm{fy} \mathrm{y}_1+\mathrm{c}\right)\) = (xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c)²
⇒ (x² + y² – 6x + 4y – 2) [(3)² + (2)² + 2(-3)(3) + 2(2) (2) – 2] = [x(3) + y(2) – 3(x + 3) + 2(y + 2) + (-2)]²
⇒ (x² + y² – 6x + 4y – 2) (9 + 4- 18 + 8- 2) = (3x + 2y – 3x – 9 + 2y + 4 – 2)²
⇒ (x² + y² – 6x + 4y – 2) = (4y – 7)²
⇒ x² + y² – 6x + 4y – 2 = 16y² + 49 – 56y
⇒ x² – 15y² – 6x + 60y – 51 = 0

Question 62.
Find the pair of tangents from (4, 10) to the circle x² + y² = 25.
Solution:
Given the equation of the circle is x² + y² = 25
Here, g = 0, f = 0, c = -25
Let the given point P(x₁, y₁) = (4, 10)
The equation of the pair of tangents drawn from (4, 10) to x² + y² = 25 is SS₁₁ = \(\mathrm{S}_1{ }^2\)
⇒ (x² + y² + 2gx + 2fy + c) \(\left(x_1^2+y_1^2+2 g x_1+2 f y_1+c\right)\) = [xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c]²
⇒ (x² + y² – 25)((4)² + (10)² + 2(0)(4) + 2(0)(10) – 25) = [x(4) + y(10) + 0(x + 4) + 0(y + 10) – 25]²
⇒ (x² + y² – 25) (16 + 100 – 25) = (4x + 10y – 25)²
⇒ (x² + y² – 25) (91) = 16x² + 100y² + 625 + 80xy – 500y – 200x
⇒ 91x² + 91y² – 2275 = 16x² + 100y² + 625 + 80xy – 500y – 200x
⇒ 75x² – 80xy – 9y² + 200x + 500y – 2900 = 0

Question 63.
Find the angle between the pair of tangents drawn from (1, 3) to the circle x² + y² – 2x + 4y – 11 = 0. [(TS) Mar. ’16]
Solution:
Given equation of the circle is x² + y² – 2x + 4y – 11 = 0

Question 64.
m Find the equation of a circle with centre (1, 4) and radius 5. [Mar. ’17 (AP)]
Solution:
Given centre C(h, k) = (1, 4)
radius r = 5
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 1)² + (y – 4)² = (5)²
⇒ x² + 1 – 2x + y² + 16 – 8y = 25
⇒ x² + y² – 2x – 8y – 8 = 0

Question 65.
Obtain the parametric equation of the circle 4(x² + y²) = 9. [Mar. ’17 (TS)]
Solution:
Given equation of the circle is 4(x² + y²) = 9
⇒ x² + y² = \(\frac{9}{4}\)
This is of the form x² + y² = r²
centre C = (h, k) = (0, 0)
radius r = \(\frac{3}{2}\)
∴ The parametric equations of the circle are
x = h + r cos θ = 0 + \(\frac{3}{2}\) cos θ = \(\frac{3}{2}\) cos θ
y = k + r sin θ = 0 + \(\frac{3}{2}\) sin θ = \(\frac{3}{2}\) sin θ
0 ≤ θ ≤ 2π

Question 66.
Find the centre and radius of the circle 3x² + 3y² + 6x – 12y – 1 = 0.
Solution:
Given equation of the circle is 3x² + 3y² + 6x – 12y – 1 = 0
⇒ x² + y² + 2x – 4y – \(\frac{1}{3}\) = 0 …….(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0, we get

Question 67.
Find the equation of the circle whose extremities of a diameter are (1, 2), (4, 6).
Solution:
Let A(x₁, y₁) = (1, 2) and B(x₂, y₂) = (4, 6) are the two given points.
∴ The equation of the required circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 1) (x – 4) + (y – 2) (y – 6) = 0
⇒ x² – 4x – x + 4 + y² – 6y – 2y + 12 = 0
⇒ x² + y² – 5x – 8y + 16 = 0

Question 68.
Find the number of common tangents to the circles x² + y² + 6x + 6y + 14 = 0, x² + y² – 2x – 4y – 4 = 0.
Solution:
Given equations of the circles are
x² + y² + 6x + 6y + 14 = 0 ……..(1)
x² + y² – 2x – 4y – 4 = 0 ………(2)
For the circle (1), centre, C₁ = (-3, -3)

Now, r₁ + r₂ = 2 + 3 = 5 = √25
∴ C₁C₂ > r₁ + r₂
∴ Number of common tangents = 4

Question 69.
Find the equation of the circle whose centre is (-1, 2) and which passes through (5, 6). [Mar. ’18 (TS)]
Solution:
Given centre c(h, k) = (-1, 2) and P(5, 6) be a point on the circle

Radius r = CP = \(\sqrt{(5+1)^2+(6-2)^2}\)
= \(\sqrt{36+16}\)
= √52
Equation of the required circle is (x – h)² + (y – k)² = r²
⇒ (x + 1)² + (y – 2)² = (√52)²
⇒ x² + 1 + 2x + y² + 4 – 4y = 52
⇒ x² + y² + 2x – 4y – 47 = 0

Question 70.
Write the parametric equations of the circle 2x² + 2y² = 7. [Mar. ’19 (TS)]
Solution:
Given the equation of the circle is 2x² + 2y² = 7
⇒ x² + y² = \(\frac{7}{2}\)
This is of the form x² + y² = r²
centre C = (h, k) = (0, 0)
radius r = \(\frac{\sqrt{7}}{\sqrt{2}}\)
∴ The parametric equations of the circle are
x = h + r cos θ = 0 + \(\frac{\sqrt{7}}{\sqrt{2}}\) cos θ = \(\frac{\sqrt{7}}{\sqrt{2}}\) cos θ
y = k + r sin θ = 0 + \(\frac{\sqrt{7}}{\sqrt{2}}\) sin θ = \(\frac{\sqrt{7}}{\sqrt{2}}\) sin θ
0 ≤ θ ≤ 2π

Question 71.
Prove that the equation of the circle with centre C(h, k) and radius r is (x – h)² + (y – k)² = r².
Solution:
Let P(x, y) be a point P lies in the circle then PC = r

⇒ PC = r
⇒ \(\sqrt{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2}\) = r
squaring on both sides
⇒ \(\left(\sqrt{(x-h)^2+(y-k)^2}\right)^2=r^2\)
⇒ (x – h)² + (y – k)² = r²
∴ The locus of P is (x – h)² + (y – k)² = r²
∴ The equation of the circle is (x – h)² + (y – k)² = r²

Question 72.
Prove that the equation of the circle with centre O(0, 0) and radius ‘r’ is x² + y² = r².
Solution:
Let P(x, y) be a point.
P lies in the circle then PO = r

\(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) = r
Squaring on both sides
∴ x² + y² = r²

Question 73.
Find the equation of the circle with centre (cos α, sin α) and radius 1.
Solution:
Given centre C(h, k) = (cos α, sin α)
radius r = 1
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – cos α)² + (y – sin α)² = (1)²
⇒ x² + cos²α – 2x cos α + y² + sin²α – 2y sin α = 1
⇒ x² + y² – 2x cos α – 2y sin α + 1 – 1 = 0
⇒ x² + y² – 2x cos α – 2y sin α = 0

Question 74.
Find the equation of the circle with centre \(\left(\frac{-1}{2},-9\right)\) and radius 5.
Solution:
Given centre C(h, k) = \(\left(\frac{-1}{2},-9\right)\)
radius r = 5
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ \(\left(\mathrm{x}+\frac{1}{2}\right)^2+(\mathrm{y}+9)^2=(5)^2\)
⇒ x² + x + \(\frac{1}{4}\) + y² + 81 + 18y – 25 = 0
⇒ x² + y² + x + 18y + 56 + \(\frac{1}{4}\) = 0
⇒ 4x² + 4y² + 4x + 72y + 1 + 224 = 0
⇒ 4x² + 4y² + 4x + 72y + 225 = 0

Question 75.
Find the equation of the circle with centre \(\left(\frac{5}{2}, \frac{-4}{3}\right)\) and radius 6.
Solution:
Given centre C(h, k) = \(\left(\frac{5}{2}, \frac{-4}{3}\right)\)
radius r = 6
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ \(\left(x-\frac{5}{2}\right)^2+\left(y+\frac{4}{3}\right)^2=(6)^2\)
⇒ \(x^2+\frac{25}{4}-5 x+y^2+\frac{16}{9}+\frac{8 y}{3}=36\)
⇒ \(\frac{36 x^2+225-180 x+36 y^2+64+96 y}{36}\) = 36
⇒ 36x² + 36y² – 180x + 96y + 289 = 1296
⇒ 36x² + 36y² – 180x + 96y – 1007 = 0

Question 76.
Find the equation of the circle with centre (1, 7) and radius \(\frac{5}{2}\).
Solution:
Given centre C(h, k) = (1, 7)
radius r = \(\frac{5}{2}\)
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 1)² + (y – 7)² = \(\left(\frac{5}{2}\right)^2\)
⇒ x² + 1 – 2x + y² + 49 – 14y = \(\frac{25}{4}\)
⇒ 4x² + 4 – 8x + 4y² + 196 – 56y = 25
⇒ 4x² + 4y² – 8x – 56y + 175 = 0

Question 77.
Find the equation of the circle with centre (0, 0) and radius 9.
Solution:
Given centre C(h, k) = (0, 0)
radius r = 9
∴ The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 0)² + (y – 0)² = (9)²
⇒ x² + y² – 81 = 0

Question 78.
Find the centre and radius of the circle 2x² + 2y² – 3x + 2y – 1 = 0.
Solution:
Given equation of the circle is 2x² + 2y² – 3x + 2y – 1 = 0

Question 79.
Find the equation of the circle having (7, -3), (3, 5) as the endpoints of a diameter.
Solution:
Let A(x₁, y₁) = (7, -3) and B(x₂, y₂) = (3, 5) are the two given points.
∴ The equation of the required circle is (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 7)(x – 3) + (y + 3)(y – 5) = 0
⇒ x² – 7x – 3x + 21 + y² – 5y + 3y – 15 = 0
⇒ x² + y² – 10x – 2y + 6 = 0

Question 80.
Find the equation of the circle having (1, 1), (2, -1) as the endpoints of a diameter.
Solution:
Let A(x₁, y₁) = (1, 1) and B(x₂, y₂) = (2, -1) are the two given points.
∴ The equation of the required circle is (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 1) (x – 2) + (y – 1) (y + 1) = 0
⇒ x² – 2x – x + 2 + y² – 1 = 0
⇒ x² + y² – 3x + 1 = 0

Question 81.
Find the equation of the circle having (0, 0), (8, 5) as the endpoints of a diameter.
Solution:
Let A(x₁, y₁) = (0, 0) and B(x₂, y₂) = (8, 5) are the two given points.
The equation of the required circle is (x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 0) (x – 8) + (y – 0) (y – 5) = 0
⇒ x² – 8x + y² – 5y = 0
⇒ x² + y² – 8x – 5y = 0

Question 82.
Obtain the parametric equations of the circle x² + y² – 4x – 6y – 12 = 0.
Solution:
Given the equation of the circle is
x² + y² – 4x – 6y – 12 = 0 ……..(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
We get 2g = -4 ⇒ g = -2
2f = -6 ⇒ f = -3
c = -12
Centre C(h, k) = (-g, -f) = (2, 3)
Radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)
= \(\sqrt{(-2)^2+(-3)^2+12}\)
= √25
= 5
∴ The parametric equations of the circle are
x = h + r cos θ = 2 + 5 cos θ
y = k + r sin θ = 3 + 5 sin θ, 0 ≤ θ ≤ 2π.

Question 83.
Show that the power of a point P(x1, y1) w.r.t the circle S = 0 is S11.
Solution:
Let S = x² + y² + 2gx + 2fy + c = 0 be the given circle and P(x₁, y₁) be any point in the plane, then centre C(-g, -f),

Question 84.
Prove that the length of the tangent drawn from an external point P(x₁, y₁) to the circle S = 0 is \(\sqrt{\mathbf{S}_{11}}\).
Solution:
Let S = x² + y² + 2gx + 2fy + c = 0 be the given circle

Centre C = (-g, -f) and Radius r = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\)
Let the tangents drawn from P touches the circle at A.
∴ ∠PAC = 90°
⇒ OP² = PA² + CA²

Question 85.
Locate the position of the point P(1, 5) with respect to the circle x² + y² – 2x – 4y + 3 = 0.
Solution:
Given equation of the circle is x² + y² – 2x – 4y + 3 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0,
We get g = -1, f = -2, c = 3
Given point P(x₁, y₁) = (1, 5)
Now S₁₁ = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (1)² + (5)² + 2(-1)(1) + 2(-2)(5) + 3
= 1 + 25 – 2 – 20 + 3
= 29 – 22
= 7
Since S₁₁ > 0, then the point P(1, 5) is outside the given circle.

Question 86.
Locate the position of the point P(4, 2) with respect to the circle 2x² + 2y² – 5x – 4y – 3 = 0.
Solution:
Given equation of the circle is 2x² + 2y² – 5x – 4y – 3 = 0
⇒ \(x^2+y^2-\frac{5}{2} x-2 y-\frac{3}{2}=0\)
Comparing this equation with x² + y² + 2gx + 2fy + c = 0

= 20 – 14 – \(\frac{3}{2}\)
= 6 – \(\frac{3}{2}\)
= \(\frac{9}{2}\) > 0
Since S₁₁ > 0, then the point P(4, 2) lies outside the circle.

Question 87.
Locate the position of the point P(2, -1) w.r.t the circle x² + y² – 2x – 4y + 3 = 0.
Solution:
Given equation of the circle is x² + y² – 2x – 4y + 3 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0
We get g = -1, f = -2, c = 3
Given point P(x₁, y₁) = (2, -1)
Now S₁₁ = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)
= (2)² – (-1)² + 2(-1)(2) + 2(-2)(-1) + 3
= 4 + 1 – 4 + 4 + 3
= 8 > 0
∵ S₁₁ > 0, then the point P(2, -1) lies outside the circle.

Question 88.
Locate the position of the point (2, 4) w.r.t the circle x² + y² – 4x – 6y + 11 = 0.
Solution:
Given equation of the circle is x² + y² – 4x – 6y + 11 = 0
Comparing this equation of the circle is x² + y² + 2gx + 2fy + c = 0
We get g = -2, f = -3, c = 11
Given point P(x₁, y₁) = (2, 4)
Now S₁₁ = \(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}\)
= (2)² + (4)² + 2(-2)(2) + 2(4)(-3) + 11
= 4 + 16 – 8 – 24 + 11
= 31 – 32
= -1 < 0
∵ S₁₁ < 0, then the point P(2, 4) lies inside the given circle.

Question 89.
Find the power of the point (5, -6) with respect to the circle x² + y² + 8x + 12y + 15 = 0.
Solution:
Given equation of the circle is x² + y² + 8x + 12y + 15 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = 4, f = 6, c = 15
Point P(x₁, y₁) = (5, -6)
∴ The power of P(5, -6) w.r.t the given circle is S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{gx}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)
S₁₁ =(5)² + (-6)² + 2(4)(5) + 2(6)(-6) + 15
= 25 + 36 + 40 – 72 + 15
= 116 – 72
= 44

Question 90.
Find the power of the point (2, 3) with respect to the circle x² + y² – 2x + 8y – 23 = 0.
Solution:
The given equation of the circle is x² + y² – 2x + 8y – 23 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -1, f = 4, c = -23
Point P(x₁, y₁) = (2, 3)
∴ The power of P(2, 3) w.r.t the given circle is S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{fy}_1+\mathrm{c}\)
S₁₁ = (2)² – (3)² + 2(-1)(2) + 2(4)(3) – 23
= 4 + 9 – 4 + 24 – 23
= 10

Question 91.
Find the power of the point (2, 4) with respect to the circle x² + y² – 4x – 6y – 12 = 0.
Solution:
The given equation of the circle is x² + y² – 4x – 6y – 12 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -2, f = -3, c = -12
Point P(x₁, y₁) = (2, 4)
∴ The power of P(2, 4) w.r.t the given circle is S₁₁ = \(\mathrm{x}_1^2+\mathrm{y}_1^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{f} \mathrm{y}_1+\mathrm{c}\)
S₁₁ = (2)² + (4)² + 2(-2)(2) + 2(-3)(4) – 12
= 4 + 16 – 8 – 24 – 12
= -24

Question 92.
Find the length of tangent from P(-2, 5) to the circle x² + y² – 25 = 0.
Solution:
The given equation of the circle is x² + y² – 25 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = 0, f = 0, c = -25
Point P(x₁, y₁) = (-2, 5)
∴ The length of the tangent = \(\sqrt{\mathrm{S}_{11}}\)

Question 93.
Find the equation of the tangent at (7, -5) of the circle x² + y² – 6x + 4y – 12 = 0.
Solution:
Given equation of the circle is x² + y² – 6x + 4y – 12 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -3, f = 2, c = -12
The given point P(x₁, y₁) = (7, -5)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(7) + y(-5) – 3(x + 7) + 2(y – 5) – 12 = 0
⇒ 7x – 5y – 3x – 21 + 2y – 10 – 12 = 0
⇒ 4x – 3y – 43 = 0

Question 94.
Find the equation of the tangent at (3, 4) of the circle x² + y² – 4x – 6y + 11 = 0.
Solution:
The given equation of the circle is x² + y² – 4x – 6y + 11 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -2, f = -3, c = 11
The given point P(x₁, y₁) = (3, 4)
The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(3) + y(4) + (-2)(x + 3) – 3(y + 4) + 11 = 0
⇒ 3x + 4y – 2x – 6 – 3y -12 + 11 = 0
⇒ x + y – 7 = 0

Question 95.
Find the equation of the normal at (3, -4) of the circle x² + y² + x + y – 24 = 0.
Solution:
The given equation of the circle is x² + y² + x + y – 24 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = \(\frac{1}{2}\), f = \(\frac{1}{2}\), c = -24
The given point P(x₁, y₁) = (3, -4)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(3) + y(-4) + \(\frac{1}{2}\)(x + 3) + \(\frac{1}{2}\)(y – 4) – 24 = 0
⇒ 3x – 4y + \(\frac{x}{2}+\frac{3}{2}+\frac{y}{2}\) – 2 – 24 = 0
⇒ 6x – 8y + x + 3 + y – 4 – 48 = 0
⇒ 7x – 7y – 49 = 0
⇒ x – y – 7 = 0
The slope of the tangent is m = \(\frac{-a}{b}=\frac{-1}{-1}\) = 1
The slope of the normal at P = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{1}\) = -1
∴ The equation of the normal at P is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ y + 4 = -1(x – 3)
⇒ y + 4 = -x + 3
⇒ x + y + 1 = 0

Question 96.
Find the equation of the normal at (3, 5) of the circle x² + y² – 10x – 2y + 6 = 0. [(AP) Mar. ’18]
Solution:
The given equation of the circle is x² + y² – 10x – 2y + 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -5, f = -1, c = 6
The given point P(x₁, y₁) = (3, 5)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(3) + y(5) + (-5)(x + 3) – 1(y + 5) + 6 = 0
⇒ 3x + 5y – 5x – 15 – y – 5 + 6 = 0
⇒ -2x + 4y – 14 = 0
⇒ x – 2y + 7 = 0
The slope of the tangent at P is
m = \(\frac{-a}{b}=\frac{-1}{-2}\) = 1
The slope of the normal at
P = \(\frac{-1}{m}=\frac{-1}{\frac{1}{2}}\) = -2
∴ The equation of the normal at P(3, 5) is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ y – 5 = -2(x – 3)
⇒ y – 5 = -2x + 6
⇒ 2x + y – 11 = 0

Question 97.
Find the equation of the normal at (1, 3) of the circle 3(x² + y²) – 19x – 29y + 76 = 0.
Solution:
The given equation of the circle is 3x² + 3y² – 19x – 29y + 76 = 0
⇒ \(x^2+y^2-\frac{19}{3} x-\frac{29}{3} y+\frac{76}{3}=0\)
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get \(\mathrm{g}=\frac{-19}{6}, \quad \mathrm{f}=\frac{-29}{6}, \mathrm{c}=\frac{76}{3}\)
The given point P(x₁, y₁) = (1, 3)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ \(x(1)+y(3)-\frac{19}{6}(x+1)-\frac{29}{6}(y+3)+\frac{76}{3}=0\)
⇒ 6x + 18y – 19x – 19 – 29y – 87 + 152 = 0
⇒ -13x – 11y + 46 = 0
⇒ 13x + 11y – 46 = 0
The slope of the tangent at P is
m = \(\frac{-a}{b}=\frac{-13}{11}\)
The slope of the normal at
P = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{\frac{-13}{11}}=\frac{11}{13}\)
∴ The equation of the normal at P is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ y – 3 = \(\frac{11}{33}\) (x – 1)
⇒ 13y – 39 = 11x – 11
⇒ 11x – 13y + 28 = 0

Question 98.
Find the equation of the normal at (1, 2) of the circle x² + y² – 22x – 4y + 25 = 0.
Solution:
The given equation of the circle is x² + y² – 22x – 4y + 25 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = -11, f = -2, c = 25
The given point P(x₁, y₁) = (1, 2)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(1) + y(3) – 11(x + 1) – 2(y + 2) + 25 = 0
⇒ x + 2y – 11x – 11 – 2y – 4 + 25 = 0
⇒ -10x + 10 = 0
⇒ x – 1 = 0
The slope of the tangent at P is m = \(\frac{-1}{0}\)
The slope of the normal at
P = \(\frac{-1}{m}=\frac{-1}{\frac{-1}{0}}=0\)
∴ The equation of the normal at P(1, 2) is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ y – 2 = 0(x – 1)
⇒ y – 2 = 0

Question 99.
Find the equation of the circle with centre (-3, 4) and touching y-axis.
Solution:
Given centre C(h, k) = (-3, 4).
The equation of the y-axis is x = 0.

Since the required circle is touching the y-axis, the y-axis is the tangent to the required circle.
∴ Radius r = The perpendicular distance from the centre C(-3, 4) to the y-axis (x = 0)
∴ r = d = \(\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}\)
= \(\frac{|1(-3)+0(4)+0|}{\sqrt{(1)^2+(0)^2}}=\frac{|-3|}{1}\)
= 3
∴ The required equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + 3)² + (y – 4)² = (3)²
⇒ x² + 9 + 6x + y² + 16 – 8y = 9
⇒ x² + y² + 6x – 8y + 16 = 0

Question 100.
Find the equation of tangent and normal at (3, 2) of the circle x² + y² – x – 3y – 4 = 0. [(AP) May ’19]
Solution:
The given equation of the circle is x² + y² – x – 3y – 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = \(\frac{-1}{2}\), f = \(\frac{-3}{2}\), c = -4
The given point P(x₁, y₁) = (3, 2)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(3) + y(2) – \(\frac{1}{2}\)(x + 3) – \(\frac{3}{2}\)(y + 2) – 4 = 0
⇒ 6x + 4y – x – 3 – 3y – 6 – 8 = 0
⇒ 5x + y – 17 = 0
The slope of the tangent at P is
m = \(\frac{-5}{1}\) = -5
The slope of the normal at
P = \(\frac{-1}{m}=\frac{-1}{-5}=\frac{1}{5}\)
∴ The equation of the normal at P is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ -2 = \(\frac{1}{5}\) (x – 3)
⇒ 5y – 10 = x – 3
⇒ x – 5y + 7 = 0

Question 101.
Find the equation of tangent and normal at (1, 1) of the circle 2x² + 2y² – 2x – 5y + 3 = 0.
Solution:
The given equation of the circle is 2x² + 2y² – 2x – 5y + 3 = 0
⇒ \(x^2+y^2-x-\frac{5}{2} y+\frac{3}{2}=0\) …..(1)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0
We get \(\mathrm{g}=\frac{-1}{2}, \mathrm{f}=\frac{-5}{4}, \mathrm{c}=-\frac{3}{2}\)
The given point P(x₁, y₁) = (1, 1)
∴ The equation of the tangent at P is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(1) + y(1) – \(\frac{1}{2}\)(x + 1) – \(\frac{5}{4}\)(y + 1) + \(\frac{3}{2}\) = 0
⇒ 4x + 4y – 2x – 2 – 5y – 5 + 6 = 0
⇒ 2x – y – 1 = 0
The slope of the tangent at P is m = \(\frac{-2}{-1}\) = 2
The slope of the normal at P is \(\frac{-1}{\mathrm{~m}}=\frac{-1}{2}\)
∴ The equation of the normal at P is
\(y-y_1=\frac{-1}{m}\left(x-x_1\right)\)
⇒ \(y-1=\frac{-1}{2}(x-1)\)
⇒ 2y – 2 = -x + 1
⇒ x + 2y – 3 = 0

Question 102.
Show that the circle S = x² + y² + 2gx + 2fy + c = 0 touches the
(i) X-axis if g² = c
(ii) Y-axis if f² = c
Solution:
Given the equation of the circle is
S = x² + y² + 2gx + 2fy + c = 0
(i) The intercept made by S = 0 on X-axis = \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)
Since, the circle touches the X-axis, then
\(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\) = 0
⇒ \(\sqrt{g^2-c}\) = 0
⇒ g² – c = 0
⇒ g² = c

(ii) The intercept made by S = 0 on Y-axis is \(2 \sqrt{\mathrm{f}^2-\mathrm{c}}\)
Since, the circle touches the Y-axis, then
⇒ \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\)
⇒ \(\sqrt{f^2-c}\) = 0
⇒ f² – c = 0
⇒ f² = c

Question 103.
If the parametric values of two points A and B lying on the circle x² + y² – 6x + 4y – 12 = 0 are 30° and 60° respectively, then find the equation of the chord joining A and B.
Solution:
Given equation of the circle is x² + y² – 6x + 4y – 12 = 0
Comparing the given equation with x² + y² + 2gx + 2fy + c = 0

⇒ 2x – 6 + 2y + 4 = 5√3 + 5
⇒ 2x + 2y – 2 – 5 – 5√3 = 0
⇒ 2x + 2y – (7 + 5√3) = 0

Question 104.
Find the chord of contact of (0, 5) w.r.t the circle x² + y² – 5x + 4y – 2 = 0.
Solution:
Given equation of the circle is x² + y² – 5x + 4y – 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = \(\frac{-5}{2}\), f = 2, c = -2
Let, the given point P(x₁, y₁) = (0, 5)
∴ The equation of the chord of contact is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(0) + y(5) – \(\frac{5}{2}\)(x + 0) + 2(y + 5) – 2 = 0
⇒ 0 + 5y – \(\frac{5}{2}\)x + 2y + 10 – 2 = 0
⇒ 10y – 5x + 4y + 20 – 4 = 0
⇒ -5x + 14y + 16 = 0
⇒ 5x – 14y – 16 = 0

Question 105.
Find the chord of contact of (2, 5) w.r.t the circle x² + y² – 5x + 4y – 2 = 0.
Solution:
Given equation of the circle is x² + y² – 5x + 4y – 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = \(\frac{-5}{2}\), f = 2, c = -2
Let, the given point P(x₁, y₁) = (2, 5)
∴ The equation of the chord of contact is S₁ = 0
⇒ xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ x(2) + y(5) – \(\frac{5}{2}\)(x + 2) + 2(y + 5) – 2 = 0
⇒ 2x + 5y – \(\frac{5}{2}\)x – 5 + 2y + 10 – 2 = 0
⇒ 2x – \(\frac{5}{2}\)x + 7y + 3 = 0
⇒ 4x – 5x + 14y + 6 = 0
⇒ -x + 14y + 6 = 0
⇒ x – 14y – 6 = 0

Question 106.
Show that the points (4, 2) and (3, -5) are conjugate points w.r.t the circle x² + y² – 3x – 5y + 1 = 0.
Solution:
Given equation of the circle is x² + y² – 3x – 5y + 1 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get \(\mathrm{g}=\frac{-3}{2}, \mathrm{f}=\frac{-5}{2}, \mathrm{c}=1\)
Let, the given point are A(x₁, y₁) = (4, 2), B(x₂, y₂) = (3, -5)
Now S₁₂ = x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c = 0

Since S₁₂ = 0, the given points are conjugate w.r.t the given circle.

Question 107.
Find the value of ‘k’ if the points (1, 3) and (2, k) are conjugate w.r.t the circle x² + y² = 35.
Solution:
Given the equation of the circle is x² + y² = 35
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = 0, f = 0, c = -35
Let, the given points are A(x₁, y₁) = (1, 3), B(x₂, y₂) = (2, k)
Since the given points are conjugate, then S₁₂ = 0
⇒ x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c = 0
⇒ 1(2) + 3(k) + 0(1 + 2) + 0(3 + k) – 35 = 0
⇒ 2 + 3k + 0 + 0 – 35 = 0
⇒ 3k – 33 = 0
⇒ 3k = 33
⇒ k = 11

Question 108.
Find the value of ‘k’ if the points (4, 2) and (k, -3) are conjugate points w.r.t the circle x² + y² – 5x + 8y + 6 = 0.
Solution:
Given equation of the circle is x² + y² – 5x + 8y + 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0
We get g = \(\frac{-5}{2}\), f = 4, c = 6
Let, the given points are A(x₁, y₁) = (4, 2) and B(x₂, y₂) = (k, -3)
Since the given points are conjugate, then S₁₂ = 0
⇒ x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + c = 0
⇒ 4(k) + 2(-3) – \(\frac{5}{2}\)(4 + k) + 4(2 – 3) + 6 = 0
⇒ 4k – 6 – \(\frac{20}{2}-\frac{5 \mathrm{k}}{2}\) – 4 + 6 = 0
⇒ \(\frac{-5 k}{2}\) + 4k – 14 = 0
⇒ \(\frac{-5 k+8 k-28}{2}\) = 0
⇒ 3k – 28 = 0
⇒ 3k = 28
⇒ k = \(\frac{28}{3}\)

Question 109.
Find the angle between the tangents drawn from (3, 2) to the circle x² + y² – 6x + 4y – 2 = 0.
Solution:
Given equation of the circle is x² + y² – 6x + 4y – 2 = 0

Question 110.
Discuss the relative position of the pair of circles x² + y² – 4x – 6y – 12 = 0, x² + y² + 6x + 18y + 26 = 0.
Solution:
Given equations of the circles are
x² + y² – 4x – 6y – 12 = 0 …….(1)
x² + y² + 6x + 18y + 26 = 0 ………(2)
for the circle (1), centre c₁ = (2, 3)

Now, r₁ + r₂ = 5 + 8 = 13
∴ c₁c₂ = r₁ + r₂
∴ The given circles touch each other externally.

Question 111.
Discuss the relative position of the pair of circles x² + y² – 2x + 4y – 4 = 0, x² + y² + 4x – 6y – 3 = 0.
Solution:
Given equations of the circles are
x² + y² – 2x + 4y – 4 = 0 …….(1)
x² + y² + 4x – 6y – 3 = 0 ………..(2)
For the circle (1), centre c₁ = (1, -2)

Question 112.
Find the number of possible common tangents that exist for the pairs of circles x² + y² + 6x + 6y + 14 = 0, x² + y² – 2x – 4y – 4 = 0.
Solution:
Given equations of the circles are
x² + y² + 6x + 6y + 14 = 0 ……..(1)
x² + y² – 2x – 4y – 4 = 0 ……….(2)
For the circle (1), Centre c₁ = (-3, -3)


Now, r₁ + r₂ = 2 + 3 = 5 = √25
∴ c₁c₂ > r₁ + r₂
∴ Given circles, each circle lies completely outside the other.
∴ The number of common tangents = 4.

Question 113.
Find the number of possible common tangents that exist for the pairs of circles x² + y² – 4x – 2y + 1 = 0, x² + y² – 6x – 4y + 4 = 0.
Solution:
Given equations of the circles are
x² + y² – 4x – 2y + 1 = 0 ……..(1)
x² + y² – 6x – 4y + 4 = 0 ……….(2)
For the circle (1), Centre c₁ = (2, 1)

Now, |r₁ – r₂| = |2 – 3| = |-1| = 1
r₁ + r₂ = 2 + 3 = 5 = √25
∴ |r₁ – r₂| < c₁c₂ < r₁ + r₂
∴ The given circles intersect at two points.
∴ The number of common tangents = 2.

Question 114.
Find the number of possible common tangents that exist for the pairs of circles x² + y² – 4x + 2y – 4 = 0, x² + y² + 2x – 6y + 6 = 0.
Solution:
Given equations of the circles are
x² + y² – 4x + 2y – 4 = 0 ……..(1)
x² + y² + 2x – 6y + 6 = 0 ……..(2)
For the circle (1), Centre c₁ = (2, -1)


Now, r₁ + r₂ = 3 + 2 = 5
∴ c₁c₂ = r₁ + r₂
∴ The given circles touch each other externally.
∴ The number of common tangents = 3.

Question 115.
Find the number of possible common tangents that exist for the pairs of circles x² + y² + 4x – 6y – 3 = 0, x² + y² + 4x – 2y + 4 = 0.
Solution:
Given equations of the circles are
x² + y² + 4x – 6y – 3 = 0 ……..(1)
x² + y² + 4x – 2y + 4 = 0 ……….(2)
For the circle (1), Centre c₁ = (-2, 3)

Now, |r₁ + r₂| = |4 + 1| = 5
∴ c₁c₂ < |r₁ + r₂|
∴ In the given circles one circle lies completely inside the other.
∴ The number of common tangents = 0.

Question 116.
Find the equation of the pair of tangents from (10, 4) to the circle x² + y² = 25.
Solution:
The given equation of the circle is x² + y² = 25.
Here g = 0, f = 0, c = -25
Let, the given point P(x₁, y₁) = (10, 4)
∴ The equation of the pair of tangents drawn from (10, 4) to the circle x² + y² = 25 is
SS₁₁ = \(\mathrm{S}_1{ }^2\)
⇒ (x² + y² + 2gx + 2fy+ c) \(\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{g} \mathrm{x}_1+2 \mathrm{fy}_1+\mathrm{c}\right)\) = (xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c)²
⇒ (x² + y² – 25) [(10)² + (4)² + 2(0)(10) + 2(0)(4) – 25] = [x(10) + y(4) + 0(x + 10) + 0(y + 4) – 25]²
⇒ (x² + y² – 25) (100 + 16 – 25) = (10x + 4y – 25)²
⇒ (x² + y² – 25) (91) = (10x + 4y – 25)²
⇒ 91x² + 91y² – 2275 = 100x² + 16y² + 625 + 80xy – 200y – 500x
⇒ 9x² + 80xy – 75y² – 500x – 200y + 2900 = 0

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Complex Numbers Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Complex Numbers Important Questions Very Short Answer Type

Question 1.
Find the additive inverse of (√3, 5). [March ’90]
Solution:
Let z = (√3, 5)
The additive inverse of z is – z = – (√3, 5) = (- √3, – 5)

Question 2.
If z = (cos θ, sin θ), find (z – \(\frac{1}{z}\)). [TS – Mar. 2019]
Solution:
Given that z = (cos θ, sin θ)
The multiplicative inverse of z is
z^-1 = \(\frac{1}{z}=\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)\)
= \(\left(\frac{\cos \theta}{\cos ^2 \theta+\sin ^2 \theta}, \frac{-\sin \theta}{\cos ^2 \theta+\sin ^2 \theta}\right)\)
= \(\left(\frac{\cos \theta}{1}, \frac{-\sin \theta}{1}\right)\)
= (cos θ – sin θ)
z – \(\frac{1}{z}\) = (cos θ, sin θ) – (cos θ, – sin θ)
= (cos θ – cos θ, sin θ + sin θ)
= (0, 2 sin θ)

Question 3.
Find the multiplicative inverse of (sin θ, cos θ).
Solution:
Let z = (sin θ, cos θ)
The multiplicative inverse of z is
z^-1 = \(\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)\)
z^-1 = \(\left(\frac{\sin \theta}{\cos ^2 \theta+\sin ^2 \theta}, \frac{-\cos \theta}{\cos ^2 \theta+\sin ^2 \theta}\right)\)
= \(\left(\frac{\sin \theta}{1}, \frac{-\cos \theta}{1}\right)\)
= (sin θ, – cos θ)

Question 4.
Express \(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\) in the form a + ib.
Solution:

Question 5.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\). [TS – Mar. 2015]
Solution:

Question 6.
Express (1 – i)³ (1 + i) in the form a + ib.
Solution:
Let z = (1 – i)³ (1 + i)
= (1 – i)² (1 – i) (1 + i)
= (1 + i² – 2i) (1 – i²)
= (1 – 1 – 2i) (1 – (- 1)) (∵ i² = – 1)
= (- 2i) (2)
= – 4i = 0 + i (- 4)
It is in the form a + ib.

Question 7.
Find the multiplicative inverse of 7 + 24i. [TS – Mar. 2016, AP – May 2015]
Solution:
Let z = 7 + 24i
z^-1 = \(\)
∴ The multiplicative inverse of complex number z = 7 + 24i is
z = \(\left(\frac{7-24 i}{7^2+(24)^2}\right)\)
= \(\left(\frac{7-24 \mathrm{i}}{49+576}\right)=\frac{7-24 \mathrm{i}}{625}\) .

Question 8.
If 4x + i(3x – y) = 3 – 6i where x and y are real numbers, then find the values of\ x and y.
Solution:
Given that 4x + i(3x – y) = 3 – 6i
Comparing real and imaginary parts on both sides
4x = 3
x = \(\frac{3}{4}\)

3x – y = – 6
3 (\(\frac{3}{4}\)) – y = – 6
y = \(\frac{9}{4}\) + 6
= \(\frac{9+24}{4}\)
= \(\frac{33}{4}\)
∴ x = \(\frac{3}{4}\), y = \(\frac{33}{4}\)

Question 9.
Find the complex conjugate of (3 + 4i) (2 – 3i). [May ’14, Mar. 1997, AP – Mar. ’18, May ’16]
Solution:
Let z = (3 + 4i) (2 – 3i)
= 6 – 9i + 8i – 12i²
= 6 – i + 12
z = 18 – i
∴ The conjugate of z is \(\overline{\mathrm{z}}=\overline{(18-\mathrm{i})}\) = 18 + ¡

Question 10.
Find the square root of – 5 + 12i. [Board Paper]
Solution:

Question 11.
Find the square root of – 47 + i. 8√3. [March ’13(old), May ’12, ’09, May ’10, ’03]
Solution:

Question 12.
Simplify i² + i⁴ + i⁶ + …………… + (2n + 1) terms.
Solution:
= i² + i⁴ + i⁶ + …………… + (2n + 1) terms
= i² + (i²)² + (i²)³ + …………….. + (2n + 1) terms
= – 1 + 1 – 1 + ……………. + (2n + 1) terms = – 1

Question 13.
Simplify i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (- i)²⁶.
Solution:
i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (- i)²⁶.
= (i²)⁹ – 3 (i²)³ . i + i² (1 + (i²)²) (i²)¹³
= (- 1)⁹ – 3 (- 1)³ . i + (- 1) (1 + (- 1)²) (- 1)¹³
= – 1 + 3i + 2 = 1 + 3i

Question 14.
If (a + ib)² = x + iy, find x² + y². [TS – May 2015]
Solution:
Given that,
(a + ib)² = x + iy
a² + i²b² + 2abi = x + iy
a² – b² + i(2ab) =x + iy
Comparing real and imaginary parts on both sides
x = a² – b², y = 2ab
∴ x² + y² = (a² + b²)² + 4a²b² = (a² + b²)²

Question 15.
Find the least positive integers n, satisfying \(\left(\frac{1+i}{1-i}\right)^n\) = 1.
Solution:

If n = 4
⇒ i⁴ = 1
∴ The required least positive integer, n = 4.

Question 16.
Write z = – √7 + i √21 in the polar form. [AP – May 2015; March 11]
Solution:

Let, z = – √7 + i √21
= r (cos θ + i sin θ)
then r cos θ = – √7, r sin θ = √21
r = \(\sqrt{x^2+y^2}=\sqrt{(-\sqrt{7})^2+(\sqrt{21})^2}\)
= \(\sqrt{7+21}=\sqrt{28}=2 \sqrt{7}\)
Hence,
2√7 cos θ = – √7
cos θ = \(-\frac{1}{2}\)
2√7 sin θ = √21
2√7 sin θ = √3 . √7
sin θ = \(\frac{\sqrt{3}}{2}\)
∴ θ lies in the Q2.
∴ θ = \(\frac{2 \pi}{3}\)
∴ z = – √7 + i√21
= 2√7 (cos \(\frac{2 \pi}{3}\) + i sin \(\frac{2 \pi}{3}\)).

Question 17.
Express – 1 – i in polar form with principle value of the amplitude.
Solution:

Let, – 1 – i = r (cos θ + i sin θ)
Then r cos θ = – 1; r sin θ = – 1
r = \(\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = – 1
cos θ = \(\frac{-1}{\sqrt{2}}\)
∴ θ lies in the Q3.
∴ θ = – \(\frac{3 \pi}{4}\)

√2 sin θ = – 1
sin θ = \(\frac{-1}{\sqrt{2}}\)

∴ – 1 – i = √2 (cos (- \(\frac{3 \pi}{4}\)) + i sin (- \(\frac{3 \pi}{4}\)))

Question 18.
Express – 1 – i√3 in the modulus amplitude form.
Solution:

Let – 1 – i√3 = r (cos θ + i sin θ)
Then r cos θ = – 1, r sin θ = – √3
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\)
= \(\sqrt{(-1)^2+(-\sqrt{3})^2}\)
= \(\sqrt{1+3}=\sqrt{4}\) = 2
Hence, 2 cos θ = – 1, 2 sin θ = – √3
cos θ = – \(\frac{1}{2}\), sin θ = \(\frac{-\sqrt{3}}{2}\)
∴ θ lies in the Q3.
∴ θ = – \(\frac{2 \pi}{3}\)
∴ For the given complex number modulus = 2
Principle amplitude = – \(\frac{2 \pi}{3}\)
∴ – 1 – i √3 = 2 [cos (- \(\frac{2 \pi}{3}\)) + i sin (- \(\frac{2 \pi}{3}\))]

Question 19.
If z₁ = – 1 and z₂ = – 1 then find Arg(z₁z₂).
Solution:
Given, z₁ = – 1
= cos π + i sin π
∴ Arg z₁ = π
z₂ = – i
= cos (- \(\frac{\pi}{2}\)) + i sin (- \(\frac{\pi}{2}\))
∴ Arg z₂ = – \(\frac{\pi}{2}\)
Now,
Arg (z₁z₂) = Arg z₁ + Arg z₂
= π – \(\frac{\pi}{2}\) = \(\frac{\pi}{2}\)

Question 20.
If z₁ = – 1, z₂ = i, then find Arg (\(\frac{z_1}{z_2}\)).
[May 14 11, March 09 Board Paper TS-Mar. 16; AP – Mar. 18, 17]
Solution:
Given z₁ = – 1
= cos π + i sin π
∴ Arg z₁ = π
z₂ = i
= cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
∴ Arg z₂ = \(\frac{\pi}{2}\)
Now, Arg \(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)\) = Arg z₁ – Arg z₂
= π – \(\frac{\pi}{2}\) = \(\frac{\pi}{2}\)

Question 21.
If (cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ, then find the value of θ.
Solution:
Given,
(cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ
cis (2α) . cis (2β) = cis θ
cis (2α + 2β) = cis θ
cos (2α + 2β) + i sin (2α + 2β) = cos θ + i sin θ
Comparing real parts on both sides, we get
cos θ = (2α + 2β)
θ = 2 (α + β)

Question 22.
If √3 + i = r (cos θ + i sin θ) then find the value of θ in radian measure.
Solution:

Given that,
√3 + i = r (cos θ + i sin θ)
Then r cos θ = √3, r sin θ = 1
r = \(\sqrt{x^2+y^2}=\sqrt{(\sqrt{3})^2+(1)^2}\)
= \(\sqrt{3+1}=\sqrt{4}\) = 2
Hence,
2 cos θ = √3
cos θ = \(\frac{\sqrt{3}}{2}\)
2 sin θ = 1
sin θ = \(\frac{1}{2}\)
∴ θ lies in the Q1.
∴ θ = \(\frac{\pi}{6}\)

Question 23.
If x + iy = cis α . cis β, then find the value of x² + y². [AP – Mar. 18]
Sol.
Given that,
x + iy = cis α . cis β
= cis (α + β)
= cos (α + β) + i sin (α + β)
Now, comparing real parts on both sides.
we get x = cos (α + β)
Comparing imaginary parts on both sides
we get y = sin (α + β)
Now,
x² + y² = cos² (α + β) + sin² (α + β) = 1.

Question 24.
If (√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib) then show that a² + b² = 4. [AP – Mar.2016] [AP – Mar. 2019]
Solution:
Given that,
(√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib)

Squaring on both sides,
a² + b² = 4

Question 25.
If z = x + iy and |z| = 1, then find the locus of z. [TS – Mar. 2019]
Solution:
Given, z = x + iy
|z| = 1
|x + iy| = 1
\(\sqrt{x^2+y^2}\) = 1
Squaring on both sides,
x² + y² = 1
∴ Locus of z is x² + y² = 1.

Question 26.
If the amplitude of (z – 1) is \(\frac{\pi}{2}\) then find the locus of z. [TS – May 2015]
Solution:
Let, z = x + iy
Now, z – 1 = x + iy – 1
= (x – 1) + iy
If z = x + iy then θ = tan^-1 (\(\frac{y}{x}\))
Given that,
the amplitude of (z – 1) is \(\frac{\pi}{2}\)
tan^-1 (\(\frac{y}{x-1}\)) = \(\frac{\pi}{2}\)
\(\frac{y}{x-1}\) = tan \(\frac{\pi}{2}\)
\(\frac{y}{x-1}\) = ∞ = \(\frac{1}{0}\)
x – 1 = 0
x = 1
∴ Locus of z is x = 1.

Question 27.
If the Arg \(\overline{\mathbf{z}}_1\) and Arg z₂ are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, then find (Arg z₁ + Arg z₂). [AP-May, Mar. 2016]
Solution:
If z = x + iy then θ = tan^-1 (\(\frac{y}{x}\))
If \(\overline{\mathrm{Z}}\) = x – iy then θ = tan^-1 (\(\frac{-y}{x}\))
= – tan^-1 (\(\frac{y}{x}\))
Given that,
Arg \(\overline{\mathbf{z}}_1\) = \(\frac{\pi}{5}\)
Arg z₁ = – \(\frac{\pi}{5}\)
Arg z₂ = \(\frac{\pi}{3}\)
Now,
Arg z₁ + Arg z₂ = – \(\frac{\pi}{5}\) + \(\frac{\pi}{3}\)
= \(\frac{-3 \pi+5 \pi}{15}=\frac{2 \pi}{15}\).

Question 28.
If z = \(\frac{1+2 i}{1-(1-i)^2}\) then find Arg(z).
Solution:
z = \(\frac{1+2 i}{1-(1-i)^2}\)
= \(\frac{1+2 \mathrm{i}}{1-\left(1+\mathrm{i}^2-2 \mathrm{i}\right)}\)
= \(\frac{1+2 i}{1-1-i^2+2 i}\)
= \(\frac{1+2 \mathrm{i}}{1+2 \mathrm{i}}\) = 1
∴ z = 1 = 1 + i(0)
∴ Arg z = tan^-1 (\(\frac{y}{x}\))
= tan^-1 (\(\frac{0}{1}\))
= tan (0) = 0°

Question 29.
Simplify \(\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\) and find its modulus.
Solution:
Given,

Question 30.
Simplify \(\frac{(1+i)^3}{(2+i)(1+2 i)}\) and find its modulus.
Solution:
Given, \(\frac{(1+i)^3}{(2+i)(1+2 i)}\)

Question 31.
If (1 – i) (2 – i) (3 – i) ……….. (1 – ni) = x – iy, then prove that 2 . 5 . 10 …………….. (1 + n²) = x² + y².
Solution:
Given that,
(1 – i) (2 – i) (3 – i) (1 – ni) = x – iy

Squaring on bothsides,
2 . 5 . 10 . …………… (1 + n²) = x² + y².

Question 32.
If the real part of \(\frac{z+1}{z+i}\) is 1, then find the locus of z.
Solution:
Let, z = x + iy
Now,

Given that,
the real part of \(\frac{z+1}{z+i}\) is 1.
⇒ \(\frac{x^2+x+y^2+y}{x^2+(y+1)^2}\) = 0
x² + y² + x + y = x² + y² + 2y + 1
x – y – 1 = 0
∴ The locus of z is x – y – 1 = 0.

Question 33.
If |z – 3 + i| = 4 determine the locus of z. [March 14, May 08]
Solution:
Let z = x + iy
Given, |z – 3 + i| = 4
|x + iy – 3 + i| = 4
|(x – 3) + i (y + 1)| = 4
\(\sqrt{(\mathrm{x}-3)^2+(\mathrm{y}+1)^2}\) = 4
Squaring on both sides
(x – 3)² + (y + 1)² = 16
x² + 9 – 6x + y² + 2y + 1 – 16 = 0
x² + y² – 6x + 2y – 6 = 0
∴ The locus of z is x² + y² – 6x – 2y – 6 = 0.

Question 34.
If |z + ai| = |z – ai|, then find the locus of z.
Solution:
Let, z = x + iy
Given, |z + ai| = |z – ai|
|x + iy + ai| = |x – iy – ai|
|x + i(y + a)| = |x + i(y – a)|
\(\sqrt{(x)^2+(y+a)^2}=\sqrt{(x)^2+(y-a)^2}\)
Squaring on both sides
x² + (y + a)² = x² + (y – a)²
(y + a)² – (y – a)² = 0
4ya = 0
y = 0
∴ The locus of z is y = 0.

Question 35.
Find the equation of the straight line joining the points represented by (- 4 + 3i), (2 – 3i) in the Argand plane.
Solution:
Let the two complex numbers be represented in the Argand plane by the points P, Q respectively.
Then P = (- 4, 3), Q = (2, – 3)
∴ The equation of \(\overline{\mathrm{PQ}}\) is
y – y₁ = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) (x – x₁)
y – 3 = \(\frac{-3-3}{2+4}\) (x + 4)
y – 3 = \(\frac{-6}{6}\) (x + 4)
y – 3 = – 1 (x + 4)
y – 3 = – x – 4
x + y + 1 = 0.

Question 36.
z = x + iy represents a point in the Argand plane. Find the locus of z such that z = 2.
Solution:
Given, z = x + iy
=> P = (x, y)
|z| = 2
|x + iy| = 2
\(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) = 2
Squaring on both sides
x² + y² = 4
x² + y² – 4 = 0
∴ Locus of P is x² + y² – 4 = 0
The locus represents a circle with centre (0, 0) and radius 2 units.

Question 37.
The point ‘P’ represents a complex number z in the Argand plane. If the amplitude of z is \(\frac{\pi}{4}\), determine the locus of P.
Solution:
Let z = x + iy
⇒ P = (x, y)
Given that,
Amp(z) = \(\frac{\pi}{4}\)
tan^-1 (\(\frac{y}{x}\)) = \(\frac{\pi}{4}\)
\(\frac{y}{x}\) = tan \(\frac{\pi}{4}\)
\(\frac{y}{x}\) = 1
y = x
x = y
∴ Locus of P is x = y i.e., x – y = 0.
The locus represents a line passing through origin and making an angle of 45° with the (+) ve direction of X – axis.

Question 38.
Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i in the Argand plane.
Solution:

Let the two complex numbers be represented in the Argand plane by the points P, Q respectively.
Then P = (7, 7), Q = (7, – 7)
Since, C is the midpoint of \(\overline{\mathrm{PQ}}\) then
C = \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right)\)
= \(\left(\frac{7+7}{2}, \frac{7-7}{2}\right)=\left(\frac{14}{2}, \frac{0}{2}\right)\) = (7, 0)
Slope of \(\overline{\mathrm{PQ}}\) is m = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-7-7}{7-7}=\frac{-14}{0}\)
∵ \(\overline{\mathrm{AB}}\) is ⊥ to \(\overline{\mathrm{PQ}}\) then
slope of \(\overline{\mathrm{AB}}\) = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{\frac{-14}{0}}\) = o
The equation of the straight line \(\overline{\mathrm{AB}}\) is
y – y₁ = \(-\frac{1}{m}\) (x – x₁)
y – 0 = 0 (x – 7)
⇒ y = 0.

Question 39.
Find the equation of the straight line joining the points – 9 + 6i, 11 – 4i in the Argand plane.
Solution:
Let the two complex numbers be represented in the Argand plane by the point P, Q respectively.
Then P = (- 9, 6), Q = (11, – 4)
∴ The equation of \(\overline{\mathrm{PQ}}\) is
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
y – 6 = \(\frac{-4-6}{11+9}\) (x + 9)
y-6 = \(\frac{-10}{20}\) (x + 9)
y – 6 = \(\frac{-1}{2}\) (x + 9)
2y – 12 = – x – 9
x + 2y – 3 = 0.

Question 40.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i – 2 – 2i, – 2√3 + 2√3i are the vertices of an equilateral triangle. [AP – May 15, 07, TS-Mar. 18]
Solution:
Let the three complex numbers be represented in the Argand plane by the points P, Q, R respectively.
Then P = (2, 2), Q = (- 2, – 2). R = (- 2√3, 2√3)
Now,
PQ = \(\sqrt{(2+2)^2+(2+2)^2}\)
= \(\sqrt{(4)^2+(4)^2}=\sqrt{16+16}=\sqrt{32}\)

QR = \(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)
= \(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}=\sqrt{32}\)

PR = \(\sqrt{(2+2 \sqrt{3})^2+(2-2 \sqrt{3})^2}\)
= \(\sqrt{4+12+8 \sqrt{3}+4+12-8 \sqrt{3}}=\sqrt{32}\)

∴ PQ = QR = PR
∴ Given points form an equilateral triangle.

Question 41.
If z = 2 – 3i, then show that z² – 4z + 13 = 0. [TS- Mar. 18, Mar. 08, AP – Mar. 2019]
Solution:
Given that z = 2 – 3i
z – 2 = – 3i
Squaring on both sides
(z – 2)² = (- 3i)²
z² + 4 – 4z = 9i²
z² – 4z + 4 = – 9
∴ z² – 4z + 13 = 0

Question 42.
Find the multiplicative inverse of (3, 4).
Solution:
(\(\frac{3}{25}\), \(\frac{4}{25}\))

Question 43.
Write \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\) in the form of a + ib.
Solution:
\(\frac{-8}{13}\) + i (0)

Question 44.
Write the complex number (1 + 2i)³ in the form a + ib. [TS – Mar. 2017]
Solution:
– 11 – 2i

Question 45.
Find the multiplicative inverse of √5 + 3i.
Solution:
\(\frac{\sqrt{5}}{14}-\frac{3}{14} i\)

Question 46.
Write the conjugate of (2 + 5i) (- 4 + 6i).
Solution:
– 38 + 8i.

Question 47.
Find the square roots of 3 + 4i. [March ’13 (old). May ’12, ’09, May ’10, ’03]
Solution:
± (2 + i)

Question 48.
Find the square roots of 7 + 24i. [TS – May 2016, Mar. ‘14]
Solution:
± (4 + 3i)

Question 49.
Find the square roots of – 8 – 6i.
Solution:
± (1 – 3i)

Telangana TSBIE TS Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology

Very Short Answer Type Questions

Question 1.
What factors constitute dairying?
Answer:
Breeding, feeding and management of milch animals, production, processing and marketing of their milk and milk products on economic basis constitute dairying.

Question 2.
Mention any two advantages of inbreeding.
Answer:
Two advantages of inbreeding :

1. Inbreeding increases homozygosity.
2. It helps in the accumulation of superior genes and elimination of less desirable genes.

Question 3.
Distinguish betwen out-cross and cross-breed.
Answer:
Outcrossing is the crossing of unrelated pure breeding animals of different traits with in the same breed whereas cross-breeding is the mating of animals of different breeds.

Question 4.
Define the terms layer and broiler. [Mar. ’17,’15 (A.P.)]
Answer:

1. The birds which are raised exclusively for the production of eggs are called Layers.
2. The birds which are raised only for their meat are called Broilers.

Question 5.
What is apiculture? [Mar. ’20, 17, May 17 (A.P); Mar. ’15 (T.S); Mar. ’14]
Answer:
Apiculture or Beekeeping is the maintenance of hives of honey bees for the production of honey and wax. Beekeeping is an age old cottage industry.

Question 6.
Distinguish between a drone and worker in a honey bee colony.
Answer:

1. Drones are robust, large winged small numbered, short lived and fed with bee bread by nurse workers. They are developed from unfertilized ova by arrhenotoky (male parthenogenesis).
2. Worker bees are multifaceted sterile females which develop from the fertilised eggs and peform diverse functions. They live for two or three months. They are very small in size.

Question 7.
Define the term Fishery.
Answer:
Exploitation of fish and other related aquatic organisms is called Fishery.

Question 8.
Differentiate aquaculture and pisciculture.
Answer:

1. Aquaculture or culture fishery involves rearing and management of selected aquatic organisms under regulated conditions and their subsequent harvesting after the stipulated time.
2. The term PISCICULTURE is used when the organisms cultured and exclusively fin fishes.

Question 9.
Explain the term hypophysation. [March 2015 (T.S.)]
Answer:
Hypophysation or induced breeding is followed in artificial breeding. Pituitary extracts containing FSH and LH or ovaprim are injected into brood fish to induce release of spawn for seed production.

Question 10.
List out any two Indian carps and two exotic carps. [March 2018 (A.P.); May / June 2014]
Answer:

1. Two Indian carps are catla Catla (catla) and Cirrhinus mrigala (mrigai).
2. Two exotic carps are Cyprinus carpio (Chinese carp), Tilapia (grass carp).

Question 11.
Mention any four fish by-products. [March 2019]
Answer:
Four fish by products are

1. Shark and cod liver oils containing vit A and vit D.
2. Oil from Sardine and Salmon – good source of omega – 3 fatty acids.
   (prevent cancer cel! growth, lowers blood cholesterol)
3. Fish guano – Fertiliser prepared from scrap fish.
4. Shagreen.

Question 12.
How many amino acids and polypeptide chains are present in insulin? [March 2019]
Answer:
Human insulin is made up of 51 aminoacids arranged in two polypeptide chains chain A (21 amino adds) and chian B (30 – amino acids).

Question 13.
Define the term ’vaccine’.
Answer:
The term vaccine was coined by Edward Jenner. A vaccine is a biological preparation that improves immunity to a particular disease.

Question 14.
Mention any two features of PCR.
Answer:

1. Polymerase Chain Reaction (PCR) is a powerful technique to identify many other genetic disorders such as haemophilia, phenylketonuria etc.
2. PCR helps to detect very Sow amounts of DNA by amplification of the small DNA fragment.

Question 15.
What does ADA stand for? Deficiency of ADA causes which disease?
Answer:
ADA stands for Adenosine De Aminase. ADA deficiency causes severe combined immune deficiency (SCID). It is caused by the deletion or dysfunction of the gene encoding for the enzyme ADA.

Question 16.
Define the term transgenic animal.
Answer:
Animals that have their own geneome and had their DNA manipulated to possess and express an extra gene are known as transgenic animals.

Question 17.
What is popularly called ‘Guardian Angel of Cell’s Genome’? [March 2020]
Answer:
The protein p53 plays an important role with reference to the G1 check point in the regulation of cell division cycle. It guards the integrity of the DNA. Hence it is often called the Guardian Angel of Cell’s Genome.

Question 18.
List out any four features of cancer cells. [May 2007 (A.P.)]
Answer:
Characters of cancer cells :

1. These cells are characterised by indefinite growth.
2. These cells are with out contact inhibition.
3. These cells are characterised by evading apoptosis (no death).
4. These cells a typical of parent autonomons and aggressive.

Question 19.
How do we obtain radiographs?
Answer:
A beam of X-rays is produced by an X-ray generator and is projected on the body parts. X-rays that pass through the body parts are recorded on a photographic film or observed on a fluorescent screen. Photographs developed using X-rays are known as radiographs or skiagraphs.

Question 20.
What is tomogram?
Answer:
The X-ray detector of the CT-scanner can see hundreds of different levels of density and tissues in a solid organ. The data is transmitted to a computer which builds up 3-D cross sectional picture of the part of the body and displays the picture on the screen. This recorded image is called tomogram.

Question 21.
MRI scan in harmless-justify. [March 2018 (A.P.); May/June 2014]
Answer:
MRI (Magnetic Resonance Imaging) scan is harmless because MRI does not use ionizing radiation as involved in X-rays and is generally a very safe procedure.

Question 22.
What is electrocardiography and what are the normal components of ECG? [March 2015 (A.P.)]
Answer:
Electrocardiography (ECG) is a commonly used, non invasive procedure for recording electrical changes in the heart.
A normal ECG consists 1) Waves 2) Intervals 3) Segments and 4) Complexes.

Question 23.
What does prolonged P-R interval indicate?
Answer:
Prolonged P-R interval indicates delay in conduction of impulses from S-A node to the A-V node. P-R interval is prolonged in bradycardia (slow beating of heart) and shortened in tachycardia (fast beating of heart).

Question 24.
Differentiate between primary and secondary antibodies.
Answer:

1. Primary antibodies react with the antigens of interest.
2. Secondary antibodies react with primary antibodies.

Question 25.
Which substances in a sample are detected by direct and indirect ELISA respectively? [March 2014]
Answer:
Direct ELISA – ELISA used to detect antigens.
Indirect ELISA – ELISA done to detect antibodies.

Short Answer type Questions

Question 1.
What are the various methods employed in animal breeding to improve livestock?
Answer:
Methods of animal breeding : There are broadly two methods of breeding :

1. INBREEDING
2. OUTBREEDING.

1) Inbreeding :
When crossing is done between animals of the same breed it is called inbreeding. It refers to mating of more closely related individuals within the same breed of individuals in a lineage. The breeding strategy is the identification and mating of superior males and superior females of the same breed.

Inbreeding is of two types :
1) Close breeding,
2) Line breeding.,
Close breeding is mating between male parent (sire) and female offspring and / or female (dam) with male offspring. Line breeding (cousin mating) is the selective breeding of animals for a desired feature by mating them within a closely related line (but not as close as close breeding). It leads to upgrading (to improve the quality of livestock by selective breeding for desired characteristics) of a desired commerical character.

2) Outbreeding :
Out-breeding is the breeding of the unrelated animals; it is the cross between different breeds. Out-breeding is of three types

1. Out-crossing
2. Cross-breeding
3. Interspecific hybridisation.

1. Out-crossing :
It is the practice of mating of animals within the same breed, but having no common ancestors on either side of the pedigree for 4-6 generations. The offspring of such a mating is known as an out-cross. It is the best breeding method for animals that are below average in milk production, growth rate (in beef cattle) etc. At times a single out-cross often helps to overcome inbreeding depression.

2. Cross-breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross-breed. Cross-breeding allows the desirable qualities of two different breeds to be combined. The progeny (cross breeds) are not only used for commercial production but also inbreeding and selection to develop stable breeds which may be superior to existing breeds.

3. Interspecific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents. For example when a male donkey (jack / ass) is crossed with a female horse (mare), it leads to the production of a mule.

Question 2.
Define the term ‘breed’. What are the objectives of animal breeding?
Answer:
Animal Breeding :
Animal breeding is an important aspect of animal husbandry which aims at increasing the yield of animals and improving the desirable qualities of the produce. A breed is a group of animals related by descent and similar in most characters like appearance, features, size, configuration, etc. The following are the desirable qualities for which we breed animals :

1. Disease resistance,
2. Increase in the quality and quantity of milk, meat, wool, etc.
3. Fast growth rate,
4. Enhanced productive life by improving the genetic merit of livestock,
5. Early maturity and
6. Economy of feed.

Question 3.
Explain the role of animal husbandry in human welfare.
Answer:
Animal Husbandry :
The strategies adopted for enhancing food production are bound to play a major role in meeting the requirement of food for the ever increasing world’s population in the near future. The biological principles that are applied to animal husbandry will become crucial in our efforts to increase the food production.

Animal husbandry is the agricultural practice of breeding and raising livestock (all domesticated animals reared for the benefit of man). It includes buffaloes, cows, pigs, horses, cattle, sheep, camels, goats etc. However the term livestock is often used for farm animals. If extended, it also includes poultry farming and fisheries.

Even though the estimated world’s livestock population in India and China together is more than 70%, their contribution to world’s farm produce is only 25%, i.e., the productivity per unit is very low. The average annual milk yield is about 170 liters per cow in India. Contrary to it, the average annual milk yield is about 4,100 liters per cow in Netherlands.

Because of its low productivity the Indian cow is known as ‘teacup cow’. So, newer technologies have to be applied to achieve improvement in quality and productivity. Modern methods of breeding, MOET (multiple ovulation and embryo transfer) and production of transgenic animals must be taken up on a large scale in addition to conventional practices and care.

Question 4.
List out the various steps involved in MOET.
Answer:
Multiple Ovulation and Embryo Transfer (MOET) :
The following are the steps invovled in MOET.

1. A cow is administered hormones, with FSH-like activity.
2. This induces follicular maturation and super ovulation (In super ovulation- instead of one egg, which they normally produce per cycle, they produce 6-8 eggs).
3. The animal (cow) is either mated with an elite buli or artificially inseminated.
4. The embroys are at 8-32 celled stages are recovered non-surgically and transferred to surrogate mother (an animal that develops the offspring of another animal in its womb).

Now the genetic mother is ready for another round of super ovulation.

Question 5.
Write short notes on controlled breeding experiments.
Answer:
Controlled Breeding Experiments :
They are carried out using artificial insemination and multiple ovulation and embryo transfer technology (MOET).

Artificial insemination (Al) is the technique in which semen is collected from superior bulls and introduced into the famale reproductive tract when the female is in ‘heat’. This semen can be used immediately or can be frozen and used at a later period. It can be transported in a forzen form to the place where a female is housed. In this way desirable crosses can be made. The major advantage of Al over natural mating is that it permits the dairy farmer to use top proven sires (males) for genetic improvement of his herd and control of venereal diseases. Al is also tremendous value in making optimal use of different sires and enables dairy farmer to breed individual cows to selected sires according to their breeding goals.

The breeding centre at SALON in Rae Bareli is at present the breeder and producer of top quality frozen semen of pure exotic breeds.

Multiple Ovulation and Embryo Transfer (MOET) :
The following are the steps invovled in MOET.

1. A cow is administered hormones, with FSH-like activity.
2. This induces follicular maturation and super ovulation (In super ovulation-instead of one egg, which they normally produce per cycle, they produce 6-8 eggs).
3. The animal (cow) is either mated with an elite bull or artificially inseminated.
4. The embroys are at 8-32 celled stages are recovered non-surgically and transferred to surrogate mother (an animal that develops the offspring of another animal in its womb).

Now the genetic mother is ready for another round of super ovulation.

Question 6.
Explain the important components of poultry management.
Answer:
Important Components of Poultry Management:
1) Selection of disease free and suitable breeds :
The selected breed should get acclimatised to a wide range of climatic conditions. Hybrid layers used in India are BV-300, Hyline, Poona pearls, etc. Commercial broiler strains used in India include Hubbard, Vencobb, etc.

2) Feed management (proper feed and water) :
Balanced diet is a must to maximise the yield. Brooder / chick mash, grower mash, prelayer mash and layer mash are fed to layers at different ages. Likewise pre starter mash, starter mash and finish mash are the feed given to broilers. ‘Safe water’ should be supplied through waterers at all times.

3) Health care :
Vaccination against viral diseases (Ranikhet, Marek’s, and Gumboro) and using antibiotics to treat bacterial diseases {Fowl cholera, Infectious coryza, Chronic Respiratory Disease (CRD)} make the poultry birds disease free. Fungal diseases affecting poultry are Brooder’s pneumonia. Aflatoxicosis and Thrush.

4) In addition to the above, hygiene, proper and safe farm conditions ensure better produce.

Question 7.
Discuss in brief about ‘Avian Flu’. [March 2020]
Answer:
AVIAN FLU (BIRD FLU) is an important disease affecting poultry birds and man has to be very watchful about this disease as it is very dangerous to him too. Causative organism : Bird flu is caused by an ‘avian flu virus’, the H5N1. The virus that causes the bird infection infects human too. It can start a worldwide epidemic (Pandemic disease).

Mode of infection : Infection may be spread simply by touching contaminated surfaces. Birds infected by this type of influenza, continue to release the virus as in their faeces and saliva for as long as 10 days.

Symptoms :
Infection by the avian influenza virus H5N1 in humans causes typical flu-like symptoms, which might include : cough (dry or with phlegm), diarrhoea, difficulty in breathing, fever, headache, malaise, muscle aches and sore throat.

Prevention:

1. Avoiding consumption of undercooked chicken meat reduces the risk of exposure to avian flu.
2. People who work with birds should use protective clothing and special breathing masks.
3. Complete culling of infected flock by burying or burning them.

Question 8.
Explain in brief about queen bee. [May/June 2014]
Answer:
QUEEN :
It is the individual in the colony; It is a fertile, diploid female, one per bee hive and the egg layer of the colony. She lives for about five years and her only function is to lay eggs. The queen bee during its nuptial flight receives sperms from a drone and stores in the spermatecae and lays two types of eggs, the fertilised and unfertilised. All fertilised eggs develop into females. All the larvae developing from the fertilised eggs are fed with the royal jelly (vitamin and nutrient rich secretion from the glands in the hypopharynx of the nurse workers) for the first 4 days only. Afterwards royal jelly is fed only to the bee that is bound to develop into next queen, whereas the other larvae fed on bee bread (honey and pollen) become workers (sterile females).

Question 9.
Honey bees are economically important-justify.
Answer:
Economic importance of Honey bees; The bee products like Honey, wax, propolis and bee venom are used in various ways.

1. Honey is a rich source of fructose, water, glucose, minerals and vitamins.
2. Bee’s wax is used in the preparation of cosmetics, polishes of various kinds and candles.
3. Propolis is used in the treament of inflammation and superficial burns.
4. Bee’s venom, which is extracted from the sting of worker bees, is used in the treatment of rheumatoid arthritis.
5. Pollination : Bees are the pollinators of our crop plants such as sunflower, brassica, apple and pear.

Question 10.
What are the various factors required for bee keeping?
Answer:
Factors / requirements for successful Beekeeping :

1. Knowledge of nature and habits of honey bees.
2. Selection of suitable location (termed Apiary or Bee yard) for keeping the beehives.
3. Raising a hive with the help of a queen and small group of worker bees.
4. Management of beehives during different seasons.
5. Handling and collection of honey and bee wax.

Question 11.
Fisheries have carved a niche in Indian economy-explain.
Answer:
Economic Importance of Fishery :
1. As Food :
Fish meat, in general, is a good source of proteins, vitamins (A and D), minerals and rich in iodine. Tunas, shrimps and crabs are not only edible but have export value also.

2. By-products:
A) Shark and cod liver oils are good sources of vitamins A and D. Oil from Sardine and Salmon are good sources of omega 3 fatty acids, which have multiple functions (reduce cholesterol, help prevent cancer cell growth etc.)
B) Fish guano : Fertilizer prepared from ‘scrap fish’.
C) Other fish by-products are shagreen, Isinglass (substance obtained from dried swim bladders of mostly cat fish, used in clarification of wines) etc.

Fisheries have carved a niche in the Indian economy. We now talk about ‘Blue Revolution’ as being implemented on lines similar to ‘Green Revolution’.

In addition to pisciculture, the culture of prawns, crabs and pear! oysters enable us earn foreign exchange worth millions of dollars from their exports.

Question 12.
Explain in brief structure of Insulin. [March 2015 (A.P.)]
Answer:

Structure of insulin :
Human insulin is made up of 51 amino acids arranged in two polypeptide chains – chain A (21 amino acids) and chain B (30 amino acids), which are linked together by disulphide linkages. In mammals, including humans, insulin is synthesised as a prohormone (like a pro-enzyme, which needs to be processed before it becomes fully mature and functional hormone) which contains an extra stretch called the c peptide. This c peptide is not present in the mature insulin and is removed during maturation into insulin.

The main challenge for the production of insulin in the laboratory using rDNA technique was getting insulin assembled into its mature form.

Question 13.
Define vaccine and discuss about types of vaccines.
Answer:
Vaccines :
The term ‘vaccine’ was coined by Edward Jenner. He immunised a boy against small pox by inoculating him with a relatively less dangerous cow pox virus. The technique of attenuating or weakening of a microbe was developed by Pasteur.

A vaccine is a biological preparation that improves immunity to a particular disease. A vaccine typically contains the disease causing microorganism and is often made from weakened of killed forms of the microbe. The toxins or one of the surface proteins of the microorganisms are also used in preparing vaccines.

The following are some important Biotechnologically produced vaccines.
1. Attenuated Whole Agent Vaccines :
They contain disabled (made iess virulent) live microorganisms. Mostly they are antiviral. Examples: Vaccines against yellow fever, measles, rubella, and mumps and the bacterial disease such as typhoid.

2. Inactivated Whole Agent Vaccines :
They contain “killed microbes’ (virulent before killing). Examples : Vaccines against influenza, cholera, bubonic plague, polio, hepatitis A, rabies and Sabin’s oral polio vaccine.

3. Toxoids :
They contain ‘toxoids’ which are inactivated ‘exotoxins’ of certain microbes. Examples : The vaccines against Diphtheria and Tetanus.

Thus, vaccines are used in the prevention of diseases as they induce artificially acquired active immunity.

Question 14.
Write in brief the types of gene therapy.
Answer:
Gene therapy is the insertion of genes into an individual’s cells and tissues to treat a disease such as a heriditary disease in which a deleterious mutant allele is replaced with a functional one.

Types of Gene Therapy :
Two basic types of gene therapy can be applied to humans, germ line and somatic line.

Germ line gene therapy :
In this type of therapy, functional genes (normal genes) are introduced into sperms or ova and are thus integrated into their genomes. Therefore the change or modification becomes heritable. Due to various technical and ethical reasons, the germ line gene therapy remained at the ‘infant stage’ for the time being.

Somatic line therapy :
In this type of therapy, functional genes are introduced into somatic cells of a patient. The approach is to correct a disease phenotype by treating some somatic cells in the affected person. The changes effected in this type of GT are non-heritable.

Somatic line therapy can be either ex-vivo or in vivo. In ex-vivo, cells are modified outside the body and then transplanted back. In in-vivo, genes are changed in cells, while they are still inside the body.

Question 15.
List out any four salient features of cancer cells.
Answer:
Salient features of Cancer cells :

1. These cells are characterised by indefinite growth.
2. These cells are without contact inhibition.
3. Divide eratically, with increased cell division rate.
4. These cells are characterised by evading apoptosis (no death).
5. These cells are with mutated genes.
6. These cells are atypical of parent autonomous and aggressive.
7. Antigens on the surface are abnormal.
8. These are with unusual number of chromosomes.
9. These cells are spherical due to less number of microfilaments.
10. The cells detach and exhibit metastasis as cadherins either partly or entirely missing.
11. Unlimited growth potential due to over abundance of telomerase, enzyme. [Note : You can select any 4 salient features from the given 11 points.]

Question 16.
Explain the different types of cancers. [Mar. ’18(A.P.); Mar. ’15 (T.S.); Mar. ’14]
Answer:
Types of cancers :
There are different types of cancers such as carcinomas (cancers of epithelial tissues / cells which are most common as epithelial cells divide more often), sarcomas (cancers of connective tissues), leukemias (cancers of bone marrow cells resulting in understrained production of WBC – a liquid tumor), lymphomas (cancers of the lymphatic system). Certain types of cancers are called ‘familial cancers’ (cancer that occurs in families; genetic based) and others ‘sporadic cancers’ (non-hereditary cancers occurring without any family history). Some types of cancers are caused by ‘tumor forming RNA viruses’ (oncoviruses), e.g. Rous sarcoma virus which causes ‘avian sarcoma’.

Question 17.
Write about the procedure involved in MRI. [Mar. 2017, May ’17 (A.P.)]
Answer:
MRI scan (Magnetic Resonance Imaging) is a diagnostic Radiology Technique. MRI is a non invasive medical imaging technique that helps physicians diagnose certain anatomical abnormalities or pathological conditions.

MRI scanner and procedure :
MRI scanner is a giant circular magnetic tube. The patient is placed on a movable bed that is inserted into the magnet. Human body is mainly composed of water molecules which contain two hydrogen nuclei /protons, each. The magnet creates a strong ‘magnetic field’ that makes these protons align with the direction of the magnetic field (protons are not aligned under normal conditions). A second radiofrequency electromagnetic field is then turned on for a ‘brief period’. The ‘protons’ absorb some energy from these ‘radio waves’. When this ‘second radio frequency emitting field’ is turned off, the protons release energy at a radiofrequency which can be detected by the MRI scanner (the protons return to their ‘equilibrium state’ from the ‘energized state’ at different ‘relaxation’ rates).

Question 18.
Write briefly about different waves and intervals in an ECG. [March 2019]
Answer:
Waves :
The waves in a normal record are named: P, Q, R, S, and T, in that order. A typical ECG tracing of a normal heartbeat (or cardiac cycle) consists of I. a ‘P’ wave, II. a ‘QRS complex’ of ‘waves’, III. a T wave.

P wave :
It represents the ‘atrial depolarization or atrial systole’. P wave shows that the impulse is passing through the atria. The normal duration of a P wave is -0.1 sec.

QRS complex of ‘waves’:
(Ventricular depolarization / ventricular systole) i) Q wave is a small negative wave, ii) R wave is a tall positive wave, iii) S wave is a negative wave. The normal duration of QRS complex of waves is about 0.08 – 0.1 sec.

T wave is a positive wave. It represents the ventricular repolarisation. Its duration is 0.2 sec.

Intervals:

1. P-R interval is the interval between the onset of P wave and the onset of Q wave. P-R interval is normally 0.12 – 02 sec.
2. Q-T interval is the interval between the onset of Q wave and the end of the T wave. It represents the electrical activity in the muscle of the ventricles (ventricular depolarisation). ‘QT Interval’ is dependent on the ‘heart rate’ (the ‘faster’ the ‘heart rate’ – the ‘shorter’ the interval). It lasts for about 0.4 sec.
3. R-R interval signifies the duration of one ‘cardiac cycle’ and it lasts for about 0. 8 sec. (60/72 = 0.8 sec.).

Question 19.
Discuss briefly the process of indirect ELISA.
Answer:
Indirect ELISA :
It is used to detect antibodies. The blood of the person undergoing the ‘assay’ (for example the HIV test) is allowed to clot and the cells are centrifuged out to obtain the clear serum with antibodies (called primary antibodies).

Protocol:

1. It is used to detect antibodies.
2. A known ‘antigen’ is added to the ‘well’ (adsorbed).
3. Patient’s antiserum (serum with specific antibodies) is added.
4. The ‘antibodies’ in the patient’s ‘antiserum’ (primary / complementary antibodies) bind to the antigens coated on the surface of the ‘well’.
5. Enzyme linked antihuman serum globulins (anti HISGs) are added. They bind to the antibody which is already bound to the antigen.
6. Enzyme’s substrate is added and the reaction produces a visible colour change which can be measured by a spectrophotometer.

Question 20.
Write short note on EEG.
Answer:
EEG :
Electroencephalography is the process of recording the electrical activity of the brain (graphical recording called electroencephalogram) with the help of an EEG machine and some ‘electrodes’ placed all over the scalp. Electro¬encephalograph is a very useful tool in diagnosing neurological and sleep disorders. The changed EEG patterns in the case of ‘epilepsy’ are conveniently studied with the help of an EEG. Brain shows continuous electrical activity of innumerable neurons.

The intensity and pattern of electrical activity depends on wakefulness, sleep, coma, certain pathological and phychological conditions. The main diagnostic application of EEG in neurological studies is the diagnosis of epilepsy (seizures). EEG shows distinct abnormal pattern in the case of epilepsy. EEG is also useful in the diagnosis of ‘coma’ and ‘brain death’. EEG studies are useful in analyzing sleep disorders (such as insomnia).

Waves of EEG :
The waves recorded by an EEG consist of

1. Synchronized waves which are common in normal healthy people and
2. In certain neurological conditions the waves are desynchronized (irregular wave pattern). The wave pattern can be broadly classified into ALPHA, BETA, THETA and DELTA wave patterns. The nature of the waves depends on the intensity of activity of the different parts of the cerebral cortex.

Long Answer Type Questions

Question 1.
Write in detail about out breeding.
Answer:
Out breeding :
Out-breeding is the breeding of the unrelated animals; it is the cross between different breeds. Out-breeding is of three types

1. Out-crossing
2. Cross-breeding
3. Interspecific hybridisation.

1. Out-crossing :
It is the practice of mating of animals within the same breed, but having no common ancestors on either side of the pedigree for 4-6 generations. The offspring of such a matting is known as an out-cross. It is the best breeding method for animals that are below average in milk production, growth rate (in beef cattle) etc. At times a single out-cross often helps to overcome inbreeding depression.

2. Cross-breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross-breed. Cross-breeding allows the desirable qualities of two different breeds to be combined. The progeny (cross breeds) are not only used for commercial production but also inbreeding and selection to develoop stable breeds which may be superior to existing breeds. For example, Hisardale is a new breed of sheep developed in Punjab by crossing ‘Bikaneri ewes’ and ‘Marino rams’.

3. Interspecific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents. For example when a male donkey (jack / ass) is crossed with a female horse (mare), it leads to the production of a mule (sterile). Similarly when a male horse (stallion) is crossed with a female donkey (jennet), hinny (sterile) is produced. Mules have considerable economic value.
Jack / ass X mare = mule; Stallion X Jenne thinny.

Question 2.
Explain in detail clinical inferences from ECG.
Answer:
ECG may mean electrocardiogram / electrocardiograph, but most commonly used for electrocardiogram. Electrocardiography is a commonly used, non-invasive procedure for recording electrical changes in the heart. The graphic record, which is called an electrocardiogram (ECG or EKG), shows the series of waves that relate to the electrical impulses which occur during each cardiac cycle. An electrocardiograph is a device which records the electrical activity of the heart muscle (depolarisations and repolarisations).

What is electrocardiography?
Electrocardiography is the technique by which the electrical activities of the heart are studied. Sensors (electrodes) are placed at specific parts of the body and linked to the ECG machine. ECG is recorded using 12 ‘LEADS’ (sensors from limbs and chest). Obtaining an electrocardiogram typically takes a few minutes, after which the electrodes are removed.

Clinical Inferences from ECG :

1. Enlarged P wave, indicates enlarged atria.
2. Variations in the duration, amplitude and morphology of the QRS complex indicate disorders such as bundle branch block (block of conduction of impulses through the branches of the bundle of His).
3. If the duration of the P-R interval is prolonged, it indicates delay in conduction of impulses from S-A node (pace maker) to the A-V node. P-R interval is prolonged in ‘bradycardia’ (slow beating of the heart) and shortened in ‘tachycardia’ (fast beating of the heart).
4. Prolonged Q-Tinterval indicates myocardial infarction (Ml) and hypothyroidism. Shortened Q-T interval indicates ‘hypercalcemia’.
5. Elevated S-Tsegment indicates myocardial infarction.
6. Tall T wave indicates hyperkalemia; small, flat or inverted T wave indicates hypokalemia.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution

Very Short Answer Type Questions

Question 1.
What are panspermia?
Answer:
According to Cosmozoic theory or Panspermia, life might have existed all over the universe in the form of resistant spores called cosmozoa or panspermia. They might have reached the earth accidentally.

Question 2.
Define prebiotic soup. Who coined this term?
Answer:
The molecules of ammonia, hydrocarbons and water underwent condensation, oxidation, reduction and polymerisation due to energy sources to produce complex molecules like sugars, amino acids, fatty acids, purines, pyramidines and later nucleosides and nucleotides. All these reactions occurred in the ocean, which was described as the hot dilute soup or prebiotic soup by Haldane.

Question 3.
How did eukaryotes evolve?
Answer:
Eukaryotes evolved probably by two processes, a) Prokaryotes lived in the ancestral eukaryotes symbiotically and evolved into organelles such as mitochondria and plastids. b) The endomembrane system of eukaryotes might have evolved by the infolding of plasma membrane of the ancestral prokaryotes.

Question 4.
What are the components of the mixture used by Urey & Miller in their experiments to simulate the primitive atmosphere?
Answer:
The components used by Urey and Miller for their simulation experiment are water vapour, methane, ammonia and hydrogen.

Question 5.
Mention the names of any four connecting links that you have studied.
Answer:
The four connecting links are :

1. Eusthenopteron between fishes and amphibians.
2. Seymouria between amphibians and reptiles.
3. Archaeopteryx between reptiles and birds.
4. Cynognathus between reptiles and mammals.

Question 6.
Define Biogenetic Law, giving an example. [March 2020]
Answer:
Biogenetic law or Theory of recapitulation was proposed by Ernst Haeckel. It states that Ontogeny repeats phylogeny which means the developmental history of an organism repeats the evolutionary history of its ancestor, e.g. : Tad pole larva of frog resembles fish both externally and internally. It possesses a tail, gills and 2 chambered heart like that of a fish. Later is metamorphoses into adult frog.

Question 7.
Define atavism with an example. [March 2020]
Answer:
Sudden appearance of some vestigial organs in a better developed condition as in the case of the tailed human baby is called atavism. .

Question 8.
Cite two examples to disprove Lamarck’s inheritance of acquired characters.
Answer:

1. Well developed muscles of athletes are not inherited to their children.
2. Making perforations to pinna for wearing ornaments has been in practice in India for the past several centuries. However no girl child is born with ready made perforations in their pinna.

Question 9.
Who influenced Darwin much in formulating the idea of Natural Selection?
Answer:

1. Thomas Malthus (An essay on the principles of populations)
2. Sir Charles Lyell (Principles of Geology)
3. Alfred Russel Wallace (On the tendency of varieties to depart from original types)

Question 10.
What is common between Darwinism and Lamarckism?
Answer:
Lamarckism is the first scientific assumption that recognised the “adoption to the environment as a primary product of evolution. Darwinism also says that during struggle for existence, the organisms with beneficial variations alone will survive.

Question 11.
What is meant by genetic load? Give an example.
Answer:
The existence of deleterious genes within the populations is called genetic load, e.g.: Gene for sickle cell anaemia, (homozygenes individuals for sickle cell gene (Hbs Hbs) usually die early due to anaemia. Those heterozygous (HbA Hbs) can live reasonably healthy and exhibit resistance to malaria. So this disadvantageous gene is carried).

Question 12.
Distinguish between allopatric and sympatric speciations.
Answer:

1. If speciation takes place due to geographical isolation, it is called allopatric speciation.
2. If speciation takes place in the organisms which live in the same habitat, capable of interbreeding, but do not interbreed due to some isolation mechanisms is called sympatric speciation.

Question 13.
Mention the scientific names of ape like and man like earlier primates. Which man like primate first used hides to cover the bodies?
Answer:

1. Ape like earlier primate – Dryopthecus Man like earlier primate – Ramapithecus
2. Man like primate first used hides to cover their bodies is Homoneanderthalensis.

Short Answer Type Questions

Question 1.
Distinguish between homologous and analogous organs. [Mar. 18, 17; May 17 (A.P); Mar. 15 (A.P & T.S) May/June; Mar. 14]
Answer:
Homologous organs :
The organs which have similar structure and origin but not necessarily the same function are called homologous organs. The evolutionary pattern that describes the occurrence of similarity in origin and internal structure is called homology. Such organs show adaptive radiation, hence ‘divergent evolution’, e.g. the appendages of vertebrates such as the flippers of whale, wings of bat, forelimbs of horse, paw of cat and hand of man, have a common pattern in arrangement of bones eventhough their external form and function may vary to suit their mode of life. It explains that all vertebrates might have had a common ancestor.

Analogous organs :
The organs which have dissimilar structure and origin but perform the same function are called analogous organs. Analogous organs suggest ‘convergent evolution’, e.g. wings of a butterfly and wings of a bird.

Question 2.
Write a short note on the theory of mutations. [Mar. ’15 (A.P.); May/June ’14]
Answer:
Mutation theory :
It was proposed by Hugo de Vries, a Dutch botanist who coined the term ‘mutation’. Mutations are sudden, random inheritable changes that occur in organisms. He found four different forms in Oenothera lamarckiana (commonly called ‘evening primrose’) such as O. brevistylis-smaW style, O. levifolia-smooth leaves. O. gigas- the giant form, O. nanella- the dwarf form (mutant varieties). T.H. Morgan studied the inheritance pattern of mutations in Drosophila melanogaster. Darwin called mutations (large variations) sports of nature or saltations, whereas Bateson called them discontinuous variations.

Salient Features of Mutation theory:

1. Mutations occur from time to time in naturally breeding populations.
2. They are discontinuous and are not accumulated over generations.
3. They are full-fledged, and so there are no ‘intermediate forms’.
4. They are subjected to Natural Selection.

Question 3.
Explain Darwin’s theory of Natural Selection with industrial melanism as an experimental proof. [Mar. ’18, 17; May ’17 (A.P.); Mar. ’15 (T.S.) Mar. ’14)]
Answer:
Experimental verification of Natural Selection – Industrial melanism :
An important practical proof for the operation of Natural Selection is the classical case of industrial melanism, exhibited by peppered moth – Biston betularia. These moths were available in two colours, grey and black. Prior to industrial revolution, the grey moths were abundant. During the industrial revolution, the black forms were more and the grey forms were less in the industrial cities like Birmingham. Biologists proposed that with the industrial revolution, more soot was released due to the burning of coal, which resulted in the darkening of the barks of trees.

Grey moths on the dark bark were easily identified and predated more by birds. Hence the number of grey moths decreased and that of the black moths increased in the population. It means Nature offered ‘positive selection’ pressure to the black (melanic) forms. Bernard Kettlewell, a British ecologist, tested this hypothesis experimentally. He collected both the grey and the black forms of Biston betularia for his experiment.

He released them in two sets of equal numbers; one set in Birmingham, a polluted urban area, and the other set in Dorset, an unpolluted rural area. After a few days he recaptured them. Of those moths recaptured from Birmingham, there were more black forms. Among those recaptured from Dorset there were more grey forms. The reason for such a difference is: the melanic forms could not be easily spotted by predator birds as their body colour merged with the dark colour of the bark of trees in Birmingham area. In the rural areas (Dorset) the grey forms had better survival chance as their body colour merged with the light coloured surroundings. This explains the differential survival of the moths due to Natural Selection. It will be interesting to know that there was a reversal in the selection process after the introduction of pollution check laws in the urban areas.

Question 4.
Discuss the role of different patterns of selections in evolution.
Answer:
Selection is a process by which the organisms that are physically. Physiologically and behaviourally better adapted to the environment, survive and reproduce. Selection is an operative process. Selections are 3 types.
a) Stabilising or Centripetal selection :
It is the selective elimination of phenotypically extreme individuals from the two ends of the phenotypic distribution and preserving those that are in the mean of the phenotypic distribution.

b) Directional selection :
It operates in response to gradual changes in environmental conditions. Directional selection works by constantly removing individuals from one end of the phenotypic distribution.

c) Disruptive or Centrifugal selection :
It is a rarest form of selection and is very important in bringing about evolutionary change. As a result of increased competition, selection pressure acting within the population may push the phenotypes away from the population mean towards the ends of the population. This can split the population into two or more sub-population called species populations. Each population may give rise to a new species. It is also called as adaptive radiation.

Question 5.
Write a short note on Neo-Darwinism. [March 2020]
Answer:
Modern synthetic theory of Evolution or Neo-Darwinism :
Weismann’s germplasm theory, de Vries’ mutation theory and Mendel’s laws of inheritance helped a lot in understanding the origin and inheritance of variations. The scientists such as Huxley, Haeckel, Simpson, etc., supported Darwinism. Later Fisher, Sewall Wright, Mayr explained Natural Selection in the light of post-Darwinian discoveries (Synthetic theory / Genetical theory / Neo-Darwinism). According to this theory, five basic factors are involved in the process of organic evolution. They are (i) Gene mutations, (ii) Chromosomal mutations, (iii) Genetic recombinations, (iv) Natural Selection and (v) Reproductive isolation.

i) Gene mutations :
Changes in the structure of a gene (DNA molecule).are called gene mutations or point mutations. They alter the phenotypic characters of the individuals. Thus, gene mutations tend to produce ‘variations’ in the offspring.

ii) Chromosomal mutations :
Changes in the structure of chromosomes (due to deletion, addition, duplication, inversion or translocation) are called chromosomal mutations. They also bring about variations in the phenotype of organisms which lead to the occurrence of variations in the offspring.

iii) Genetic recombinations :
Recombinations of genes due to crossing over during meiosis are also responsible for bringing about genetic variability among the individuals of the same species, thus, contributing to the occurrence of heritable variations.

iv) Natural Selection :
Natural selection does not produce any genetic changes but once genetic changes occurred, it favours some genetic changes while rejecting others. Hence it is considered the driving force of evolution.

v) Reproductive isolation :
The absence of gene exchange between populations is called the reproductive isolation. It plays a great role in giving rise to new species and preserving the species integrity.

Question 6.
In a population of 100 rabbits which is in Hardy-Weinberg equilibrium, 24 are homozygous long-eared. Short ears are recessive to Along ears. There are only two alleles for this gene. Find out the frequency of recessive allele in the population.
Answer:
Number of rabbits in the population with H.W. equilibrium = 100
Number of dominant homozygous long eared rabbits = 24
Frequency of homozygous dominant long eared rabbits, p² = \(\frac{1}{100}\) × 24 = 0.24
Frequency of dominant allele, p = 0.49
Frequency of recessive allele, q = 1 – 0.49 = 0.51

Question 7.
What is meant by genetic drift? Explain genetic drift citing the example of Founder Effect. [March 2019]
Answer:

Genetic Drift :
The change in the frequency of a gene that occurs merely by chance and not by selection, in small populations, is called genetic drift or Sewall Wright effect. Suppose, for a gene with two alleles, the frequency of a particular allele is 1% (q = 0.01), the probability of losing that allele by chance from the small population is more. The end result is either Fixation (p or q = 1) or Loss (p or q = 0) of that allele. The probability of reaching the end point depends on the size of the population. Genetic drift tends to reduce the amount of genetic variation within the population mainly by removing the alleles with low frequencies. It can be exemplified by the Founder Effect and Bottleneck Effect.

Founder effect :
If a small group of individuals from a population start a new colony in an isolated region, those individuals are called the founders of the new population. The allelic frequencies of their descendants are similar to those of the founders rather than to their ancestral parent population, e.g. presence of O^{+ ve} blood group in nearly 100% of the Red-lndians. It means the forefathers of the Red Indian tribe were predominantly O^{+ ve} and they isolated themselves reproductively from other populations.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material 6th Lesson Genetics Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material 6th Lesson Genetics

Very Short Answer Type Questions

Question 1.
What is pleiotropy?
Answer:
The phenomenon of multiple effects of a single gene is called pleiotropy i.e., the same gene is activated in several different tissues producing different phenotypic effects.

Question 2.
What are the antigens causing ‘ABO’ blood grouping? Where are they present?
Answer:
Antigens are present on the plasma membrane of the RBCs. They are also called iso agglutinogens. They are antigen A antigen B are responsible for ABO blood grouping.

Question 3.
What are the antibodies of ‘ABO’ blood grouping? Where are they present?
Answer:
Isoagglutinins (or) Antibodies are present in the blood plasm. Iso aggulutinin A and Iso agglutinin B are the antibodies of ABO blood grouping.

Question 4.
What are multiple alleles?
Answer:
When more than two allelic forms occur at a same locus on the homologous chromosomes of an organism, they are called multiple alleles. When more than two alleles exist in a population of specific organism, the phenomenon is called multiple allelism.

Question 5.
What is erythroblastosis foetalis?
Answer:
When father is Rh⁺ and mother is Rh^– in the second Rh⁺ baby onwards, due to immunological in compatibility between mother and growing foetus, the RBC of Rh⁺ foetus are destroyed. This is called erythroblastosis foetalis.

Question 6.
A child has blood group ‘O’. If the father has blood group ‘A’ and mother blood group ‘B’. Work out the genotypes of the parents and the possible genotypes of the other offspring.
Answer:
Child blood group is ‘O’ i.e., the genotype will be ‘OO’. Father blood group phenotype A hence it must AO genotype. Mother blood group phenotype B hence it must BO genotype. Possible genotypes of other offsprings are AB, BO, AO phenotypes are AB, B, A.

Question 7.
What is the genetic basis of blood types in ABO system in man?
Answer:
ABO blood group phenotypes are specified by Isoagglutinogen (I) gene with three alleles I^A, I^B and I° located on chromosome 9 (nine). I^A I^A and I^A I° genotypes specify group. I^B I^B and I^B I° genotype specify B group. I^A I^B genotype specified AB group and I°I° specifies ‘O’ group phenotypes.

Question 8.
What is polygenic inheritance?
Answer:
Many characters, such as human skin colour and height, an either – or classification is impossible because the characters vary in the population in gradations along a continuum. These are called quantitative characters. Quantitative variation usually indicates polygenic inheritance, an additive/ cumulative effect of two or more genes on a single phenotypic character.

Question 9.
Compare the importance of Y – chromosome in human being and Drosophila.
Answer:

1. In human beings XX – XY type is seen. A pair of X chromosomes is present in female and XY is present in male. During spermatogenesis among males, two types of gametes 50% sperms with X and 50% with ‘Y’- chromosomes. It is evident that Y – chromosome decides the sex of child (if XX female if Xy male).
2. In sex determination of Drosophila the sex of an individual is determined by the ratio of number of its chromosomes and that of its autosomal sets, the ‘Y’ chromosome taking no part in the determination of sex. The ratio is termed as sex index.

Question 10.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:
If the two sex chromosomes / allosomes are different (XY or ZW) or it contains only one allosome (XO or ZO) the individual is heterogametic and if the two sex chromosomes / allosomes are similar (XX or ZZ) the individual is homogametic.

Question 11.
What is haplo – diploidy?
Answer:
In Hymenopterous insects like honey bees, sex is determined by the number of sets of chromosomes (haploid or diploid) in a bee, the fertilized eggs (diploid) develop into females and the unfertilized eggs (haploid) develop into males. This method of sex determination is haplodiploidy.

Question 12.
What are Barr bodies?
Answer:
Darkly stained, highly condensed and heterochromatinized X-chromosome present in the somatic cells like buccal mucosa, fibroblasts of female human beings is called Barrbody or Sex chromatin body.

Question 13.
What is Klinefelter’s syndrome?
Answer:
This genetic disorder is caused by trisomy of 23^rd pair. The karyotype is 47, XXY. A Klinefelter male possesses an additional X – chromosome along with normal XY. At the same time, feminine sexual development is not entirely suppressed. Slight enlargement of breast (gynecomastia) and hips are often rounded.

Question 14.
What is Turner’s syndrome?
Answer:
The karyotype is 45, X is due to monosomy 23^rd pair where one X – chromosome is last. A turner female does not show Barr bodies in her somatic cells. The symptoms are short stature, gonadal dysgenesis, webbed neck and broad shield like chest with widely spaced nipples.

Question 15.
What is Down syndrome?
Answer:
Down syndrome is a genetic condition that causes delays in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21 (Trisomy of 21^st set). The Karyotype is designated as TRISOMY 21 (47, XX, + 21). Characters are short statured, round head, furrowed tongue and partially opened mouth. Mental development and physical development is retarded.

Question 16.
What is Lyonisation?
Answer:
X – chromosome inactivation is also called Lyonisation. (It is proposed by Mary Lyon and Liane Russell). It is a process by which one of the two copies of the X – chromosome present in the body cells of female mammals is activated.

Question 17.
What is sex – linked inheritance?
Answer:
The inheritance of a trait that is determined by a gene located on one of the sex chromosomes is called sex – linked inheritance.

Question 18.
Define hemizygous condition.
Answer:
Genes that are present on the X or Y chromosome are called sex linked genes. The genes located on X – chromosome, whose alleles are absent on the Y – chromosome are called X – linked genes. Male human beings are hemizygons that is genes alleles are not present on Y-chromosome. Sometimes alleles are absent on X – chromosome of male. They are called holandric genes or Y – linked genes.

Question 19.
What is crisscross inheritance?
Answer:
The criss cross pattern or inheritance (skip gene – ration in heritance) in which a gene responsible for the white eyes is transmitted from a male parent to a male grandchild through carrier female of the first generation.

Question 20.
Why are sex – linked recessive characters more common in the male human beings?
Answer:
Sex linked recessive characters more common in male human beings because a mutation in a gene on the chromosome causes the phenotype to be expressed in males who are necessarily hemizygous for the recessive allele and they have only one X – chromosomes.

Question 21.
Why are sex – linked dominant characters more common in the female human beings ?
Answer:
X – linked dominant inheritance is a mode of genetic inheritance by which a dominant gene is carried on the X – chromosome. X linked dominant inheritance indicates that a gene responsible for a genetic disorder is located on the chromosome, and only one copy of the allele is sufficient to cause the disorder when inherited from a parent who has the disorder X – linked dominant traits are more prominent in woman.

Question 22.
What are sex limited characters?
Answer:
Sex limited genes are autosomal genes present in both males and female. Their phenotypic expression is limited to only one sex due to internal hormonal environment, e.g.: Beard in man, development of breast and secretion of milk in woman etc., are sex limited traits.

Question 23.
What are sex influenced characters?
Answer:
Sex influenced genes are the autosomal genes present in both males and females. In sex influenced inheritance, the genes behave differently in the two sexes, probably because of sex hormones providing different cellular environments in males and females. Cases of sex influenced inheritance include pattern baldness in humans, horn formation in certain breeds of sheep. (Dorset Horn Sheep)

Question 24.
How many base pairs are observed in human genome? What is the average number of base pairs in a human gene?
Answer:
The human genome contains 3164.7 million nucleotide bases. The average gene consists of 3000 bases, but sizes vary greatly with the largest known human gene being the one that codes for the protein called dystrophin.

Question 25.
What is ‘junk DNA’?
Answer:
Some DNA is involved in regulating the expressions of the genes that code for specific proteins. The remaining non-functional DNA is called Junk DNA.

Question 26.
What are VNTRs?
Answer:
No two people (other than identical twins) have exactly the same sequence of bases in their DNA. Restriction Fragment Length Polymorphism RFLPs-(pronounced riflips) are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as Genetic markers.

Question 27.
List out any two applications of DNA finger printing technology.
Answer:

1. DNA finger printing is a technique by which the DNA of an individual can be compared with that found in a sample or another individual (a suspect in a crime)
2. DNA finger printing is used particularly paternity / maternity testing and for forensic work.
3. Taxonomical applications – to study of phylogeny.

Short Answer Type Questions

Question 1.
Briefly mention the contribution of T.H. Morgan to genetics.
Answer:
Experimental verification of the “Chromosomal theory of inheritance” by Thomas Hunt Morgan and his colleagues, led to discovering the basis for the variation that sexual reproduction produced. For his work, Morgan selected a species for fruit fly, Drosophila melanogaster, which can be grown on simple synthetic medium in the laboratory. It completes its life cycle in about two weeks, and a single mating could produce a large number of progeny.

Further, it has many types of morphological, hereditary variations that can be seen under a low power microscope. Another advantage of the fruit fly is that it has only four pairs of chromosomes, which are easily distinguishable under a light microscope. There are three pairs of autosomes and one pair of sex chromosomes. Female fruit flies have a pair of homologous X – chromosomes, and males have one X – chromosome and one Y – chromosome.

Question 2.
What is pedigree analysis? Suggest how such can analysis, can be useful.
Answer:
Pedigree analysis is useful in many ways like it helps to work out the possible genotypes from the knowledge of the respective phenotypes. It helps to study the pattern of inheritance of a dominant or a recessive trait. The possible genetic makeup of a person for a trait can also be known with the help of the pedigree chart. Some of the important standard symbols used in the pedigree analysis are shown in the figure.

Question 3.
How is sex determined in human beings? [March 2020, 2018 (A.P.); March 2015 (T.S.)]
Answer:
Sex determination in Humans : It has already been mentioned that the sex determining mechanism in case of humans is XX – XY type. Out of 23 pairs of chromosomes present, 22 pairs are exactly same in both males and females; these are the autosomes. A pair of X- chromosomes is present in the female, where as the presence of an X and Y – chromosome are determinant of the male characteristic. During spermatogenesis among males, two types of gametes are produced. 50 percent of the total sperm produced carry the X – chromosome and the rest 50 percent has Y – chromosome besides the autosomes. Females, however, produce only one type of ovum with an X – chromosome.

There is an equal probability of fertilisation of the ovum by the sperm carrying either X or Y chromosome. In case the ovum is fertilised by a sperm carrying X – chromosome, the zygote develops into a female and the fertilisation of ovum with Y – chromosome carrying sperm results into a male offspring. Thus, it is evident that it is the genetic makeup of the sperm that determines the sex of the child. It is also evident that in each pregnancy there is always 50 percent probability of either a male or a female child.

Question 4.
Describe erythroblastosis foetolis. [March 2019, ’17, May ’17 (A.P.); Mar. ’14]
Answer:
Destruction of RBC of Rh positive foetus by anti Rh antibodies produced by Rh negative mother due to immunological incompatibility is called Erythroblastosis foetalis or Haemolytic disorder of newborn (HDNB). This is due to genetically incompatible marriage involving Rh positive father and Rh negative mother. At the time of birth the Rh positive foetal blood mixes with the Rh negative blood of mother, through the ruptured placenta.

The Rh antigens sensitize the mother to produce anti Rh antibodies (IgG antibodies) and memory cells. This first Rh positive by is unaffected because it is delivered by the time mother is sensitized. During the next pregnancy bearing Rh positive foetus these antibodies increase in concentration due to memory cells and cross the placenta, enter the blood of baby and destroy the RBC. Haemolytic anaemia is the symptom in this disorder.

Question 5.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Autosomal disorders are two types.
1) Sex – Limited inheritance :
in contrast to X – linked inheritance, patterns of gene expression may be affected by the sex of an individual even when the genes are not on the X – chromosome. Sex – limited genes are autosomal genes present in both males and females. Their phenotypic expression is limited to only one sex due to internal hormonal environment, e.g. beard in man, development of breast and secretion of milk in woman etc., are sex limited traits.

2) Sex – influenced Inheritance :
Sex – influenced genes are the autosomal genes present in both males and females. In sex -influenced inheritance, the genes behave differently in the two sexes, probably because the sex hormones provide different cellular environments in males and females. Thus, the heterozygous genotype may exhibit one phenotype in males and the contrasting one in females. Cases of sex – influenced inheritance include pattern baldness in humans, horn formation in certain breeds of sheep (e.g. Dorset Horn sheep).

Question 6.
Describe the genetic basis of ABO blood grouping.
Answer:
Bernstein discovered that these blood group phenotypes were inherited by the interaction of three ‘autosomal alleles’ of the gene named I, located on chromosome 9. I^A, I^B and i (or I°) are the three alleles of the gene I. The antibodies ‘anti – A’ and ‘anti – B’ are called isoagglutinins (also called isohaemagglutinins) which are usually IgM type. The isoagglutinins of an individual cause agglutination reactions with the antigens of another individual. The alleles I^A and I^B are responsible for the production of the respective antigens ‘A’ and ‘B’. The allele i does not produce any antigen. The alleles lA and lB are dominant to the allele i, but co-dominant to each other (I^A = I^B > i). A child receives one of the three alleles from each parent, giving rise to six possible genotypes
Table : Genetic control of the human ABO blood groups

and four possible blood types (phenotypes). The genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B and ii. The phenotypic expressions of I^AI^A and I^Ai are ‘A’ – type blood, the phenotypic expressions of I^AI^A and I^Bi are ‘B’ – type blood, and that of I^AI^B is ‘AB’ – type blood. The phenotype of ii (I°I°) is ‘O’ – type blood.

Question 7.
Describe male heterogamety.
Answer:
Male Heterogamety : In this method of sex determination, the males (heterogametic) produce dissimilar gametes while females (homogametic) produce similar gametes. Male heterogamety is of two kinds, XX – XO type and XX – XY type.

i) XX – XO type :
In some insects such as bugs, grasshoppers and cockroaches, females are with two X – chromosomes and males are with one X – chromosome in each somatic cell. McClung discovered this type in grasshoppers. The unpaired X – chromosome determines the male sex. The karyotype of the female (homogametic) is AAXX and that of the male (heterogametic) is AAXO. All the ova contain ‘AX’ complement of chromosomes and the sperms are of two types. One half of the sperms have ‘AX’ complement and the other half have ‘A1 complement of chromosomes. The sex of the offspring depends on the type of sperm that fertilizes the ovum.

ii) XX – XY type :
In human beings and some insects such as Drosophila, both females and males have the same number of chromosomes. The karyotype of the female is AAXX and that of the male is AAXY. Females are ‘homogametic’ with ‘XX’ chromosomes.

They produce similar ova having one X – chromosome each. Males are ‘heterogametic’ with X and Y – chromosomes. They produce two kinds of sperms; one half of them with X – chromosome and tbe other half with Y – chromosome. The sex of the offspring depends on the fertilizing sperm. The XX – XY type is also found in most other mammals.

Question 8.
Describe female heterogamety.
Answer:

Female Heterogamety :
In this method of sex determination, the males produce ‘similar gametes’ while females produce ‘dissimilar gametes’. Female heterogamety is of two kinds, ZO – ZZ type and ZW – ZZ type.

i) ZO – ZZ type : In moths and some butterflies, female is heterogametic with one z – chromosome (ZO) and male is homogametic with two Z – chromosomes (ZZ). The karyotype of female is AAZO and male is AAZZ. Females produce two kinds of ova, half of them with a Z – chromosome and the other half with no sex chromosome. Males produce similar type of sperms. The sex of the offspring depends on the type of ovum that is fertilized.

ZW – ZZ type :
In birds, reptiles, some fishes, etc., the females are heterogametic with ZW – allosomes and males are homogametic with ZZ – allosomes. The karyotype of the female is AAZW and that of the male is AAZZ. All sperms are similar with the allosome – Z. Ova are of two different kinds; one half of the ova are with the allosome – Z and the other half with the allosome – W. The sex of the offspring depends on the typ£of ovum that is fertilized.

Question 9.
Describe the Genic Balance Theory of sex determination. [March 2015 (A.P.)]
Answer:
Genic balance theory of sex determination was proposed by Calvin Bridges with reference to sex determination in Drosophila. He proposed that both the X- chromosomes and autosomes together play a role in sex determination in Drosophila, where as Y-chromosome has no role. This theory explains that genes for maleness are located on autosome and for femaleness on X-chrombsorne in Drosophila. Y-chromosome in lacks male – determining factor but contains gametic information essential to male fertility. Sex in Drosophila is determined by ratio of X-chromosome to the number of haploid sets of autosomes.

The chromosomal complement, X/A ratio and sex of Drosophila are tabulated below.

Question 10.
Explain the inheritance of sex linked recessive character in human being.
Answer:

Colour Blindness :
It is a sex -linked recessive disorder. Retina of the eye in man contains the cells sensitive to red and green colours, this phenotypic trait is genetically controlled. Its alleles are located on the X-chromosome. When a woman with normal vision (homozygous) marries a colour – blind man, all the sons and daughters are normal, but daughters are carriers (heterozygous),. If a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and another half of sons are colour – blind. Colour – blind trait is inherited from a male parent to his grand sons through carrier daughter, which is an example of crisscross pattern of inheritance.

Question 11.
Describe the experiment conducted by Morgan to explain sex linkage.
Answer:
Thomas H. Morgan (Father of Modern Genetics ) discovered sex linkage in Drosophila melanogaster. Morgan wanted to analyze the behavior of the two alleles of a fruit fly’s eye – colour gene. When he crossed a white eyed (mutant) male to a normal (wild) red eyed female, in the F₁ generation all the males and females were red eyed. When the F₁ generation ‘red eyed female’ was crossed to a ‘red eyed male’, in the F₂ generation all the females were red eyed and 50 percent of the males were ‘white eyed’.

The white eyed trait from the male is inherited to the male of the F₂ generation through the ‘carrier daughter’ of the F₁ generation. This pattern of inheritance is called crisscross pattern of inheritance (skip generation inheritance) in which a gene responsible for the white eyes is transmitted from a male parent to a male grandchild through carrier female of the first generation.

In a reciprocal cross (to test the role of parental sex on inheritance pattern), in which a white eyed female was crossed to a red eyed male, the results were different. The first generation male offspring had white eyes while the female offspring had red eyes. The reason was that the allele responsible for the white eye is sex – linked (more specifically X – linked, as it occurs on the X – chromosome) and recessive. Males always inherit the X – linked recessive traits from the female parents.

Morgan’s discovery that transmission of the X – chromosome in Drosophila correlates with the inheritance of an eye – colour trait was the first solid evidence indicating that a specific gene is associated with a specific chromosome.

Question 12.
Explain the inheritance of sex influenced characters in human beings.
Answer:
Sex influenced genes are autosomal genes in both males and females, whose phenotypic expression is different in different sexes, dominant in one sex and recessive in the other.

Pattern baldness is a sex influenced trait in human being, in which a fringe of hair is present low on the head. The gene for baldness ‘B’ is dominant in males and recessive in females in heterozygous condition.

When mother is homozygous bald and father is homozygous nonbald in the progeny all the daughters are nonbald and all sons are bald.

If both the parents are heterozygous bald, nonbald daughters are 1 : 3, and sons are 3 : 1. In the progeny bald, non-bald ratio is 1 : 1.

Question 13.
A man and woman of normal vision have one son and one daughter. Son is colour – blind and his son is with normal vision. Daughter is with normal vision, but one of her sons is colour – blind and the other is normal. What are the genotypes of the father, mother, son and daughter?
Answer:
Color blindness in human being is because of recessive X – linked gene, and show criss – cross inheritance.

Question 14.
A colour – blind man married a woman who is the daughter of a colour – blind father and mother homozygous normal vision. What is the probability of their daughters being colour – blind?
Answer:
The woman he married is a carrier (heterozygous) since her father is color blind but mother is homozygous normal. When a color-blind man marries a carrier woman in the progeny half daughters are carriers and half daughter are color blind.

So the probability of their daughters being color blind is 50%.

Question 15.
A heterozygous bald man who is non – haemophilic, married a woman who is homozygous for the non – bald trait and is haemophilic. What is the probability of her male children becoming bald and haemophilic?
Answer:

When these two marry they produce two kinds of sons, half heterozygous bald and haemophilic and half nonbald and haemophilic.

So the probability of their male children being bald and haemophilic is 50%.

Question 16.
A woman’s father shows ‘IP’but her mother and husband are normally pigmented. What will be the phenotypic ratio of her children?
Answer:
In continentia pigmentii is ‘X’ linked dominant truit. It is more in females than in males, because they have two chances to inherit this allele.

As that womans, mother is normal and father is incontinentia pigmenti, she is heterozygous for incontinentia pigmentii. As her husband is normal in their progeny half daughters are heterozygous incontinentia pigmentii, half daughters are normal, half sons are incontinentia pigmentii and half sons are normal, so the phenotype ratio is 1 : 1 : 1 : 1.

Question 17.
Write the salient features of ‘HGP’.
Answer:
Salient Features of Human Genome: Some of the salient observations drawn from human genome project are as follows :

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being the one that codes for the protein called dystrophin.
3. The total number of genes is estimated at 30,000. Almost all (99.9%) nucleotide bases are exactly the same in the people.
4. The functions are unknown for over 50% of the genes discovered.
5. Less than 2 percent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has the highest number of genes (2,968), and the Y – chromosome has the fewest genes (231).
9. Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – single nucleotide polymorphism, pronounced as snips) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease – associated sequences and tracing human history.

Question 18.
Describe the steps involved in DNA finger printing technology.
Answer:
The following is the process (protocoJ) of DNA (genetic) fingerprinting.

1. First step is obtaining DNA sample from substances like blood, semen, hair roots, bone or saliva.
2. The DNA is cut at specific sites into small fragments using restriction enzymes and then amplified by rDNA or PCR methods.
3. Double stranded DNA is split into single stranded DNA using alkaline chemicals.
4. The DNA fragments are applied to one end of thin agarose gel and separated into individual bands, with fragments in each one progressively smaller in size, by electrophoresis.
5. A thin dylon membrane covered by paper towels is placed over the gel, as the
   paper towels absorb the moisture from gel, DNA is transferred into nylon membrane, this process is blotting.
6. A radio active DNA probe is introduced, which binds with specific complemental DNA sequences on nylon membrane. The excess DNA probe is washed away.
7. When a photographic / X – ray film is placed on the nylon membrane radio active probes will expose the film producing a pattern of thick and thin bands. This pattern of bars is DNA (genetic) fingerprint.

Long Answer Type Questions

Question 1.
What are multiple alleles ? Describe multiple alleles with the help of ABO blood groups in man. [March 2020, 2018 (A.P.); March 2014; May/June ’14]
Answer:
Multiple alleles and human blood groups :
Generally a gene has two alternative forms / versions called alleles. They are present at the same locus in a pair of homologous chromosomes. Two alleles of a gene can form three genotypes in a diploid organism. Sometimes a gene may have more than two alleles. When more than two allelic forms occur at the same locus on the homologous chromosomes of an organism, they are called mutiple alleles when more than two alleles exist in a population of a specific organism, the phenomenon is called mutiple allelism.

As mentioned above ‘multiple alleles’ cannot be observed in the genotype of a diploid individual, but can be observed in a population. The number of genotypes that can occur for multiple alleles is given by the expression n (n + 1) /2 where, n = number of alleles. A well known example of multiple allelism in man is the expression of ABO blood types by three alleles of a single gene which can produce six genotypes.

ABO Blood Types :
The ABO blood group system was proposed by Karl Landsteiner. He was awarded the Nobel Prize in Physiology or Medicine in 1930 for his work. The phenotypes (blood types) A, B, AB and O types are characterized by the presence or absence of ‘antigens’ on the plasma membrane of the RBCs. The A and B antigens are actually carbohydrate groups (sugar polymers) that are bound to lipid molecules (fatty acids) protruding from the membrane of the red blood cell. They are also called isoagglutinogens because they cause blood cell agglutination in the case of incompatible blood transfusions.

‘Blood type A’ persons have antigen A on their RBCs and anti – B antibodies in the plasma. ‘Blood type B’ persons have antigen B on their RBCs and anti – A antibodies in the plasma. ‘Blood type AB’ person have antigens ‘A’ and ‘B’ on theRBCs and no antibodies in the plasma. ‘Blood type O’ persons have no antigens on their RBCs and both ‘anti – A, and ‘anti – B’ antibodies are present in the plasma.

Bernstein discovered that these phenotypes were inherited by the interaction of three ‘autosomal alleles’ of the gene named I, located on chromosome 9. I^A, l^B and i (or I°) are the three alleles of the gene I. The antibodies ‘anti – A’ and ‘anti – B’ are called isoagglutinins (also called isohaemagglutinins) which are usually IgM type. The isoagglutinins of an individual cause agglutination reactions with the antigens of another individual. The alleles l^A and l^B are responsible for the production of the respective antigens ‘A’ and’B’. The allele i does nto produce any antigen. The alleles l^A and l^B are dominant to the allele i, but co-dominant to each other (l^A = l^B > i). A child receives one of the three alleles from each parent, giving rise to six possible
Table: Genetic control of the human ABO blood groups

genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B and ii. The phenotypic expressions of I^AI^A and I^Ai are ‘A’ – type blood, the phenotypic expressions of I^BI^b and I^Bi are ‘B’ – type blood, and that of I^AI^B is ‘AB’ – type blood. The phenotype of ii (I°I°) is ‘O’ – type blood.

Question 2.
Describe chromosomal theory of sex determination.
Answer:
Sex Determination :
The mechanism of sex determination has always been a puzzle to geneticists. In fact, the cytological observations made in a number of insects led to the development of the concept of genetic / chromosomal basis of sex determination.

Sex Chromosomes :
In most of the animals a pair of chromosomes is responsible for the determination of sex. These two chromosomes are called sex chromosomes or allosomes. The chromosomes other than the sex chromosomes are called autosomes. The first indication that sex chromosomes were distinct from the other chromosomes came from the experiments conducted by Henking. He could trace a specific nuclear structure all through spermatogenesis in wasps, and it was also observed by him that 50 percent of the spermatozoa received this structure after spermatogenesis, whereas the other 50 percent did not receive it.

Henking gave the name X – body to this structure, but he could not explain its significance. Further investigations by other scientists led to the conclusion that the X – body of Henking was in fact a chromosome and that is why it was given the name X – chromosome. Stevens and Wilson first identified Y – chromosome as a sex determining chromosome in the mealworm, Tenebrio molitor. The revealed that the chromosomal basis of sex depended on the presence or absence of the Y – chromosome.

Heterogametic Sex Determination :
Heterogametic sex refers to the sex of a species in which the sex chromosomes are not similar.’The process of sex determination by allosomes is called genetic or chromosomal sex determination. In the heterogametic sex determination, one of the sexes produces ‘similar’ gametes and the other sex (heterogametic sex) produces ‘dissimilar / unlike gametes. The sex of the young one is determined at the time of syngamy (fertilization). It depends on which gamete of the two dissimilar gametes unites with the other gamete produced by the ‘homogametic parent’.

Male Heterogamety :
In this method of sex determination, the males (heterogametic) produce dissimilar gametes while females (homogametic) produce similar gametes. Male heterogamety is of two kinds, XX – XO^type and XX – XY type.

XX – XO type :
In some insects such as bugs, grasshoppers and cockroaches, females are with two X – chromosomes and males are with one X – chromosome in each somatic cell. McClung discovered this type in grasshoppers. The unpaired X – chromosome determines the male sex. The karyotype of the female (homogametic) is AAXX and that of the male (heterogametic) is AAXO. All the ova contain ‘AX’ complement of chromosomes and the sperms are of two types. One half of the sperms have ‘AX’ complement and the other half have ‘A’ complement of chromosomes. The sex of the offspring depends on the types of sperm that fertilizes the ovum.

XX – XY type :
In human beings and some insects such as Drosophila, both females and males have the same number of chromosomes. The karyotype of the female is AAXX and that of the male is AAXY. Females are ‘homogametic’ with ‘XX’ chromosomes.

They produce similar ova having one X – chromosome each. Males are ‘heterogametic’ with X and Y – chromosomes. They produce two kinds of sperms; one half of them with X – chromosome and the other half with Y – chromosome. The sex of the offspring depends on the fertilizing sperm. The XX – XY type is also found in most other mammals.

Female Heterogamety :
In this method of sex determination, the males produce ‘similar gametes’ while females produce ‘dissimilar gametes’. Female heterogamety is of two kinds, ZO – ZZ type and ZW – ZZ type.

ZO – ZZ type :
In moths and some butterflies, female is heterogametic with one Z-chromosome (ZO) and male is homogametic with two Z – chromosomes (ZZ). The karyotype of female is AAZO and male is AAZZ. Females produce two kinds of ova, half of them with a Z – chromosome and the other half with no sex chromosome. Males produce similar type of sperms. The sex of he offspring depends on the type of ovum that is fertilized.

ZW – ZZ type :
In birds, reptiles, some fishes, etc., the females are heterogametic with ZW – allosomes and males are homogametic with ZZ – allosomes. The karyotype of the female is AAZW and that of the male is AAZZ. All sperms are similar with the allosome – Z. Ova are of two different kinds; one half of the ova are with the allosome – Z and the other half with the allosome – W. The sex of the offspring depends on the type of ovum that is fertilized.

Sex Determination in Humans :
It has already been mentioned that the sex determining mechanism in case of human is XX – XY type. Out of 23 pairs of chromosomes present, 22 pairs are exactly same in both males and females; these are the autosomes. A pair of X- chromosome is present in the female, whereas the presence of an X and Y – chromosome are determinant of the male characteristic. During spermatogenesis among males, two types of gametes are produced. 50 percent of the total sperm produced carry the X – chromosbme and the rest 50 percent has Y – chromosome besides the autosomes. Females, however, produce produce only one type of ovum with an X – chromosome.

There is an equal probability of fertilisation of the ovum by the sperm carrying either X or Y- chromosome. In case the ovum js fertilised by a sperm carrying X – chromosome, the zygote develops into a female and the fertilisation of ovum with Y – chromosome carrying sperm results into a male offspring. Thus, it is evident that it is the genetic makeup of the sperm that determines the sex of the child. It is also evident that in each pregnancy there is always 50 percent probability of either a male or a female child.

Question 3.
What is crisscross inheritance? Explain the inheritance of one sex linked recessive character in human beings. [Mar. ’19, 17, May ’17 (A.P.); Mar. ’15 (A.P. & T.S.)]
Answer:
The crisscross pattern of inheritance (skip generation inheritance) is one in which a gene responsible for the sex linked recessive character is transmitted from a male parent to a male grand child through a carrier female of the first generation. Colour blindness is the best example for criss – cross inheritance in human being.

Colour blindness :
It is a sex-linked recessive disorder. Retina of the eye in man contains the cells sensitive to red and green colours. This phenotypic trait is genetically controlled. Its alleles are located on the X – chromosome; When a woman with normal vision (homozygous) marries a colour – blind man, all the sons and daughters are normal, but daughters are carriers (heterozygous). If a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and another half of sons are colour – blind. Colour – blind trait is inherited from a male parent to his grandsons through carrier daughter, which is an example of crisscross pattern of inheritance.

Haemophilia :
Haemophilia A is recessive X – linked genetic disorder involving lack of the functional clotting Factor – VIII and represents 80% of haemophilia cases. Haemophilia B is also a recessive X – linked genetic disorder involving lack of the functional clotting Factor IX. When a person with hemophilia is injured, bleeding is prolonged because a firm clot is slow to form. Haemophilia follows the characteristic crisscross pattern of inheritance like that of colour – blindness.

Duchenne Muscular dystrophy :
Duchenne muscular dystrophy (DMD) is a recessive X – linked form of muscular dystrophy, affecting around 1 in 3,600 boys. The disease is characterized by a progressive weakening of the muscles and loss of coordination. Affected individuals rarely live past their early 20s. The disorder is caused by a mutation in the dystrophin gene (the largest known gene in humans) located on the X – chromosome, which codes for the protein dystrophin, an important structural component within muscle tissue (connects sarcolemma and the outer most layer of muscle filaments and supports muscle fiber strength).

If the mother is known to be a carrier of this gene, about X – linked recessive inheritance half of her male children are expected to Colour blindness be affected. All female children born to a carrier mother are expected to be normal, since the possibility of their being homozygous for this sex – linked recessive gene is virtually non – existent.

Question 4.
Write an essay on common genetic disorders.
Answer:
Genetic disorders :
A number of disorders in human beings have been found to be associated with the inheritance of changed “or altered genes or chromosomes. Genetic disorders may broadly be grouped into two categories – Mendelian disorders and Chromosomal disorders.

Mendelian disorders :
Medelian disorders are genetic diseases showing Mendelian pattern of inheritance, caused by a single mutation in the structure of DNA, which causes a single basic defect with pathologic consequences, in some cases. Mendelian disorders are also called monogenic diseases. Monogenic diseases run in families and can be dominant or recessive and autosomal or sex linked (allosomic). The pattern of inheritance of Mendelian disorders can be traced in a family with the help of pedigree charts and their analyses. The most common and prevalent Mendelian disorders are Haemophilia, Cystic fibrosis, Sickle – cell anaemia, Colour blindness, phenylketonuria, Thalassemia, DMD, Albinism, etc.

Haemophilia:
Haemophilia A (caused by deficiency of clotting factor VIII) and Haemophilia B (caused by deficiency of clotting factor IX) are X – linked recessive disorders that impair the body’s ability to control clotting or coagulation of blood. Haemophilia C is an autosomal recessive disorder involving lack of the functional clotting ‘factor XI’. Haemophilia is also called bleeder’s disease. Haemophilia A and B follow the characteristic crisscross pattern of inheritance like that of colour – blindness. In this disease, a single protein that is a part of the cascade of reactions involved in the clotting of blood is affected.

Haemophilia is more likely to occur in males than in females. This is because female have two X -chromosomes while males have only one, and so the defective gene on the X will certainly express in the male who carries it. As it is caused by a recessive allele on the X chromosome, a female human being has to be ‘double recessive’ to express haemophilia. Because the chance of a female having two defective copies of the gene (alleles) is very remote, the females are mostly asymptomatic carriers of the disorder. The ‘allele’ is typically passed on from an affected father to 50% of his grand sons through his ‘carrier daughters’. The family pedigree of Queen Victoria shows a number of haemophilic descendents, as she was a carrier for the disease.

Sickle – cell anaemia :
Sickle – cell anaemia is an autosomal recessive genetic blood disorder characterized by red blood cells that assume an abnormal, rigid, sickle-shape in hypoxia conditions (at high altitudes or under physical stress, for instance). Sickled cells may clump and clog small blood vessels, often leading to other symptoms throughout the body, including physical weakness, pain, organ damage, and even paralysis.

This disease is controlled by a single pair of alleles, HbA and Hbs found on the chromosome 11. The homozygous individuals for sickle – cell anaemia (Hb^A Hb^s) express the diseased phenotype. Heterozygous individuals (Hb^A Hb^s) appear ‘unaffected’ but they are still, carriers of the disease. Even though two sickle cell alleles are necessary to cause sickle cell anaemia, one dose can affect the phenotype. Persons ‘heterozygous’ to sickle cell trait can usually lead a healthy life but in prolonged periods of reduced oxygen content in the blood may suffer from symptoms of SCD as both normal and sickle cell haemoglobin are formed in them.

Micrograph of the red blood cells and the amino acid composition of the relevent portion of β – chain of haemoglobin; (a) From a normal individual; (b) From an individual with sickle – cell anaemia.

Sickle cell anaemia is caused by a point mutation in the DIMA that codes for the beta globin polypeptide chains of the haemoglobin molecule, causing the replacement of the glutamic acid in the sixth position by valine. The heterozygous individuals are relatively resistant to the most severe effects of malaria such as those of ‘falciparum malaria’ also (although they are not resistant to malaria infection) – an effect called heterozygote advantage. The heterozygous individuals carry the deleterious alleles in their genomes (genetic load).

Phenylketonuria (PKU) :
Phenylketonuria was discovered by A. Foiling. This is an autosomal recessive, metabolic genetic disorder caused by a mutation in the gene (PAH. Phenylalanine hydroxylase gene) located on the chromosome 12 for the hepatic enzyme ‘phenylalanine hydroxylase’, rendering it nonfunctional. The affected individual lacks the above mentioned enzyme that converts the amino acid phenylalanine into tyrosine. When phenylalanine hydroxylase’s activity is reduced, phenylalanine accumulates and is converted into phenylpyruvate and other derivatives. Accumulation of these substances in the brain causes mental retardation. Adherence to a low phenylalanine diet prevents major mental retardation.

Colour blindness :
Colour blindness (colour vision deficiency) is a sex – linked recessive disorder. It is the inability or decreased ability to see certain colours or perceive differences between some colours. This phenotypic trait is due to mutation in certain genes located on X – chromosome. The most common inherited forms of colour blindness are Protanopia (red colour blindness), Deuteranopia (green colour blindness) and Tritanopia (blue colour blindness – autosomal).

The son of a woman who carries the allele has a 50 percent chance of being colour – blind. The mother herself is not colour – blind because the allele is recessive. That means its effect is suppressed by her matching dominant normal allele. A daughter will not normally be colour blind, unless her mother is ‘colour – blind’ or a ‘carrier’ and her father is colour – blind. The Ishihara colour test, which consists of a series of pictures of coloured spots, is most often.used to diagnose red – green colour blindness.

Thalassemia :
Thalassemia is an autosome linked recessive blood disroder. This disease is caused by the excessive destruction or degradation of red blood cells due to formation of abnormal haemoglobin molecules, because of a defect through a genetic mutation or deletion (a type of chromosomal mutation). Normally, haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains arranged into a hetero tetramer. In the case of thalassemia, patients have defects in either the alpha or beta globin chain (unlike sickle cell anaemia, which is caused due to a specific change in the beta chain), causing production of abnormal haemoglobin molecules resulting in anaemia which is characteristic of the disease. The effected people make less haemoglobin and fewer RBC in the circulating blood, hence anaemia.

Thalassemias are classified based on which chain of haemoglobin molecule is affected. In Alpha thalassemia, the production of alpha globin chain is affected. Alpha thalassemia is controlled by two closely linked genes HBAI and HBA2 on chromosome 16 of each parent and it is caused due to mutation or deletion of one or more of the four alpha gene “alleles”. The more genes affected, the less alpha globin molecules are produced. In Beta thalassemia (Cooley’s Anaemia) production of beta globin chain is affected. The Beta thalassemia is controlled by single gene HBB on chromosome 11 of each parent and occurs due to mutation of one or both alleles. It is the most common type of thalassemia. In this disorder the alpha chains which are produced in excess bind to RBCs and damage them.

Cystic fibrosis :
Cystic fibrosis is an autosomal recessive genetic disorder. CF is the result of mutations affecting a gene on the long arm of chromosome 7 that influences salt and water movement across epithelial cell membranes. The genetic defect causes increased sodium and chloride content in sweat and increased resorption of sodium and water from respiratory eptihelium. The extracellular chloride causes the mucus that coats certain cells to become more viscous and sticky. The mucus builds up in organs such as lungs, pancrias, Gl tract etc., and leads to further complications and may lead to death by the age five, if untreated.

Chromosomal disorders :
Chromosomal disorders are caused by errors in the ‘number’ or ‘structure’ of chromosomes. Chromosomal anomalies usually occur when there is an error in cell division. Aneuploidy is a chromosomal aberration where there is again or loss of one or more chromosomes in a ‘set’. It is caused by non-disjunction of chromosomes. The result of this error is origin of cells with a deviation from the normal number of chromosomes – aneuploidy.

A) Allosomal disorders :
Klinefelter’s syndrome :
This genetic disorder is caused by trisomy 23rd pair. The karyotype is 47, XXY. A Klinefelter male possesses an additional X – chromosome along with the normal XY. The principal effects include hypogonadism and reduced fertility. At the same time, feminine sexual development is not entirely suppressed. Slight enlargement of the breasts (gynecomastia) is common, and the hips are often rounded. The somatic cells of a Klinefelter male exhibit Barr bodies in their nuclei.

Turner’s syndrome :
The Karyotype is 45, X. It is due to monosomy 23rd pair, where one X – chromosome is lost. A Turner female does not show Barr bodies in her somatic cells. The symptoms are short stature, gonadal dysgenesis, webbed neck and broad shield like chest with widely spaced nipples.

B) Autosomal disorders :
Most of the following disorders are common in children born to woman who conceive babies rather late in their reproductive phase.

Down syndrome (Trisomy 21) :
Down syndrome is a genetic condition that causes delays in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21 (trisomy of 21st set). The karyotype is designated as Trisomy 21 (47, XX, + 21).The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Physical, psychomotor and mental development is retarded.

Edwards syndrome (Trisomy 18) :
Edwards syndrome (47, XX, + 18) is a chromosomal abnormality characterized by the presence of an extra copy of the genetic material on the 18th chromosome, either in whole (trisomy 18) or in part (such as due to translocations). Edwards syndrome occurs in all human populations but is more prevalent in the female offspring. The majority of people with the syndrome die during the foetal stage; infants who survive experience serious defects (cardiac abnormalities and kidney malfunction) and commonly live for short periods of time.

Patau syndrome (Trisomy 13) :
Patau syndrome, is a chromosomal condition associated with severe intellectual disability and physical abnormalities in many parts of the body. Most cases of trisomy 13 (47, XX, +13) result from having three copies of chromosome 13 in each cell in the body instead of the usual two copies. Individuals with trisomy 13 often have heart and kidney defects, brain or spinal cord abnormalities, very small or poorly developed eyes (microphthalmia), cleft palate etc. Due to the presence of several life threatening medical problems, many infants with trisomy 13 die within their first days or weeks of life.

Cri-du-Chat syndrome (5p minus syndrome):
Cri – du – chat syndrome (cat – cry) is due to a partial deletion of the short arm of chromosome 5, also called 5p monosomy. It might be considered a case of partial monosomy, but since the region that is missing is so small, it is better referred to as 5p segmental deletion. The karyotype is 46, XX, 5p“. It is a French term referring to the characteristic cat – like cry of the affected children due to problems with the larynx and nervous system. Such infants are mentally retarded, have a small head with unusual facial features. They die in infancy or early childhood.

Chronic Myelogenous (Myeloid) Leukemia (CML) :
In certain cancers such as Chronic myelogenous leukemia (also called Chronic granulocytic leukemia), a piece of the chromosome 9 and a piece of the chromosome 22 break off and ‘switch places’ (exchange places) with each other (reciprocal translocation). This results in the formation of an abnormally short chromosome 22 and abnormally long chromosome 9. The short 22nd chromosome is called Philadelphia chromosome produced by translocation which is also called Philadelphia translocation. The karyotype is 46, XXt(9; 22). The Philadelphia chromosome results in the production of an abnormal enzyme called a tyrosine kinase. Along with other abnormalities, this enzyme causes uncontrolled cell cycle progression leading to the cancer called chronic myelogenous leukemia.

Question 5.
Why is the Human Genome project called a mega project?
Answer:
Human Genome Project (HGP) was called a mega project. It was an international effort formally begun in October, 1990. The HGP was a 13 – year project coordinated by the U.S. Department of Energy and the National Institute of Health. During the early years of the HGP, the Wellcome Trust (U.K.) became a major partner, and additional contributions came from Japan, France, Germany, China and others. The project was almost completed in 2003. Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

HGP was closely associated with the rapid development of a new area in biology called Bioinformatics. Besides providing clues to understanding human biology, learning about non – human organisms’ DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation. Genomes of many non – human model organisms, such as bacteria, yeast, Caenorhabditis elegans (a free living non – pathogenic nematode), Drosophila, plants (rice and Arabidopsis), etc. have also been sequenced. In a way they helped the progress of HGP.

Goals of HGP :
Some of the important goals of HGP were as follows :

1. Identify all the approximately 20,000 – 25,000 genes in human DNA.
2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
3. Improve tools for data analysis.
4. Address the ethical, legal, and social issues (ELSI) that may arise from the project.

Methodologies :
The methods involved two major approaches. One approach focused on identifying all the genes that expressed as RNA (referred to as Expressed Sequence Tags (ESTs). The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).

What is DNA sequencing?
DNA sequencing, the process of determining the exact order of the 3 billion paired chemical building blocks (called ‘bases’ – A, T, C, and G) that make up the DNA of the 24 different human chromosomes (23 + Y in a male), was the greatest technical challenge in the Human Genome Project.

For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller size and cloned in a suitable host using specialized vectors. The cloning results in the amplification of DNA fragments which are used for sequencing the bases. The commonly used hosts are bacteria and yeast, and the vectors are called BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes). The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger.

Alignment of these sequences was humanly not possible. Therefore, specialized computer based programs were developed. These sequences were subsequently annotated and were assigned to each chromosome. The latest method of sequencing even longer fragments, by a method called Shotgun sequencing using super computers, replaced the traditional sequencing methods.

Salient Features of Human Genome :
Some of the salient observations drawn from human genome project are as follows :

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being the one that codes for the protein called dystrophin.
3. The total number of genes is estimated at 30,000. Almost all (99.9%) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50% of the genes discovered.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has the highest number of genes (2,968), and the Y – chromosome has the fewest genes (231).
9. Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – single nucleotide polymorphism, pronounced as snips) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease – associated sequences and tracing human history.

Advantages of HGP:

1. In the area of health care, identification and mapping of the genes responsible for genetic diseases helps in diagnosis, treatment and prevention of these diseases.
2. Detailed knowledge of the genomes of humans and other species will give a clearer picture of Gene expression, Cellular growth and differentiation and evolutionary biology.
3. Earlier detection of genetic predispositions to disease, rational drug design, Gene therapy is going to be easy with more knowledge on human genome.
4. A new era of Molecular Medicine, characterized by looking into the most fundamental causes of disease than treating the symptoms will be an important advantage.

Question 6.
What is DNA finger printing? Mention its applications.
Answer:
DNA Finger Printing :
Over 99% of the 3 billion nucleotide pairs in human DNA are identical among all individuals. No two people (other than identical twins) have exactly the same sequence of bases in their DNA. Restriction Fragment Length Polymorphisms (RFLPs – pronounced riflips) are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as Genetic markers. The VNTRs of two persons generally show variations. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. They are bits of chromosomes that can be cut by restriction endonucleases.

The ‘fundamental techniq’ue’ involved in DNA Finger Printing was pioneered and perfected byJeffrys of Great Britain. He observed that the gene pertaining to myoglobin of muscles contains many segments that vary in size and composition, from one person to another. For example in the following hypothetical example nucleotide base sequence, there are 6 Tandem Repeats of 16 bases each (count the first 16 and note how they are repeated). 5’GACTGCCTGCTAAGATGACTGCCTGCTAAGATGACTGCCTGCTAAGATGA CTGCCTGCTA AG ATG ACTGCCTGCTAAG ATG ACTGCCTGCTAAG AT3′

Such clusters of 10 – 100 nucleotides are called mini satellites. Such tandem repeats are characteristic of every person’s DNA. The VNTRs of two persons differ in the number of tandem repeats or the sequence of bases. Such changes are caused due to mutations and gene recombinations. For example a child might inherit a chromosome with 6 tandem repeats from the mother and the same tandem repeated 4 times from the father in a homologous chromosome. It means half of the VNTR alleles of the child resemble those of the mother and the other half those of the father. This is a ‘heterozygous condition with reference to VNTR alleles’. These tandem repeats serve as basis of a technique called DNA fingerprinting.

DNA Fingerprinting – Protocol :
1. Obtaining DNA (Isolation /Extraction) :
The first step is to obtain a sample of DNA from blood, saliva, hair roots, semen etc. If needed many copies of the DNA can be produced by PCR (cloning / DNA amplification).

2. Fragmenting DNA (Restriction Digestion) :
Treating DNA with Restriction Enzymes (Restriction endonucleases) which cut the DNA into smaller fragments by cutting it at specific sites.

3. Separation of DNA fragments by electrophoresis :
DNA fragments are applied at one end of agarose gel plate. When an electric current is applied to the gel, the DNA fragments (which are slightly negatively charged) travel across the gel (smaller and more mobile pieces travel farther). This technique of separation of DNA fragments into individual bands is called Gel Electrophoresis.

4. Denaturing DNA :
The DNA on the gel is ‘denatured’ using alkaline chemicals or by heating, (denaturing means separation / splitting of the double helix into ‘single strands’ by breaking hydrogen bonds between the two strands.)

5. Blotting :
A thin nylon membrane is placed over the ‘size fractionated DNA strands’ and covered by paper towels. As the towels draw moisture the DNA strands are transferred on to the nylon membrane by capillary action. This process is called ‘Blotting’ – more precisely Southern blotting, after the name of its inventor E.M. Southern.

6. Using probes to identify specific DNA :
A radioactive probe (DNA is labeled with a radioactive substance) is added to the DNA bands. The Probe is a single stranded DNA molecule that is ‘complementary1 to the gene of interest in the sample under study. The probe attaches by base pairing to those restriction fragments that are complementary to its sequence. The probes can be prepared by using either ‘fluorescent substances’ or ‘radioactive isotopes’.

7. Hybridisation with probe :
After the probe hybridises and the excess prob washed off, a photographic film is placed on the membrane containing ‘DNA hybrids’.

8. Exposure on film to make a Genetic / DNA Finger Print :
The radioactive label exposes the film to form an image (image of bands) corresponding to specific DNA bands. The thick and thin dark bands form a pattern of bars which constitute a Genetic fingerprint.

A given person can never have a VNTR which his parents do not have. Obtaining hybrid with radioactive probe and matching DNAs of different members of a family with biological children and adopted children, gives us an idea of how DNA Finger Prints help identification of paternity/maternity, by studying the ‘DNA Finger Prints’ of members of a Family – Biological and non-biological relationships.

The illustrations given below are the VNTR patterns for, Mrs. Rose (blue), Mr. Rao (yellow), and their four children : D1 (Mr. Rao’s biological daughter), D2 (Mr. Rao’s step -daughter, child of Mrs. Rose and her former husband (red), SI (Mr, Rao’s biological son), and S2 (Mr. Rao’s adopted son not biologically related, his parents’ DNA marked in light and dark green bands).

Applications of DNA Finger Printing :

1. Conservation of wild life – protection of endangered species. By maintaining their DNA records for identification of tissues of the dead endangered organisms.
2. Taxonomica! applications – study of phylogeny.
3. Pedigree analysis – inheritance pattern of gene through generations.
4. Anthropological studies-charting of origin and migration of human population.
5. Medico – legal cases – establishing paternity and / or maternity more accurately.
6. Forensic analysis – positive identification of a suspect in a crime.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health

Very Short Answer Type Questions

Question 1.
What are the measures one has to take to prevent contracting STDs? [March 2018,17; May 17 (A.P.)]
Answer:

1. Avoiding sex with unknown partners/ multiple partners.
2. Using condoms compulsorily during coitus.
3. Consulting qualified doctor for early detection of STDs and getting complete treatment in case of infections.

Question 2.
What in your view are the reasons for population explosion, especially in India? [Mar. ’14; May/June ’14; Mar. ’15 (T.S.)]
Answer:
Reasons for population growth explosion are :

1. Increase of growth rate due to increased health care facilities.
2. Decline in death rate, maternal mortality rate (MMR) and infant mortality rate (IMR).
3. Better living conditions protecting the people from illness or disease attack.

Question 3.
It is true that ‘MTP is not meant for population control’. Then why did the Government of India legalize MTP? [March 2019]
Answer:
MTP – Medical Termination of Pregnancy.
Government of India made an act in 1971 legalizing MTP with certain restrictions and conditions to avoid its misuse. It is because in cases where continuation of pregnancy could be harmful or even fatal either to the mother or to the foetus or for both, MTP is the inevitable solution.

Question 4.
What is ‘amniocentesis’? Name any two disorders that can be detected by amniocentesis. [March 2020, 2018, ’17; May ’17 (A.P.)]
Answer:
Amniocentesis is a diagnostic procedure to detect genetic defects in the unborn baby. The most common abnormalities that can be detected by amniocentesis are Down syndrome, Edward syndrome and Turner’s syndrome.

Question 5.
Mention the advantages of ‘lactational amenorrhea method’. [March 2019]
Answer:
Ovulation generally will not occur during the period of intense lactation by the mother following parturition (delivery). This is known as Lactational amenorrhea. Some couples utilize the contraceptive benefit of this method.

As long as the mother fully breast feeds her child, chances of conception are almost zero. In addition breast feeding offers many benefits to the infant such as enhanced immunity, protection against allergies.

Short Answer Type Questions

Question 1.
Briefly describe the common sexually transmitted diseases in human beings.
Answer:
Sexually Transmitted Diseases (STDs) :
Diseases or infections which are transmitted through sexual contact (intercourse) are collectively called sexually transmitted diseases (STDs) or venereal diseases (VDs) or reproductive tract infections (RTI). Most common STDs and their causative organisms are shown in the table below.

Question 2.
Describe the surgical methods of contraception.
Answer:
Surgical procedure to prevent pregnancy is also known as sterilization. Sterilization procedure in the male is called vasectomy and that in the female tubectomy.
i) Vasectomy :
A small part of the vas deferens on either side is removed or tied up through a small incision on the scrotum. Thus the sperms are prevented from reaching seminal vesicle and so the ‘semen’ in ’vasectomised’ males do not contain sperms.

ii) Tubectomy :
A small part of the fallopian tube on both sides is removed or tied up through a small incision made in the abdomen or through vagina. This will block the entry of ova into the fallopian tubes and thus pregnancy is prevented.

Question 3.
Write short notes on any two of the following.
a) IVF b) ICSI c) lUDs
Answer:
a) IVF :
In Vitro Fertilization and Embryo Transfer (IVF – ET) Fertilization of ovum by sperm done outside the body of a woman is called in vitro fertilization. The resultant early embryonic stage (with generally 8 biastomeres) is transferred into the mother’s uterus for further development (Embryo Transfer or Intra Uterine Transfer IUT). In this method, which is popularly known as Test Tube Baby Procedure, ova from the wife / female donor and sperms from the husband / male donor are collected, mixed and induced to form zygote under simulated conditions (almost similar conditions as that in the female body) in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced in vitro, it can be implanted in the uterus of another woman (surrogate mother) willing to carry this embryo.

b) Intracytoplasmic Sperm Injection (ICSI) :
This is another specialized procedure in which a sperm is directly injected into the ovum with the help of a microscopic needle to form an embryo in the laboratory. Later the embryo is transferred to the uterus or fallopian tube for further development. This method is employed to assist the couple where there are problems with the sperms such as decrease in sperm count.

c) Intra Uterine Devices (lUDs) :
These devices are inserted into the uterus by doctors or trained nurses through vagina. Different types of lUDs such as Nonmedicated lUDs (e.g. lippes loop), Copper releasing lUDs (Lu T, Cu 7, Multiload 375) and hormone releasing lUDs (Progestasert, LNG – 20) are available for contraception. lUDs promote ‘phagocytosis’ of sperms by white blood corpuscles within the uterus and the copper ions released suppress the motility, viability and fertilizing capacity of the spermatozoa. The hormone releasing lUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile / antagonistic to the sperms. lUDs are ideal contraceptives to females who want to delay and / or have space between children. This is a widely accepted method of contraception in India.

Question 4.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is biological inability of a person to contribute to conception. A large number of couples in the conceivable age all over the world is childless. Infertility clinics and specialized health care units could help in diagnosis and corrective treatment of some of these disorders and enable the couples to have children. Assisted Reproductive Technology (ART) offers a wide range of techniques listed below can help the childless couple. They are

1) In vitro Fertilization and Embryo Transfer (1VF – ET) 2) Zygote Intrafallopian Transfer (ZIFT) 3) Gamete intrafallopian Transfer (GIFT) 4) intracytoplasmic sperm injection (ICSI) 5) Artificial insemination (Al).

Question 5.
Is sex education necessary in schools? Why?
Answer:
Governmental and non – governmental agencies have taken various steps to educate people on reproduction – related issues using audio-visual and print media. Introduction of sex education, in schools will provide right information to the young on sex and other related issues. Proper information about the reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually tansmitted diseases such as HIV/AIDS, etc., would help people, especially those in the adolescent age group to lead a reproductively healthy life. Awareness should be created in the society on problems caused by uncontrolled population growth and social evils like sex abuse and sex related crimes etc.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System

Very Short Answer Type Questions

Question 1.
Where are the testes located in man? Name the protective coverings of each testis.
Answer:
Testes are located outside the abdomen with in a pouch called scrotum. Each testis is enclosed in a fibrous envelope, the tunica albuginea.

Question 2.
Name the canals that connect the cavities of scrotal sac and abdominal cavity. Name the structures that keep the testes in their position.
Answer:
The cavity of scrotal sac is connected to the abdominal cavity through the inguinal canal. Structure that keeps testis in their position is guberhaculum.

Question 3.
What are the functions of Sertoli cells of the seminiferous tubules and the Leydig cells in man? [Mar. ’15 (A.P. & T.S.)]
Answer:
Sertoli cells :
Nourishes the growing sperms and also produce a hormone called inhibin.

Leidig cells :
Present in seminiferous tubules produce and rogens, the most important of which is testosterone.

Question 4.
Name the copulatory structure of man. What are the three columns of tissues in it?
Answer:
The copulatory structure of man is penis. Three columns of tissues are two upper corpora cavernosa and one lower (ventral) corpora spongiosum.

Question 5.
Define spermiogenesis and spermiation.
Answer:
Development of spermatozoa from sperm mother cells in male is called spermiogenesis. After spermiogenesis sperm heads become embedded in the Sertoli cells and are finally released from the seminiferous tubules by the process called spermiation.

Question 6.
Name the yellow mass of cells accumulated in the empty follicle after ovulation. Name the hormone secreted by it and what is its function? [March 2020]
Answer:
After ovulation, the granulosa cells in the follicle proliferate and are transformed into a yellowish glandular mass called corpus luteum. It secretes progesterone hormone. This hormone is essential for maintenance of pregnancy in first few months.

Question 7.
Define gestation period. What is the duration of gestation period in the human beings?
Answer:
Intra uterine development of the embryo or foetus is called gestation period. In human being gestation period is 266 days or 38 weeks.

Question 8.
What is implantation; with reference to embryo?
Answer:
Attachment of blastocyst to the uterine mucosa till the whole of it comes to lie with in the thickness of the endometrium. This is called interstitial implantation. In human the implantation beings on the 6th day after fertilization.

Question 9.
Distinguish between epiblast and hypoblast
Answer:
The mature oocyte lies eccentrically in the follicle surrounded by some surface facing the cavity. This cells layer develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the embryonic disc is called epiblast.

Question 10.
Write two major functions, each of testis and ovary.
Answer:
Major functions of Produce
a) Testes :
Produce spermatozoa for fertilisation produce hormones which induce secondary sexual characters of males.

b) Ovary :
Produce mature ova for fertilisation. Produce before and after fertilisation estrogens and progesterone hormones.

Question 11.
Draw a labelled diagram of a sperm.
Answer:

Question 12.
What are the major components of the seminal fluid?
Answer:
Seminal fluid is alkaline, viscous fluid. 60% of the volume of seminal fluid is constituted by secretion of the seminal vesicle. Seminal fluid contains fructose, proteins, citric acid, inorganic phosphorus, potassium and prostaglandins. Prostate secretion contributes 15 – 30 percent of semen.

Question 13.
What is menstrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates is called menstrual cycle. The cyclic changes that occur in the endometrium every month are together called menstrual cycle.

LH and FSH (Gonadotropic) from pituitary, estrogens from ovarian follicle, progesterone from corpus luteum regulate menstrual cycle.

Question 14.
What is parturition? Which homones are involved in inducing parturition?
Answer:
The process of delivery of the foetus (child birth) is called parturition. Parturition is induced by oxytocin.

Question 15.
How many eggs do you think were released by the ovary of a female dog which gave birth to six puppies?
Answer:
Only six (6) ova or eggs are released by the ovary of a femaleidog which gave birth to six puppies.

Question 16.
What is neurulation?
Answer:
The neural plate invaginates towards the notochord to form a neural groove, which deepens progressively to form a tube by fusion of the lateral neural folds. The process of formation of neural tube is referred to as NEURULATION.

Question 17.
What is capacitation of sperms?
Answer:
Spermatozoa acquire the ability to fertilize the ovum only after they undergo some changes in the female genital tract. These changes are called capacitation.

Question 18.
What is compaction in the human development? [March 2015 (A.P.)]
Answer:
In the morula due to unequal cleavage smaller and larger blastomeres are formed. The morula passes through a process called compaction. Now the embryo has a superficial flat cell layer and inner cell mass. Inner cell mass gives rise to the embryo proper. This is the first sign of cell differentiation in the human embryo.

Question 19.
Distinguish between involution and ingression in the human development.
Answer:
a) Involution :
The inward growth and curling inward of a group of cells (prospective mesodermal cells), as in the formation of a gastrula from a blastula is called involution.

b) Ingression :
The inward migration of future endodermal cells from the epiblast during gastrulation is called ingression.

Question 20.
What are the four extra embryonic membranes?
Answer:
The four extra embryonic membranes are Amnion, Chorion, Allantois and Yolk Sac.

Short Answer Type Questions

Question 1.
Describe microscopic structure of testis of man.
Answer:
Each testis is enclosed in a fibrous envelope the tunica albuginea which extends inward to form septa that partition the testis into lobules. There are about 250 testicular lobules in each testes. Each lobule contains 1-3 highly coiled seminiferous tubules. A pouch of serous membrane (peritoneal layer) called tunica vaginalis covers the testis.

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH.

The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or leydig cells. Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoaTrom the seminiferous tubules to the vasa efferentia).

Question 2.
Describe the microscopic structure of ovary of woman.
Answer:
The ovaries are covered on the outside by a layer of simple cuboidal epithelium called germinal (ovarian) epithelium. This is actually the visceral peritoneum that envelops the ovaries. Underneath this layer there is a dense connective tissue capsule, the tunica albuginea. The ovarian stroma is distinctly divided into an outer cortex and an inner medulla. The cortex appears more dense and granular due to the presence of numerous ovarian follicles in various stages of development. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels, and nerve fibers.

Question 3.
Describe the Graafian follicle in woman.
Answer:
A homogenous membrane, the zona pellucida, appears between the primary oocyte and granulosa cells. The innermost layer of granulosa cells are firmly attached to zona pellucida forming the corona radiata.

A cavity (antrum) appears within the membrana granulosa. The follicular cavity increases in size. As a result, the wall of the follicle becomes relatively thin. The oocyte now lies eccentrically in the follicle surrounded by some granulosa cells. It is called cumulus oophorus. As the follicle expands the stromal cells surrounding the membrana granulosa become condensed to form a covering called the theca interna. Outside the theca interna some fibrous tissue becomes condensed to form another covering called theca externa. Now these follicles are called secondary follicles.

The cells of theca interna later secrete a hormone called oestrogen. At this stage, the primary oocyte within the secondary follicle grows in size and completes Meiosis (.It is an unequal division resulting in the formation of a large haploid secondary oocyte and a tiny first polar body (haploid). The secondary oocyte retains bulk of the cytoplasm (nutrient rich) of the primary oocyte. Then the second meiotic division begins, but stops at metaphase. The secondary follicle further changes into the mature follicle called Graafian follicle.

Question 4.
Draw a labelled diagram of the female reproductive system.
Answer:

Question 5.
Diagrammatic sectional view of the female reproductive system
Answer:

Question 6.
Describe the structure of seminiferous tubule.
Answer:

Each testis is enclosed in a fibrous envelope the tunica albuginea which extends inward to form septa that partition the testis into lobules. There are about 250 testicular lobules in each testes. Each lobule contains 1-3 highly coiled seminiferous tubules. A pouch of serous membrane (peritoneal layer) called tunica vaginalis covers the testis.

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH.

The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or leydig cells. Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoaTrom the seminiferous tubules to the vasa efferentia).

Question 7.
What is spermatogenesis? Briefly describe the process of spermatogenesis in man.
Answer:
Spermatogenesis :
In the testis, the immature male germ cells, spermatogonia produce sperms by spermatogenesis that begins at puberty. The spermatogonial stem cells (present in the seminiferous tubules) multiply by mitotic divisions and increase in numbers. Each spermatogonial stem cells is diploid and contains 46 chromosomes. Some of the spermatogonial stem cells develop into primary spermatocytes which undergo meiosis periodically. A primary spermatocyte completes the first meiotic division (Meiosis -1) leading to formation of two equal sized, haploid cells called secondary spermatocytes, Which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division (Meiosis – II) to produce four equal sized haploid spermatids.

The spermatids are transformed in to spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in the Sertoli cells, and are finally released from the seminiferous tubules by the process called spermiation. Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin releasing hormone (GnRH). LH acts on the Leydig cells and stimulates secretion of androgens. Androgens, in turn, stimulate the process of spermatogenesis.

Question 8.
What is oogenesis? Give a brief account of oogenesis in a woman.
Answer:

Oogenesis :
The process of formation of a mature female gamete is called oogenesis. Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each foetal ovary and do not multiply thereafter. These cells start division and stop the process at prophase -1 of the meiosis – I At this stage these are called primary oocytes.

Question 9.
Draw a labelled diagram of a Graafian follicle.
Answer:

Question 10.
In our society women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Answer:
One has to remember that the sex of the baby has been decided at the time of fertilization itself. Let us see how? As we know the chromosome pattern in the human female is XX and that in the male is XY. Therefore, all the haploid gametes produced by the female (ova) have the sex chromosome X, whereas the male gametes (sperms) have either X- chromosome or Y-chromosome (50 percent of sperms carry the X-chromosome while the other 50 percent carry the Y chromosome).

After fusion of the male and female gametes the zygote would carry either XX or XY depending on what type of sperm fertilised the ovum. The zygote carrying ‘XX’ would develop into a female child and that with ‘XY’ would form a male child. So, the sex of a child depends on the male parent, (heterogametic parent). So it is not correct to blame women for giving birth to daughter very often.

Question 11.
Describe the accessory glands associated with male reproductive system of man.
Answer:
Male accessory genital glands :
The male accessory glands include paired seminal vesicles, a prostate and bulbourethal glands.

Seminal vesicles :
The seminal vesicles are a pair of simple tubular glands present postero inferior to the urinary bladder in the pelvis. Each seminal vesicle opens into the corresponding vas deferens, as the vas deferens enters the prostate gland. The secretion of the seminal vesicles constitutes about 60 percent of «.ne volume of seminal fluid. It is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorous, potassium, and prostaglandins. Once this fluid joins the sperm in the ejaculatory duct, fructose acts as the main energy source for the sperm outside the body. Prostaglandins are believed to aid fertilization by causing the mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of the sperm towards the ovum with peristaltic contractions of the uterus and fallopian tubes.

Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland surrounds the prostatic urethra, and sends its secretions through several prostatic ducts. In man, the prostate contributes 15 – 30 percent of the semen. The fluid from the prostate is clear and slightly acidic. The prostatic secretion ‘activates’ the spermatozoa and provides nutrition.

Bulbourethral Glands :
Bulbourethral glands, also called Cowper’s glands, are located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen during the process of ejaculation. The fluid secreted by these glands lubricates the urethra. It is also throught to function as a ‘flushing agent’ that washes out the acidic urinary residues that remain in the urethra, before the semen is ejaculated.

Question 12.
Describe the placenta in a woman.
Answer:
Placenta :
After implantation, finger – like projections appear on the trophoblast called chorionic villi which are surrounded by the uterine tissue. The chorionic villi and uterine tissue become interdigitated with each other and jointly form a structural and functional unit called placenta between the developing embryo (foetus) and the mother. The maternal and foetal blood do not mix with each other. They are separated by the placental membranes.

The placenta consists of two essential portions :
a maternal part of the placenta derived from the endometrium of the uterus, and foetal membranes of the foetal part of the placenta. The maternal components of the placenta are : Uterine epithelium, Uterine connective tissue and Uterine capillary endothelium. The foetal components of the placenta are foetal capillary endothelium, foetal connective tissue and foetal chorionic epithelium.

The placenta of humans is called chorioallantoic placenta as allantois also fuses with he chorion in the nrocess of vascularisation. Placenta is discoidal as the villi are restricted to the dorsal surface of the blastodisc. Placenta is haemochorial as the maternal blood comes into direct contact with the foetal chorion. During parturition the placenta is cast of with loss of embryonic membranes and the encapsulating maternal tissues (decidua) causing extensive haemorrhage and thereby bleeding. So, it is also called as deciduate placenta.

Functions of Placenta :
The placenta facilitates the supply of oxygen and nutrients to the embryo and also removal of carbon dioxide and excretory / waste materials produced by the embryo. The placenta is connected to thl embryo through an umbilical cord which helps in the transport of substances to and from the embryo.

Long Answer Type Questions

Question 1.
Describe female reproductive system of a woman with the help of a labelled diagram. [May 2017 (A.P.): Mar. ’15 (A.P. & T.S.)]
Answer:
The female reproductive system consists of a pair of ovaries along with a pair of oviducts, uterus, vagina and the extemaPgenitalia located in the pelvic region. These parts of the system along with a pair of the mammary glajids are integrated structurally and functionally to support the processes of ovulation, fertilization, pregnancy, birth and child care.

Ovaries :
Ovaries are the primary female sex organs that produce the female gametes (ova) and several steroid hormones (ovarian hormones). A pair of ovaries is located one on each side of the lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of the abdominal cavity is known as the mesovarium.

The ovaries are covered on the outside by a layer of simple cuboidal epithelium called germinal (ovarian) epithelium. This is actually the visceral peritoneum that envelops the ovaries. Underneath this layer there is a dense connective tissue capsule, the tunica albuginea. The ovarian stroma is distinctly divided into an outer cortex and an inner medulla. The cortex appears more dense and granular due to the presence of numerous ovarian follicle in various stages of development. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels, and nerve fibers.

Fallopian tubes (Oviducts) :
Each fallopian tube extends from the periphery of each ovary to the uterus, and it bears a funnel shaped infundibulum. The edges of the infundibulum possess finger like projections called fimbriae, which help in collection of the ovum after ‘ovulation1. The infundibulum leads to a wider part of the oviduct called ampulla. The last part of the oviduct, isthmus has a narrow lumen and it joins the uterus. Fallopian tube is the site of fertilization. It conducts the ovum or zygote towards the uterus by peristalsis. The fallopian tube is attached to the abdominal wall by a peritoneal fold called mesosalpinx.

Uterus :
The uterus is single and it is also called womb. It is a large, muscular, highly vascular and inverted pear shaped structure present‘in the pelvis between the bladder and the rectum. The uterus is connected to the abdominal wall by the peritoneal fold called mesometrium. The lower, narrow part through which the uterus opens into the vagina is called the cervix. The cavity of the cervix is called cervical canal which along with vagina forms the birth canal.

The wall of the uterus has three layers of tissue. The external thin membranous perimetrium, the middle thick layer of smooth muscle called myometrium and inner glandular lining layer called endometrium. The endometrium undergoes cyclic changes during menstrual cycle while the myometrium exhibits strong contractions during parturition.

Vagina :
The vagina is a large, median, fibro – muscular tube that extends from the cervix to the vestibule (the space between the labia minora). It is lined by non – keratinised stratified squamous epithelium. It is highly vascular, and opens into the vestibule by the vaginal orifice.

Vulva :
The term vulva (vulva – to wrap around) or pudendum refers to the external genitals of the female. The vestibule has two apertures – the upper external urethral orifice of the urethra and the lower vaginal orifice of vagina. Vaginal orifice is often covered partially by a membrane called hymen which is a mucous membrane. Vestibule is bound by two pairs of fleshy folds of tissue called labia minora (inner) and larger pair called labia majora (outer). Clitoris is a sensitive , erectile structure, which lies at the upper junction of the two labia minora above the urethral opening. The clitoris is homologous to the penis of a male as both are supported by corpora cavernosa internally. There is a cushion of fatty tissue covered by skin and public hair present above the labia majora. It is known as mons pubis.

Accessory reproductive glands of female :
These glands include Bartholin’s glands, Skene’s glands and Mammary glands.

Bartholin’s glands :
The Bartholin’s glands (Greater vestibular glands) are two glands located slightly posterior and to the left and right of the opening of the vagina. They secrete mucus to lubricate the vagina and are homologous to the bulbourethral glands of the male reproductive system.

Skene’s glands :
The Skene’s glands (Lesser vestibular glands) are located on the anterior wall of the vagina, around the lower end of the urethra. They secrete a lubricating fluid when stimulated. The Skene’s glands are homologous to the prostate glands, of the male reproductive system.

Mammary glands :
A functional mammary gland is characteristic of all female mammals. The mammary glands are paired structures ( breasts) that contain glandular tissue and variable amount of fat. The glandular tissue of each breast is divided into 15-20 mammary lobes containing clusters of cells called alveoli. The cells of the alveoli secrete milk, which is stored in the cavities (lumens) of the alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous duct through which milk is sucked out by the baby.

Question 2.
Describe the male reproductive system of a man. Draw a labelled diagram of it. [Mar. 2020, 2019, 18,’17 (A.P.); May/.June; Mar. 14]
Answer:
The Male Reproductive System :
The male reproductive system (male genital system) consists of a number of sex organs that are a part of the human reproductive process. The sex organs which are located in the pelvic region include a pair of testes (sing : testis) along with accessory ducts, glands and the external genitalia.

Testes :
The testes (testicles) are a pair of oval pinkish male primary sex organs suspended outside the abdominal cavity with in a pouch called scrotum. The scrotum helps in maintaining the low temperature of the testes (2 – 2.5°C lower than the normal internal body temperature) necessary for spermatogenesis. The cavity of the scrotal sac is connected to the abdominal cavity through the inguinal canal. Testis is held in position in the scrotum by the gubernaculum, a fibrous cord that connects the testis with the bottom of the scrotum and a spermatic cord, formed by the vas deferens, nerves, blood vessels and other tissues that run from the abdomen down to each testicle, through the inguinal canal.

Each testis is enclosed in a fibrous envelope, the tunica albuginea, which extends inward to form septa that partition the testis into lobules. There are about 250 testicular lobules in each testis. Each lobule contains 1 to 3 highly coiled seminiferous tubules. A pouch of serous membrane (peritoneal layer) called tunica vaginalis covers the testis.

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH. The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or Leydig cells.

Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoa from the seminiferous tubules to the vasa efferentia).

Epididymis :
The vasa efferentia leave the testis and open into a narrow, tightly coiled tube called epididymis located along the posterior surface of each testis. The epididymis provides a storage space for the sperms and gives the Sperms time to mature. It is differentiated into three regions – caput epididymis, corpus epididymis and cauda epididymis. The caput epididymis receives spermatozoa via the vasa efferentia t)f the mediastinum testis (a mass of connective tissue at the back of the testis that encloses the rate testis).

Vasa deferentia :
The vas deferens or ductus deferens is a long, narrow, muscular tube. The mucosa of the ductus deferens consists of pseudostratified columnar epithelium and lamina propria (areolar connective tissue). It starts from the tail of the epididymis, passes through the inguinal canal into the abdomen and loops over the urinary bladder. It receives a duct from the seminal vesicle. The vas deferens and the duct of the seminal vesicle unite to form a short ejaculatory duct / ductus ejaculatorius. The two ejaculatory ducts, carrying spermatozoa and the fluid secreted by the seminal vesicles, converge in the centre of the prostate and open into the urethra, which transports the sperms to outside.

Urethra :
In males, the urethra is the shared terminal duct of the reproductive and urinary systems. The urethra originates from the urinary bladder and extends through the penis to its external opening called urethral meatus. The urethra provides an exit for urine as well as semen during ejaculation.

Penis :
The penis and the scrotum constitute the male external genitalia. The penis serves as urinal duct and also intromittent organ that transfers spermatozoa to the vagina of a female. The human penis is made up of three columns of tissue; two upper corpora cavernosa on the dorsal aspect and one corpus spongiosum on the ventral side. Skin and a subcutaneous layer enclose all three columns, which consist of special tissue that helps in erection of the penis to facilitate insemination. The enlarged and bulbous end of penis called glans penis is covered by a loose fold of skin (foreskin) called prepuce. The urethra traverses the corpus spongiosum, and its opening lies at the tip of the glans penis (urethral meatus).

Male accessory genital glands :
The male accessory glands include paired seminal vesicles, a prostate and bulbourethral glands.

Seminal vesicles :
The seminal vesicles are a pair of simple tubular glands present postero- inferior to the urinary bladder in the pelvis. Each seminal vesicle opens into the corresponding vas deferens, as the vas deferens enters the prostate gland. The secretion of the seminal vesicles constitutes about 60 percent of the volume of seminal fluid. It is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorus, potassium, and prostaglandins. Once this fluid joins the sperm in the ejaculatory duct, fructose acts as the main energy source for the sperm outside the body. Prostaglandins are believed to aid fertilization by causing the mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of the sperm towards the ovum with peristaltic contractions of the uterus and fallopian tubes.

Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland surrounds the prostatic urethra, and sends its secretions through several prostatic ducts. In man, the prostate contributes 15 – 30 percent of the semen. The fluid from the prostate is clear and slightly acidic. The prostatic secretion ‘activates’ the spermatozoa and provides nutrition.

Bulbourethral Glands :
Bulbourethral glands, also called Cowper’s glands, are located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen during the process of ejaculation. The fluid secreted by these glands lubricates the urethra. It is also thought to function as a ‘flushing agent’ that washes out the acidic urinary residues that remain in the urethra, before the semen is ejaculated.

Question 3.
Write an essay on different events that occur during development of a human.
Answer:
Cleavage :
The first phase of embryonic development, the cleavage is holoblastic (because of microlecithal condition of egg), radial, indeterminate and unequal. The mitotic division (cleavage) starts as the zygote moves through the isthmus of the oviduct towards the uterus. The daughter cells are called blastomeres.

Morula :
The embryo with 8 -16 blastomeres looks like a ‘mulberry’ and is called the morula. Morula is a solid mass of cells. The morula develops in the fallopian tube and reaches the uterus for further development. It is still surrounded by the zona pellucida. Due to unequal cleavage smaller and larger blastomeres are formed. The morula passes through a process called compaction. Now the embryo has a superficial flat cell layer and inner cell mass. The outer/supeficial layer becomes the trophoblast or trophectoderm. The trophoblast serves to attach the embryo to the uterine wall by forming trophoblastic villi. The inner cell mass constitutes formative cells. The inner cell mass gives rise to the embryo proper and is, therefore, also called the embryoblast. This is the first sign of cell differentiation (cells becoming different) in the human embryo.

Blastocyst :
Some fluid now passes into the morula from the uterine cavity, and partially separates the cells of the inner cell mass from those of the trophoblast. As the quantity of the fluid increases, the morula acquires the shape of a cyst. The cells of the trophoblast become flattened, and the inner cell mass comes to be attached to the inner side of the trophoblast on one side only. The morula now becomes a blastocyst.

The cavity of the blastocyst is the blastocoel or segmentation cavity or primary body cavity. The side of the blastocyst to which the inner cell mass is attached is called the embryonic or animal pole, while the opposite side is the bembryonic •pole. The cells of the trophoblast above the region of inner cell mass are called cells of Rauber.

Implantation :
The zona pellucida present around the blastocyst gradually disappears and the cells of the trophoblast stick to the uterine endometrium. The trophoblast invades the endometrium of the uterus. The blastocyst gets implanted into the uterine mucosa till the whole of it comes to lie within the thickness of the endometrium. This is called interstitial implantation. In humans, implantation begins on the 6th day after fertilisation. The process of implantation is aided by proteolitic enzymes produced by the trophoblast. The uterine mucosa also aids the process.

After the implantation of the embryo, the uterine endometrium is differentiated into what is called decidua. The portion of the decidua where the placenta is to be formed (i.e., deep to the developing blastocyst) is called the decidua basalis. The part of the decidua that separates the embryo from the uterine lumen is called the decidua capsularis. The part lining the rest of the uterine cavity is called the decidua parietalis. At the end of pregnancy the decidua is shed off, along with the placenta and membranes.

Formation of Bilaminar Embryonic Disc :
Implantation of the blastocyst is completed by the end of the second week. The inner cell mass forms into a disc called embryonic disc. Soon, the ‘cells of Rauber’ disappear exposing the embryonic disc. From the lower part of inner embryonic disc, some cells get separated by delamination and eventually form a layer of cells on the inner surface of the embryonic disc i.e., on the surface facing the cavity. This cell layer develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the embryonic disc is called epiblast. Now the embryonic disc is called bilaminar embryonic disc.

The cells of the hypoblast increase in number and spread along the inner surface of the trophoblast. This hypoblast layer below the trophoblast finally encloses a cavity called yolk sac or umbilical vesicle. Meanwhile, the thickness of the embryonic disc increases towards the caudal end. Gradually the embryonic disc becomes oval.

Gastrulation :
Gastrulation involves proliferation, differentiation and movement of cells within the embryo. Along the longitudinal axis of the embryonic disc, a primitive streak is formed. A longitudinal furrow known as primitive groove forms along the middle of the primitive streak. On either side of it are the primitive folds. Anteriorly primitive streak has a shallow primitive pit. The anterior end of the primitive streak becomes thickened. This thickened part of the streak is called the primitive knot or primitive node or Hensen’s node.

Trilaminar Embryo – Formation of Primary Germ Layers :
Ingression, the future endodermal cells from the epiblast, replaces the hypoblast and forms the endoderm of the embryo. The future mesodermal cells converge towards the primitive folds, involute through the primitive groove and reach between epiblast and endoderm. The remaining epiblast is now known as the ectoderm. This invasion of the epiblast cells into the space between the epiblast and hypoblast is called gastrulation. Thus, the ectoderm, mesoderm and endoderm are all derived from the epiblast. The process of gastrulation converts the bilaminar embryonic disc into a trilaminar embryonic disc.

Extraembryonic Membranes :
During the development of the human embryo, as in all other amniotes four extraembryonic or foetal membranes are formed. They are chorion, amnion, allantois, and yolk sac. From the blastodisc, amniotic folds called head fold, tail fold and lateral fold are developed as somatopleure. As the folds fuse they are differentiated into outer chorion and inner amnion. Between the amnion and the embryo, there is an amniotic cavity filled with amniotic fluid. Amnion protects the embryo as the amniotic fluid acts as a shock absorber and also prevents the embryo form desiccation. The chorion develops a rich supply of blood vessels and forms an intimate association with the endometrium of the uterus.

Allantois and yolk sac are derived from the splanchnopleure. Allantois is formed from the hind gut as an evagination. It stores the waste materials. Allantois and chorion are fused to form chorio allantoic membrane which constitutes the placenta. Yolk sac encloses a fluid filled cavity. It is connected to the mid gut. It has no nutritive role.

After gastrulation is complete and any extra embryonic membranes are formed, the next stage of embryonic development begins with organ formation. Organogenesis: During organogenesis, regions of the three embryonic germ layers develop into the rudiments of organs. Whereas gastrulation involves mass movements of cells, organogenesis involves more localized changes.

Formation of the Notochord and Neural Tube :
The chorda mesodermal cells converge and involute through the Hensen’s node and extend forwards as notochordal process. This is later transformed into a solid rod – the notochord, the embryonic axial skeleton which is replaced by the vertebral column. The notochordal mesoderm induces the overlying ectodermal cells to form the neural plate. This is a good example of induction. At first the neural plate is oval but later elongates oves the underlying notochord along the longitudinal axis of the embryo. The plate invaginates towards the notochord to form a neural groove, which deepens progressively to form a tube by fusion of the lateral neural folds. The process of formation of neural tube is referred to as neurulation.

Differentiation of Mesoderm and Formation of Coelom :
The intra embryonic mesoderm spreads in all directions between the outer ectoderm and inner endoderm. The longitudinal column of mesoderm adjacent to the notochord and neural tube on either side is called epimere. The mesoderm around the gut is the hypomere. The mesoderm in between these two is the mesomere. The epimere becomes segmented into cubical blocks called somites or metameres. Each somite differentiates into myotome, sclerotome and dermatome.

The sclerotome forms the vertebral column. The dermatome forms the dermis of the skin and other connective tissues. The myotome forms the voluntary muscles of the body. The mesomere forms the urinogenital organs and their ducts. The hypomere splits into outer somatic and inner splanchnic mesodermal layers. Intra embryonic coelom is formed between these two layers. It gives rise to pericardial, pleural, peritoneal cavities etc.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System

Very Short Answer Type Questions

Question 1.
Define the terms immunity and immune system.
Answer:
The over all ability of an individual to fight against the disease causing organisms is called immunity. The network of organs, cells and proteins that protect the body from harmful, infectious agents such as bacteria, viruses, animal parasites, fungi etc., is called the immune system.

Question 2.
Define the non-specific lines of defence in the body.
Answer:
The inborn resistance to diseases, possessed by all the living organisms is called innate immunity. It is a non specific type of defence and does not depend on prior contact with the micro – organisms. This is executed by providing different types of barriers.

Question 3.
Differentiate between mature B – cells and functional B – cells.
Answer:
Mature B – cells synthesize various types of antibodies which are displayed on their membrane surfaces. As these antibodies can take antigens, the mature B – cells are also called immuno – competent B cells.

These mature immuno – competent B cells reach the secondary lymphoid organs and develop into functional immune cells which later differentiate into long lived memory cells and effector plasma cells.

Question 4.
Write the names of any four mononuclear phagocytes. [March 2020, May 2017 (A.P.)]
Answer:
Mono nuclear phagocytes are 1) histocytes 2) Kupffer cells 3) microglia 4) osteoclasts 5) synovial cells.

Question 5.
What are complement proteins? [March 2017 (A.P.)]
Answer:
These are a group of inactive plasma proteins and cell surface proteins when activated, they form a membrane attack complex (MAC). Complement proteins and their activities are together called complement system.

Question 6.
“Colostrum is very much essential for the new born infants”. Justify. [May 2017 (A.P.)]
Answer:
The colostrum secreted by the mother during the initial days of lactation has abundant Ig A antibodies, to protect the infant. It is called natural passive acquired immunity.

Question 7.
Differentiate between perforins and granzymes. [Mar. ’17 (A.P.); Mar. ’15 (T.S.)]
Answer:
Cytotoxic T lymphocytes attach to the infected cells and release certain enzymes called perforins and granzymes.

Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activate certain proteins which help in the destruction of the infected cell.

Question 8.
Explain the mechanism of vaccination or immunization.
Answer:
The principle of vaccination or immunization is based on the property of the memory of the immune system. During vaccination, inactivated pathogens or antigenic proteins of pathogen are introduced into the body of host. They initiate the production of appropriate antibodies in the host and also generate memory B cells and memory T cells.

Question 9.
Mention various types of immunological disorders.
Answer:
Various types of immunological disorders are

1. Acquired immuno deficiency syndrome (AIDS)
2. Hyper sensitivity disorders (Allergies)
3. Auto immune disorders (Graves disease, Rheumatoid arthritis)
4. Graft rejections.

Question 10.
“More and more people in metrocities of India are prone to allergies Justify.
Answer:
More and more children in metro cities of India suffer front allergies leading to asthmatic attacks due to environmental pollutants. This could be mostly due to exposure to various types of pollutants in the urban atmosphere.

Question 11.
What are autoimmune disorders? Give any two examples.
Answer:
In some cases our immune system fails to recognise some of our own body proteins and treats them as foreign antigens that results in attacks on our own tissues. This leads to some very serious diseases collectively known as auto – immune diseases, e.g: Graves’ disease, Rheumatoid arthritis.

Question 12.
How can the graft rejections be avoided in patients?
Answer:
Graft reactions can be avoided in patients by tissue matching and blood group matching. Even after this, the patient has to take immuno – suppressant drugs throughout their life.

Short Answer Type Questions

Question 1.
Write short notes on B – cells. [March 2020, March 2014; May/June ’14]
Answer:
B – cells (B – Lymphocytes): The lymphocytes capable of producing antibodies and can capture circulating antigens are called B – cells. They are produced from the ‘stem cells’ in the bone marrow of adult mammals, liver of foetus and bursa of Fabricius in birds. Mature B – cells synthesize various types of antibodies which are displayed on their membrane surfaces. As these antibodies can take antigens, the mature B – cells are also called immuno – competent B – cells.

These mature immuno – competent B – cells reach the secondary lymphoid organs and develop into functional immune cells which later differentiate into ‘long lived’ memory cells and ‘effector’ plasma cells. The plasma cells produce antibodies specific to the antigen to which they are exposed. Memory cells store information about the specific antigens and show quick response, when the same type of antigen invades the body later.

Question 2.
Write short notes on immunoglobulins. [March 2019 (A.P.) March 2015 (T.S.)]
Answer:

Antibodies (Immunoglobulins) :
Whenever pathogens enter our body, the B – lymphocytes produce an army of proteins called antibodies to fight with them. They are highly specialized for binding with specific antigens. The part of an antibody that recognizes an antigen is called the paratope (antigen binding site). Based on their mobility, antibodies are of two types, namely circulating or free antibodies and surface antibodies. The circulating or free antibodies are present in the body fluids whereas the surface antibodies are present on the surface of the mature B – cells as well as the memory cells.

Structure :
The basic structure of an antibody was proposed by Rodney Porter. It is a Y shaped molecule with four polypeptide chains of which two are long, identical heavy chains (H) and two are small, identical light chains (L). Hence, an antibody is represented as H₂L₂. The two chains are linked by disulphide bonds. One end of the antibody molecule is called F_ab end (Fragmentantigen binding) and the other end is called F_c end (Fragment – crystallizable or Fragment – cell binding). Based on the structure, the antibodies are of five types, namely IgD, IgE, IgG, IgA and IgM.

Question 3.
Describe various types of barriers of innate immunity.
Answer:
The inborn resistance to diseases, possessed by all the living organisms is called innate immunity. It is a non – specific type of defence and does not depend on prior contact with the micro – organisms. This is executed by providing different types of barriers like :
a) Physical barriers :
Skin and mucous membranes are the main physical barriers. Skin prevents the entry of micro – organisms whereas the mucus membranes help in trapping the microbes entering our body.

b) Physiological barriers :
Secretions of the body like HCI in the stomach, saliva in the mouth, tears from the eyes are the main physiological barriers against microbes.

c) Cellular barriers :
Certain types of cells like polymorpho – nuclear leukocytes (PMN – neutrophils), monocytes and natural killer cells in the blood as well as macrophages in the tissues are the main cellular barriers. They phagocytose and destroy the microbes.

d) Cytokine barriers :
The cytokines secreted by the immune cells are involved in differentiation of the cells of immune system and protect the non – infected cells from further infection.

Question 4.
Explain the mechanism of humoral immunity.
Answer:

Mechanism of Humoral immunity (HI) :
In the secondary lymphoid organs, the free antigens bind to the F_ab end of the antibodies that are present on the surface of mature B – cells. They engulf and process the antigens. Then they display the antigenic fragments on their membrane with the help of Class-IIMHC molecules. Then appropriate T_H cells recognize them and interact with the antigen – MHC – II complex and release a type of interleukin, which stimulates the B – cells to proliferate and differentiate into memory cells and plasma cells. The plasma cells release specific antibodies into the plasma or extra cellular fluids. These antibodies help in opsonising and immobolising the bacteria, neutralising and cross linking of antigens leading to agglutination of insoluble antigens and precipitation of soluble antigens. They also activate the phagocytes and complement system.

Question 5.
Explain the mechanism of cell mediated immunity.
Answer:
Mechanism of cell mediated immunity (CMI) :
Antigen presenting cells process the exogenous antigens whereas the altered self – cells process endogenous antigens. Then, the processed antigenic fragments are displayed on their (APCs or ASCs) membranes. They are recognised byT -cells. Binding of T -cells to the APCs or ASCs causes the production of activated T cells and memory T cells. The activated TH cells secrete various types of interleukins which transform activated Tc cells into effector cytotoxic T – Lymphocytes (CTLs / Killer cells). They attach to the infected cells and release certain enzymes called perforins and granzymes.

Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activate certain proteins (e.g. caspases) which help irt the destruction of the infected cell (apoptosis). The NK cells are similar in their action to CTLs. However NK cells destroy the infected cells in an antibody independent manner where as the CTLs destroy the infected cells in an antibody dependent manner.

Question 6.
Explain the mechanism by which HIV multiplies and leads td AIDS.
Answer:

After getting into the body of a person, the HIV enters the TH cells, macrophages or dendritic cells. In these cells the ssRNA of HIV synthesizes a DNA strand ‘complementary’ to the viral RNA, using the enzyme reverse transcriptase. The reverse transcriptase also catalyses the formation of the second DNA strand ‘complementary’ to the first strand forming the double stranded viral DNA. This viral DNA gets incorporated into the DNA of the host cells DNA by a viral enzyme (integrase) and it is in the form of a ‘provirus’. Transcription of DNA results in the production of RNA, which can act as the ‘genome’ for the new viruses or it can be translated into viral proteins. The various components of the viral particles are ‘assembled’ and the HIV are produced.

The infected human cells continue to produce virus particles and in this way they act like HIV generating factories. New viruses ‘bud off’ from the host cell. This leads to a progressive decrease in the number of TH cells in the body of an infected person leading to the immunodeficiency in him. Even though HIV attacks cells with CD4 marker, for reasons not known, only TH cells are destroyed and not the ‘macrophages’. The gp 120 molecules on the surface of HIV attach to CD4 receptors of human cells, mostly the TH cells (gpl20 fits the CD4 marker). Attack on certain types of cells/ tissues only by viruses such as HIV is refered to as ’tissue tropism’.