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Telangana TSBIE TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Type Questions

Question 1.
Distinguish between heat and temperature. [TS Mar. ’15]
Answer:
Differences between heat and temperature :

Question 2.
Explain triple point of water.
Answer:
The temperature of a substance remains constant during its change of state.

A graph plotted between temperature (T) and pressure (P) of a substance during change of state is called “phase diagram”.

In phase diagram of water, the P – T plane is divided into three regions.

The line ‘AO’ is called fusion curve. It gives equilibrium temperature and pressure between solid and liquid states.

The line ‘CO’ is called vaporisation curve. It gives equilibrium temperature and pressure between liquid and vapour states.

The line ‘BO’ is called sublimation curve. It gives the relation between of temperature and pressure between solid and vapour states.

Triple Point:
At point ‘O’ the curves AO, BO and CO will intersect.

It gives the temperature and pressure at which solid, liquid and vapour states of water are in equilibrium.

Coordinate of triple point of water a temperature = 273.16 K, pressure = 0.006 atmos (or) 611 pascals.

Question 3.
What are the lower and upper fixing points in Celsius and Fahrenheit scales? [TS Mar. ’16; AP Mar. ’19, ’18 ’16, May ’14]
Answer:
Centigrade (Celsius) scale of temperature:
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

Question 4.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale?
Answer:
Yes. Coefficients of expansion α, α_a and α_v are not same in Celsius scale and in Fahrenheit scale. In Fahrenheit scale values of α, α_a and α_v are less than those in Celsius scale.

Since magnitude of 1°C > magnitude of 1°F this change takes place. The values of Celsius scale are 1.8 times more than the values in Fahrenheit scale.
α_F = 5/9 α_c

Question 5.
Can a substance contact on heating? Give an example. [AP Mar. ’18, ’16, May ’16; TS May, ’18, ’16]
Answer:
Yes. Some substances will contact on heating. Ex: Leather, rubber, cast Iron type metal.

Question 6.
Why gaps are left between rails on a railway track? [TS Mar. ’19; AP Mar. ’19, ’17, ’16, ’09; May ’16; June ’15]
Answer:
To allow the linear expansion rails.

In summer temperature of atmosphere increases so rails will expand. If no gap is given between rails then the rails will bend it leads to accidents. If gap is given the rails will expand into that gap and that track is safe.

Question 7.
Why do liquids have no linear and areal expansions? [TS Mar. ’19]
Answer:
Liquids have only volume expansion. No linear expansion or areal expansion. Because liquids does not have any independent shape they must be taken in a container. So we will consider only volume of liquid.

Question 8.
What is latent heat of fusion? [AP & TS May ’17]
Answer:
Latent heat of fusion (melting):
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from solid to liquid or from liquid to solid.

Question 9.
What is latent heat of vapourisation? [AP Mar. ’13]
Answer:
Latent heat of vapourisation :
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from liquid to vapour or from vapour to liquid state.

Question 10.
Why are utensils coated black? Why is the bottom of the utensils made of copper? [AP May ’18; TS Mar. ’18]
Answer:
Lower portion of the utensils is in contact with fire. Black bodies are good heat absorbers. So, a black bottom will absorb more heat.

Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

Question 11.
What is triple point of Water? Mention the values of temperature and pressure at triple point of water. [TS June ’15]
Answer:
Triple point:
The temperature and pressure where a substance can coexist in all its three states is called the “triple point”.

i.e., The substance will exist as a solid, as liquid and as vapour at that particular temperature and pressure.

For water the triple point is at a temperature of 273.16 K and at a pressure of 6.11 × 10^-3 atmospheres or nearly 610 pascals.

Question 12.
State Boyle’s law and Charles law. [AP June 15; TS Mar. 15]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) ⇒ PV = constant = K.

Charles Law:
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T ⇒ \(\frac{V}{T}\) = K (constant)

Question 13.
State Wein’s displacement law. [AP Mar. ’17]
Answer:
Wein’s Displacement Law :
The wavelength (Ain) of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature (T) of the black body.
i.e., λ_m ∝ \(\frac{1}{T}\) (or) λ_m = \(\frac{b}{P}\)
where ‘b’ is called ‘Wein’s constant”.

Question 14.
Ventilators are provided in rooms just below the roof. Why?
Answer:
Density of hot air is less. So in a room hot air goes to top layers i.e., nearer to the roof.

When ventilators are provided nearer to the roof hot air will escape easily from room. So we feel that the room is cool and circulation of air will become easy.

Question 15.
Does a body radiate heat at 0 K? Does it radiate heat at 0°C?
Answer:
According to Precost’s theory, every body above zero Kelvin will radiate heat energy to the surroundings.
So, i) A body at O’ Kelvin does not radiate heat energy.
ii) A body at 0°C i.e., at 273Kwill radiate heat energy.

Question 16.
State the different modes of transmission of heat. Which of these modes require medium? [TS May ’18]
Answer:
Transmission of heat is of three types.
1) Conduction 2) Convection 3) Radiation

For propagation of heat energy medium is required in case of conduction and convection.

Question 17.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
“The coefficient of thermal conductivity”
(K) it is the quantity of heat flowing normally per second through unit area of the substance per unit temperature gradient.

‘Temperature gradient” is defined as the change in temperature along the conductor per unit length.

Temperature gradient
Change in temperature

Question 18.
Define emissive power and emissivity.
Answer:
Emissive power:
The emissive power of a body is defined as the energy radiated by the body per second per unit area at a given temperature and wavelength.

Emissivity:
Emissivity is defined as the ratio of the emissive power of a body to that of a black body at the same temperature.

Question 19.
Is there any substance available in nature that contracts on heating? If so, give an example. [TS May ’16]
Answer:
Yes. Some substances will contact on heating.
Ex: Leather, rubber, cast Iron type metal.

Question 20.
What is greenhouse effect? Explain global warming. [AP Mar. ’15, ’13; TS Mar. & May ’16]
Answer:
Green house effect: Earth will absorb heat radiation and reradiate heat energy of longer wavelength. This longer wave length heat radiation is reflected back to earth due to green house gases such as Carbon dioxide [CO₂], Methane (CH₄) Chloroflurocarbons, Ozone (O₃), etc. As a result temperature of earth’s atmosphere is gradually increasing. This is known as “green house effect.”

Global warming:
Earth receives heat energy during day time from sun. It reradiates heat energy in the form of longer electromagnetic waves.

But due to presence of green house gases the longer electromagnetic waves were reflected back to earth. As a result temperature of earth’s atmosphere is gradually increasing.

This process will increase with the increased content of green house gases in atmosphere. As a result temperature of earth’s atmosphere increases gradually.

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body?
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of A and A + dA to the total energy flux following on the body.
Absorptive power,

Question 22.
State Newton’s law of cooling. [AP Mar. ’18, ’16, May ’18, ’17, June ’15; TS Mar. ’18, TS May ’16]
Answer:
Newton’s Law of cooling states that the rate of loss of heat of a hot body is directly pro-portional to the difference in temperature between the body and its surroundings pro-vided the difference in temperatures is small and the nature of the radiating surface remains same.

Where k is the proportionality constant

Question 23.
State the conditions under which Newton’s law of cooling is applicable. [AP May ’16; TS June ’15]
Answer:
Newton’s law of cooling is applicable

1. loss of heat is negligible by conduction and only when it is due to convection.
2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
3. temperature of the body is uniformly distributed over it.
4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

Question 24.
The roofs of buildings are often painted white during summer. Why? [TS Mar. ’17, ’15; AP May ’16]
Answer:
When roofs of buildings are coated white we will feel relatively cold during summer.

Absorptive power of white surface is less. So roofs coated white will absorb less heat energy. So less quantity of heat is transmitted into the house. So we feel less hot or cold inside the house.

Question 25.
What is thermal expansion? [TS Mar. ’16]
Answer:
The increase in interatomic distance due to thermal energy is called “thermal expansion”.

As a result the length solids or volume of liquids or pressure of gases will increase.

Question 26.
Why is it easier to perform the skating on the snow? [TS Mar. ’16]
Answer:
Due to increase of pressure melting point decreases, So it is easier to perform the skating on the snow.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Celsius (Centigrade) scale of temperature :
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

The interval between lower limit and upper limit [100 – 0 = 100] is divided into 100 equal parts and each part is called 1°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

The interval between upper fixed point and lower fixed point (212 -32 = 180) is divided into 180 equal parts and each part is called 1°F.

Relation between Fahrenheit and Celsius scale of temperatures:
In both scales, lower limit and upper limit are same. The only change is in numerical values of lower and upper limits.

In Fahrenheit lower limit = 32, upper limit = 212, difference of limits = 180

In Celsius scale lower limit = 0, upper limit = 100, difference of limits = 100

C = Temperature in Celsius scale.
F = Temperature in Fahrenheit scale.

Question 2.
Two identical rectangular strips one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating?
Answer:
When two dissimilar metals say copper and steel are riveted together that arrangement is called “bimetallic strip”.

When a bimetallic strip is heated copper strip will expand more than steel due to more expansion coefficient.

Since they are riveted they must expand as a common piece. As a result bimetallic strip will bend in the form of an arc. For the metallic strip with high a its length is more so it is on the outer side. For the strip with less a its length is less. It will be at the inner side of the arc.

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why? [TS Mar. ’19, ’17]
Answer:
In summer due to increase in temperature of atmosphere length of pendulum will increase.

Time period of pendulum T = 2π\(\sqrt{\frac{l}{g}}\)

T ∝ √l. So it will make less number of oscillations per day. So clock will run slowly in summer.

In winter temperature of atmosphere decreases. So length of pendulum decreases.

Hence time period of oscillation will also decrease. As a result, pendulum will make more oscillations per day so clocks will run fast in winter.

Question 4.
In what way is the anomalous behaviour of water advantageous to aquatic animals? [AP Mar. ’18, 17, 14; May 18. 17, 14; TS May ’18]
Answer:
In cold countries and at polar region temperature falls below 0°C at winter. So surface of water will be frozen. Due to anomalous expansion of water even though the surface of lakes, and sea are frozen water will exist at bottom layers at 4°C.

Different layers in between ice and bottom will have different temperatures like 1°C, 2°C or 3°C. In these layers, aquatic animals are able to survive even in winter.

Anomalous expansion of water helps for the survival of aquatic life at polar region and in cold countries.

Question 5.
Explain conduction, convection and radiation with examples. [TS Mar. ’18, ’16, ’15, June ’15; AP Mar. ’19, ’15, May, ’16]
Answer:
Conduction :
It is a mode of transfer of heat from one part of the body to another, from particle to particle in the direction of fall of temperature without any actual movement of the heated particles.
Ex: When one end of a metal rod is heated, its other end becomes hot. Here, the heat goes from hot end of the metal rod towards cold end, by conduction.

Convection :
It is a mode of transfer of heat from one part of the medium to another part by the actual movement of the heated particles of the medium.
Ex : Seabreeze, Tradewind, etc.

Radiation :
It is a mode of transfer of heat from the source to the receiver without any actual movement of source or receiver and also without heating the intervening medium.
Ex : Heat from sun comes to us through radiation. On standing near fire, we feel hot as heat comes to us through radiation.

Long Answer Questions

Question 1.
Explain thermal conductivity and coefficient of thermal conductivity.
A copper bar of thermal conductivity 401 W (mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross-sectional area is 1.0 × 10^-6 m^-2. What is the rate of heat conduction along the bar?
Answer:
The ability to conduct heat in solids is called ‘Thermal conductivity.”

Consider a bar with rectangular cross-section as shown in the figure. The faces ABCD and EFGH are maintained at θ₁ and θ₂ respectively (θ₁ > θ₂). Heat passes from one end to the other.

The amount of heat conducted (Q) depends on,

1. Amount of heat conducted Q is proportional to area of cross-section A perpendicular to flow.
   ∴ Q ∝ A ………. (1)
2. is proportional to temperature difference between the two ends.
   ∴ Q ∝ (θ₂ – θ₁) …………. (2)
3. is proportional to the time of flow.
   Q ∝ t ………. (3) and
4. is inversely proportional to the length of the conductor.
   Q ∝ \(\frac{1}{l}\) …………. (4)

where k = constant called coefficient of thermal conductivity.

Coefficient of thermal conductivity (k) :
It is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient.
S.I. Unit w m k^-1
Dimensional formula = [ M¹L¹T^-3θ^-1]

Problem:
Thermal conductivity of copper,
K_c = 401 W/m-k
Temperature at one end, θ₂ = 104°C
Temperature of 2nd end, θ₁ = 24°C
Length of copper bar, l = 0.1 m; Area,
A = 1.0 × 10^-6 m^-2
Rate of conduction,

Question 2.
State the explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable.
A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings. [TS May ’17, ’16; AP May ’13]
Answer:
Newtons’ Law of cooling :
The rate of loss of heat of the body is directly proportional to the difference of temperature of the body and the surroundings.

Explanation :
The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
– \(\frac{dQ}{dt}\) = k (T₂ – T₁) (sign indicates loss) …….. (1)

where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass ‘m’ and specific heat capacity ‘s’ is at temperature T₂. Let T₁ be the temperature of the surroundings. If the temperature falls by a small amount dT₂ in time dt, then the amount of heat lost is
dQ = ms dT₂
∴ Rate of loss of hfeat is given by

where K = k/ms

Conditions (under which Newton’s law of cooling is applicable):
Newton’s law of cooling is applicable

1. loss of heat is negligible by conduction and only when it is due to convection.
2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
3. temperature of the body is uniformly distributed over it.
4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

PROBLEM :
Let ‘θ_o‘ be the temperature of the surroundings.

In first case :
Initial temperature, θ₁ = 60°C
Final temperature, θ₂ = 50°C
Time of cooling, t = 5 minutes = 5 × 60 = 300s
From Newton’s law of cooling we can write,

In secoend case :
Initial temperature, θ₁ = 60°C
Final temperature, θ₂ = 40°C
Time of cooling, t = 13 minutes = 13 × 60 = 780s
Again from Newton’s law of cooling we can write,

On solving equations (1) and (2) we get, θ_o = 33.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin Fahrenheit scales are same?
Solution:
On the Kelvin and Fahrenheit scales

Question 2.
Find the increase in temperature of aluminium rod if it’s length is to be increased by 1%. (α for aluminium = 25 × 10^-6/°C) [AP Mar. ’15; June ’15]
Solution:
Coefficient of linear expansion of aluminium, α = 25 × 10^-6/°C
We know that percentage increase in length

Question 3.
How much steam at 100°C is to be passed into water of mass 100g at 20°C to raise its temperature by 5°C? (Latent heat of steam is 540 cal / g and specific heat of water is 1 cal / g°C)
Solution:
Latent heat of steam, L_s = 540 cal/g
Specific heat of water, L_w = 1 cal / g°C
Mass of water, m_w = 100g

According to method of mixture or from the principle of calorimetry we can write, Heat lost by steam = heat gained by water
i.e., m_sL_s + m_sS_w(100 – t) = m_wS_w (t – 20)
∴ m_s × 540 + m_s × 1(100 – 25)
⇒ 100 × 1 × (25 – 20)
⇒ 615m_s = 500(or)m_s = \(\frac{500}{615}\) = 0.813 g

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kg K, find the amount of heat absorbed in kJ and kcal. (J = 4.2 kJ/kcal.)
Solution:
Mass of air, m = 2 kg
Change in temperature, ∆T = 313 – 293 = 20K.
Specific heat at constant volume, C_v = 0.718 k.J/kg – K.
Heat mechanical equivalent, J = 4.2 kJ/k.cal.

∴ Heat energy absorbed,
Q = 2 × 0.718 × 10³ × 20. = 28.72 kJ
= 6.838 k calories.

Question 5.
A clock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Solution:
Temperature of correct time, t₁ = 20°C
Loss or gain of time in seconds per day = 8.212 sec.
Final temperature, t₂ = 30°C
∴ ∆t = 30 – 20 = 10
α of pendulum material = ?

In pendulum loss or gain of time in seconds per day = 43,200. α ∆t

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C? [AP May ’13]
Solution:
In first case :
Initial temperature, θ₁ = 60°C
Final temperature, θ₂ = 40°C
Time of cooling, t₁ = 7 minutes
= 7 × 60 = 420s
Temperature of surroundings, θ₀ =10°C
From. Newton’s law of cooling, we can write,

In second case:
Initial temperature, θ₁ = 40°C
Time of cooling, t₂ = 7 minutes = 420s
Again, from Newton’s law of cooling we can write,

on solving equations (1) & (2) we get, 0 = 28°C

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 µ m, what is the temperature of the radiating body? (Wein’s constant = 2.9 × 10^-3 mK)
Solution:
Wavelength corresponding to maximum intensity, λ_max = 2.65 µm = 2.65 × 10^-6 m.
Wein’s constant, b = 2.90 × 10^-3 mK.
From Wein s Law, T = \(\frac{\mathrm{b}}{\lambda_{\mathrm{m}}}=\frac{2.90 \times 10^{-3}}{2.65 \times 10^{-6}}\)
= 1094 K.

Telangana TSBIE TS Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Type Questions

Question 1.
Define average pressure. Mention it’s unit and dimensional formula. Is it a scalar or a vector? [AP Mar. ’17]
Answer:
Average pressure (P_av) :
The normal force acting per unit is called average pressure.
⇒ P_av = \(\frac{F}{A}\)
Units : In SI = Nm^-1
The dimensional formula : ML^-1T^-2
It is a scalar quantity.

Question 2.
Define Viscosity. What are its units and dimensions? [AP May ’16, ’13; TS May ’18, June ’15)
Answer:
Viscosity :
The property of a fluid which opposes the relative motion between the layers is called viscosity.

Units in SI:
Coefficient of viscosity Nm^-2 s (or) Pa – s. (or) Poiseuille
Units in C.G.S : Coefficient of viscosity = poise.
The dimensional formula = ML^-1 T^-1.

Question 3.
What is the principle behind the carburetor of an automobile? [TS Mar. ’18, ’17; AP Mar. ’19, ’15; June ’15]
Answer:
Carburetor of an automobile is based on the principle of “Bernoulli’s theorem”.

Question 4.
What is magnus effect? [AP May ’18, ’17, Mar. ’15; TS Mar. ’19, ’16]
Answer:
Magnus effect :
When a spinning ball is thrown it deviates from its usual path in flight. This effect is called Magnus effect.

Question 5.
Why are drops and bubbles spherical? [AP Mar. ’18, ’17, ’16, ’14, May ’18, ’17, ’16, ’14, ’13; TS May ’18, ’17, ’16]
Answer:
Due to property of surface tension, the surface of liquid behaves like a stretched membrane and has a tendency to acquire minimum surface area. The sphere has minimum surface area when compared to other shapes of same volume.

Therefore, drops and bubbles acquire spherical shape in order to have the minimum surface area.

Question 6.
Give the expression for the excess pressure in a liquid drop. [TS Mar. ’17]
Answer:
Excess of pressure in a liquid drop is,
p = \(\frac{2s}{r}\) where ‘s’ = surface tension and ‘r’ = the radius of the liquid drop.

Question 7.
Give the expression for the «<cess pressure in an air bubble inside the iquid. [AP Mar. ’19]
Answer:
Excess of pressure in an air buble inside the liquid is, P = \(\frac{2S}{R}\)
where S = Surface Tension, R = Radius of air bubble of liquid.

Question 8.
Give the expression for the excess pressure in a soap bubble in air. [TS Mar. ’16]
Answer:
Excess of pressure in a soap bubble is, p = \(\frac{4s}{r}\) where
‘s’ = surface tension and
‘r’ = radius of the drop.

Question 9.
What are water proofing agents and water wetting agents? What do they do?
Answer:
Water proofing agents :
The substances which are used to increase the angle of contact are called “water proofing agents”.

Wetting agents:
The substances which are used to decrease the angle of contact are called “wetting agents”.
Ex: Soaps, detergents and dying substances.

Question 10.
Why water droplets wet the glass surface and does not wet lotus leaf? [TS Mar. ’15]
Answer:
Angle of contact between water drop and glass is less than 90° so water drop will wet glass surface.
Angle of contact between water and lotus leaf is greater than 90°. So water drops cannot wet lotus leaf.

Question 11.
What is angle of contact? [AP May ’14, Mar. 16]
Answer:
Angle of contact: It is the angle between the walls of the container and the tangent drawn over the surface of the liquid. This angle must be. measured in the interior side of the liquid.

Question 12.
Mention any two examples (or) applications that Obey Bernoullis theorem and justify them. [AP Mar. ‘ 18; TS Mar. 15]
Answer:
Applications of Bernoulli’s theorem :

1. Dynamic lift on the wings of an aeroplane
2. Swinging of a spinning cricket ball is a consequence of Bernoulli’s theorem.
3. During cyclones, the roof of thatched houses will fly away. This is a consequence of Bernoulli’s theorem.

Question 13.
When water flows through a pipe, which of the layers moves fastest and slowest? [TS June ’15]
Answer:
When water is flowing through a pipe water layers in contact with bottom layers of pipe will have lowest velocity and water layer just below the top of inner layer of pipe will have highest velocity.

Question 14.
“Terminal velocity is more if surface area of the body is more.” Give reasons in support of your answer.
Answer:
Yes, Terminal velocity of a body is more when surface area of a body is more.

According to Stokes formula, terminal velocity of a smooth spherical body is,

The surface area of a spherical body A = 4πr². So when surface increases, ‘r²’ value increases. Hence from Stoke’s formula, Teminal velocity increases.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer?
Answer:
The atmospheric pressure at any point is equal to the weight of a vertical column of air of unit cross-sectional area extending from that point to the top of tfie earth’s atmosphere.

Determination of atmospheric pressure using Barometer :
A long tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer. The space above the mercury column in the tube contains only mercury vapour whose pressure p is so small that it may be neglected. Otherwise there is a perfect vacuum, which is called Torricellian vacuum.

The pressure inside the column at point A must equal the pressure at point B, which is at the same level.

‘P’ at A = Pressure at B = atmospheric pressure = Pa
Pa = ρgh …………. (1)
where ρ = density of mercury
h = height of mercury column

In this experiment, it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm).

At sea level, atmospheric pressure is the pressure exerted by 0.76 m of mercury column, i.e., h = 0.76 m
ρ = 13.6 × 10³ kg m^-3 And g = 9.8 ms^-2.
∴ Atmospheric pressure, Pa = hρg
= 0.76 × (13.6 × 10³) × 9.8
= 1.013 × 10⁵ Nn^-2 (or) Pa

A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent to 1 mm is called a torr.
1 torr = 133 Pa.

Question 2.
State Dalton’s law of partial pressures. [AP Mar. ’14]
Answer:
Dalton’s law of partial pressures :
For a mixture of non interacting ideal gases at same temperature and volume total pressure in the vessel is the sum of partial pressures of individual gases.
i.e. P = P₁ + P₂ + ……….. total pressure
P₁, P₂, ……… etc. are individual pressures of each gas.

Question 3.
What is gauge pressure and how is a manometer used for measuring pressure differences?
Answer:
Gauge Pressure :
The pressure p, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess pressure (P – Pa), at depth h is called as “gauge pressure at that point.”

Measurement of pressure difference using a Manometer :
An open tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low density liquid (such as oil) for measuring small pressure differences and high density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and other end is connected to the system whose pressure to be measured.

The pressure P at A is equal to pressure at point B. If the pressure in the vessel is more than the earth’s atmospheric pressure, then the level of liquid in arm-I of U-tube will go down upto point A and the level of liquid in arm-II of U-tube will rise up to point C. Then the pressure of air in vessel is equal to pressure at point A. Let ‘h’ be the difference of liquid levels in the two arms of U-tube. Let ρ be the density of liquid in U-tube and Pa be the atmospheric pressure.

Since, the pressure is same at all points, at the same level, so pressure at point A,
P_A = pressure at point B
= pressure at C + pressure due to column of liquid of height ‘h’.
So, P_A = P_C + hρg or P_A – P_C = hρg ………… (1)
Here, P_C = Pa = atmospheric pressure.
If P_A = P, then from eq ………. (1)
P – Pa = hρg
Here, P – Pa = P_g = gauge pressure = hρg.

Question 4.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law:
It states that “the pressure in a fluid at rest is the same at all points if they are at the same height”.

Proof of Pascal’s law:
Consider an element in the interior of a fluid at rest as shown in the figure. The element ABC – DEF is in the form of a right-angled prism.

In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of gravity is the same at all these points.

The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element.

Thus, the fluid exerts pressures P_a, P_b and P_c on this element of area corresponding to the normal forces F_a, F_b and F_c as shown in fig. on the faces BEFC, ADFC and ADEB denoted by A_a, A_b and A_c respectively. Then,
F_b sin θ₂ = F_c and F_b cos θ₂ = F_a (by equilibrium)
A_b sin θ₂ = A_c and A_b cos θ₂ = A_a (by geometry)
Thus, \(\frac{F_b}{A_b}=\frac{F_c}{A_c}=\frac{F_a}{A_a}\) ⇒ P_b = P_c = P_a
Hence, pressure exerted is same in all directions in a fluid at rest.

This proves the Pascal’s law.

Question 5.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and Hydraulic brakes are based on the Pascal’s law. The principle states that “whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions

Hydralic lift:
In a hydraulic lift, two pistons are separated by the space filled with a liquid as shown in fig.

A piston of small cross-section A₁ is used to exert a force F, directly on the liquid.

The pressure, P = \(\frac{F_1}{A_1}\) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A₂, which results in an upward force of P × A₂.

Therefore, the piston is capable of supporting large force.
F₂ = PA₂ = \(\frac{F_1A_2}{A_1}\)

By changing the force at A₁, the platform can be moved up or down. Thus, the applied force has been increased by a factor of \(\frac{A_2}{A_1}\) and this factor is the mechanical advantage of the device.

Hydraulic brakes :
Hydraulic brakes in automobiles also work on Pascal’s principle. When we apply a little force on the pedal with our foot, the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way, a small force on the pedal produces a large retarding force on the wheel.

An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.

Question 6.
What is hydrostatic paradox?
Answer:
Hydrostatic paradox :
This is useful to prove that the liquid pressure is the same at all points at the same horizontal level.

Consider three vessels A, B and C of different shapes as shown in the figure. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same though they hold different amounts of water. This is so, because water at the bottom has the same pressure below each section of the vessel.

Thus, it proves that the height of the fluid column is independent of the cross sectional or base area and the shape of the container

Question 7.
Explain how pressure varies with depth.
Answer:
Variation of pressure with depth:
Consider a fluid at rest in a container. Let point 1 is at a height h’ above a point 2 as shown in the figure. Consider a cylindrical element of fluid having area of base ‘A’ and height ‘h’. As the fluid is at rest, the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top (P₁A) acting downward, at the bottom (P₂A) acting upward.

If ‘mg’ is weight of the fluid in the cylinder, we have
(P₂ – P₁) A = mg …………. (1)
Now, if ρ is the mass density of the fluid, then mass of fluid, m = ρv
⇒ m = ρhA …………. (2)
∴ From (1) and (2)
P₂ – P₁ = ρgh …………. (3)

Pressure difference depends on the vertical distance ‘h’ between the points (1 and 2), mass density of the fluid p and acceleration due to gravity ‘g’.

If the point 1 under discussion is shifted to the top of the fluid, which is open to the atmosphere, P₁ may be replaced by atmospheric pressure (Pa) and we replace P₂ by P₂ then eq.(2) becomes
P – Pa = ρgh ⇒ P = Pa + ρgh

Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 8.
What is Torricelli’s law? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s law :
Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. The word efflux means ‘fluid outflow’.

Determination of speed of Efflux:
Consider a tank containing a liquid of density ‘ρ’ with a small hole at a height ‘y₁‘, from the bottom as shown in the figure.

The air above the liquid, whose surface is at height ‘y₂‘, is at pressure, P.

From the equation of continuity, we have
V₁ A₁ = V₂A₂ ⇒ V₂ = \(\frac{A_1}{A_2}\)V₁ …….. (1)

It the cross sectional area of the tank, A₂ is much larger than that of the hole (i.e., A₂ >> A₁), then we may consider the fluid to be approximately at rest at the top. i.e., V₂ = o.

Now, applying the Bernoulli equation at points (1) and (2) and noting that, at the hole P₁ = Pa, the atmospheric pressure, we get.
Pa + \(\frac{1}{2}\)ρv²₁ + ρgy₁ = P + ρgy₂
Taking y₂ – y₁ = h, we have

When P > > Pa and 2gh may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand if the tank is open to the atmosphere, then P = Pa and from eq (2), we get
V₁ = \(\sqrt{2gh}\) ………….. (3)

This is the speed of a freely falling body, at any point of height ‘h’ during its fall. This equation is known as “Torricelli’s law”.

Question 9.
What is Venturimeter? Explain how it is used.
Answer:
Venturi-meter :
The venturi meter is a device to measure the flow speed of incompressible fluid.

It consists of a tube with a broad diameter and a small constriction at the middle as shown in the figure.

The manometer contains a liquid of density ρ_m. The speed ν₁ of the liquid flowing through the tube at the broad neck area A is to be measured. From equation of continuity, the speed at the constriction, ν₂ = \(\frac{A}{a}\) ν₁
According to Bernoulli’s equation,

This pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.

The difference in height h measures the pressure difference.

Uses: Venturi meter is used for measuring the speed of incompressible liquid and rate of flow of liquid through pipes.

Question 10.
What is Reynold’s number? What is its significance?
Answer:
Reynold’s number R_e :
Reynold’s number is a pure number which determines the nature of flow of liquid through a pipe.
R_e = \(\frac{\rho v d}{\eta}\)
where
η = coefficient of viscosity of the liquid.
ρ = density of liquid
ν = critical velocity of the liquid flowing through the pipe.
d = diameter of the pipe.

Significance :
R_e is dimensionless number and therefore, it remains same in any system of units.

The critical Reynold’s number for the onset of turbulence is in the range 1000 to 10000, depending on the geometry of the flow. For most cases

R_e < 1000 signifies laminar flow.
1000 < R_e < 2000 is unsteady flow.
R_e < 2000 implies turbulent flow.

Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Question 11.
Explain dynamic lift with examples.
Answer:
Dynamic lift on a spinning ball :
Consider the motion of a spinning ball. Its motion consists of two parts 1) Translatory motion 2) Self rotation called spinning.

1) Translatory motion:
Due to translatory motion it passes through the medium air with a velocity say (V). Due to translatory motion the number of streamlines on the top of the ball and at the bottom of the ball are equal. So there is no resultant force on the ball due to translatory motion through the fluid. Hence dynamic lift is zero.

2) For spinning motion :
Let the ball rotates about its axis with a velocity say ∆V in clockwise direction since surface of ball is not perfectly smooth it will drag air molecules with it. So at the top layers the velocity of air is V + ∆V due to air drag. At bottom layers the velocity of air is V – ∆V.

As velocity is more at top layers pressure is less and velocity is less at bottom layers, so pressure is high at bottom layers. This is due to Bernoulli’s theorem.

Due to the pressure difference at bottom layers and top layers some upward thrust will act on the ball. So some dynamic lift will act on a spinning ball. As a result the path of a spinning ball is curved.

Question 12.
Explain Surface Tension and Surface energy. [AP Mar. ’13]
Answer:
Surface tension the force per unit length on an imaginary line drawn on the surface of the liquid and acting perpendicular to it.

Surface energy:
The work done to increase the surface area of a liquid is called surface energy.
Surface energy = Surface tension × Increase in surface area.

Question 13.
Explain how surface tension can be measured experimentally.
Answer:
To find surface tension of a liquid in laboratory torsional balance is used. It is as shown in figure. It consists of a movable metallic rod fixed on a stretched wire. The position of rod can be adjusted. A glass plate is attached at one end of the rod and weight’s pan is connected at the other end of the rod.

Procedure :
A cleaned glass plate is taken. Its length ‘l’ and thickness ‘t’ is measured. Since thickness ‘l’ is very small when compared with length t, thickness ‘t’ is ignored.

Glass plate is fixed to metallic rod. Necessary weights are placed in the pan and glass plate is made horizontal to the table. Weights in pan W₀ is measured. Pure liquid whose surface tension is to be determined is taken in a glass beaker. The liquid is poured until it just touches the glass plate. Now plate is pulled down with some force due to surface tension of liquid. Weights in the pan are gradually increased until the glass plate is just escaped from forces of surface tension. Weights W₁ are noted. The experiment is repeated for three to four times and average weight W₁ is noted.

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s theorem :
Bernoulli’s theorem states that “when a non viscous liquid flows between two points then the sum of pressure energy, kinetic energy and potential energy per unit mass is always constant at any point in the path of that liquid”.

Bernoulli’s theorem is applicable to non viscous, incompressible and irrotational liquids in streamline flow only.

Proof :
Let us consider that a liquid of density ‘ρ’ is flowing through a pipe of different area of cross sections A₁ and A₂ as shown.

Let the liquid enters at A₁ with a velocity V₁ and with a pressure P₁, density of liquid at A₁ is say ρ. Let the liquid leaves the pipe through A₂ with a velocity V₂ and pressure P₂. Density of liquid at A₂ is say ρ.

Since liquid is incompressible, p is con-stant.

At region 1 the liquid will move a distance of V₁ ∆t where ∆t is very small time interval. Similarly at region 2 the liquid will move through a distance V₂ ∆t.
Work done on fluid at region 1 = W₁ = P₁ A₁
V₁ ∆t = P₁∆V

Work done on fluid at region 2 = W₂ = P₂ A₂ V₂ ∆t = P₂ ∆V

Since same volume of liquid pass through the pipe, ∆ is constant.

∴ Work done by fluid = W₁ – W₂ = (P₁ – P₂) ∆V → 1

∵ Liquid is uncompressible ‘ρ’ is constant.

So mass of liquid entering the pipe and leaving the pipe ∆m is given by
∆m = ρA₁V₁∆t = ρ∆V
Change in potential energy of liquid
∆U = ρg∆v(h₂ – h₁) → 2
Change in kinetic energy of liquid
∆K = \(\frac{1}{2}\)ρ∆v(V²₁ – V²₂)
From work energy theorem work done = change in energy
∴ ∆W = ∆U + ∆K

i.e., sum of pressure energy, potential energy and kinetic energy of the fluid is always constant.

Limitations :
Bernoulli ‘s theorem is applicable to non-viscous and uncompressible liquids only.

Applications of Bernoulli’s theorem :

1. Dynamic lift on the wings of an aeroplane is due to Bernoulli’s theorem.
2. Swinging of a spinning cricket ball is a consequence of Bernoulli’s theorem.
3. During cyclones, the roof of thatched houses will fly away. This is a consequence of Bernoulli’s theorem.

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υ_t. Give the expression for υ_t.
Answer:
Coefficient of viscosity (η) :
The viscous force acting tangentially on unit area of the liquid when there is a unit velocity gradient in the direction perpendicular to the flow is defined as “Coefficient of viscosity.”

Coefficient of viscossity η = \(\frac{-F}{A}\frac{dx}{dv}\)
Unit Nm^-2 – s (or) pascal – second.

According to Stoke’s law, the viscous force acting on a freely falling, smooth spherical body of radius ‘a’ is proportional to the coefficient of viscosity η, radius ‘a’ and velocity ‘υ’ of the body.
∴ F ∝ ηav or F = 6 π η av, where 6π is the proportionality constant.

A rain drop of radius ‘ω’, density ρ falling under gravity through air of density a experiences a force of buoyancy equal to the weight of displaced air which is (\(\frac{1}{2}\)πa³) σg.

The weight of the rain drop acting downwards = (\(\frac{1}{2}\)πa³) ρg.

∴ Resultant force acting downwards = \(\frac{1}{2}\)πa³ρg – \(\frac{1}{2}\)πa³σg

When this force is equal to the viscous drag acting upwards, then the rain drop acquires a constant velocity called terminal velocity, v_t.

At terminal velocity viscous drag = 6πηav.

Definition:
Terminal velocity of a body falling through a liquid is defined as that constant velocity which the body acquires when it falls in a fluid.

Problems

Question 1.
Find the excess pressure inside a soap bubble of radius 5 mm. (Surface tension is 0.04 N/m). [TS May ’16]
Solution:
Radius r = 5 mm = 5 × 10^-3 m.
Surface tension S_T = 0.04 N/m
= 4 × 10^-2 N/m.
Excess pressure inside soap bubble

Question 2.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm. against the surface tension force. (Surface tension of soap solution = 2.5 × 10^-2 Nm^-1)
Solution:
Work done = Surface tension (S) × increase of area (2 × 4πr²)
∴ W = S ( 4πr² ) × 2
(since the bubble has two surfaces)

Question 3.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm? (Surface tension of methyl alcohol = 0.023 Nm^-1 and density = 0.8 gmcm^-3. Assume that the angle of contact is zero)
Solution:
Surface tension of methyl alcohol (S) = 0.023 N/m.
Density, ρ = 0.8 gr/cm³ = 800 kg/m³
Diameter of tube, D = 0.06 cm

Question 4.
What should be the radius of a capillary tube if water has to rise to a height of 6 cm in it ? (Surface tension of water – 7.2 × 10^-2 Nm^-1)
Solution:
Surface Tension of water,
S = 7.2 × 10^-2 N/m
Height of water, h = 6 cm = \(\frac{6}{100}\) m
Radius of capillary tube r = ?

Question 5.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing mercury. (Density of mercury = 13.6 × 10³ Kg m^-3 and surface tension of mercury = 0.49 Nm^-1 and angle of contact = 135°).
Solution:
Diameter of tube = 0.4 mm ;
∴ Radius, r = 0.2 mm = \(\frac{0.2}{10^3}\) m
Density of mercury = 13.6 × 10³ kg/m³
Angle of contact, 0 = 135°
Surface tension of mercury, S = 0.49 N/m.

Question 6.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm-1, find the excess pressure inside the bubble. [TS Mar. ’18, My ’16; AP June ’15; Mar. ’14]
Answer:
Diameter of soap bubble =10 mm
Radius, r = 5 mm = 5 × 10^-3m
Surface tension, S = 0.04 Nm^-1
Excess pressure inside the soap bubble, P = \(\frac{4S}{r}\)

Question 7.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R?
Solution:
Energy required to form soap bubble of radius, R = W .
∴ w = 8πR²
Work done to blow a bubble of Radius 2R = 8π(2R)² = 4w
∴ Work done to increase the radius from R to 2R = 4w – w = 3w

Question 8.
If two soap bubbles of radii R₁ and R₂ (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
When joined in isothermal condition change in temperature of system is zero. So change in internal energy of the system is zero.
Let
Surface energy of 1st bubble U₁ = 8πR2S
Surface energy of 2nd bubble U₂ = 8πR²₁S
Surface energy of new bubble U = 8πR²₂S
But 8πR²S= 8πR²₁S + 8πR²₂S
⇒ R² = R²₁ +R²₂
Radius of new bubble R² = \(\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}\)

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Very Short Answer Type

Question 1.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 8 \\
7 & 2
\end{array}\right]\) and 2X + A = B, then find X. [Mar. 15 (AP); Mar. 13, 11; May 12]
Answer:

If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X. [Mar. 17 (TS)]
Answer:
\(\left[\begin{array}{ccc}
0 & 1 & -1 \\
4 & -1 & 3 \\
5 & 2 & 3
\end{array}\right]\)

Question 2.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\), then find the values of x, y, z and a. [May 14, 06 Mar. 19(AP)]
Answer:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
From the equality of matrices,
x – 3 = 5
⇒ x = 8
2y – 8 = 2
2y = 10
y=5
z + 2 = – 2
z = – 4
a – 4 = 6
a = 10
∴ x = 8, y = 5, z = – 4, a = 10

If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\) then find the values of x, y, z and ‘a’.
Answer:
2, 2, 5, 5

If \(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]\) = \(\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\) then find the values of x, y, z and ‘a’.
Answer:
1, \(\frac{5}{2}\), 2, 0

Question 3.
Find the trace of \(\left[\begin{array}{rrr}
1 & 3 & -5 \\
2 & -1 & 5 \\
2 & 0 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{rrr}
1 & 3 & -5 \\
2 & -1 & 5 \\
2 & 0 & 1
\end{array}\right]\)
∴ Tra A = 1 – 1 + 1 = 1
The elements of the principal diagonal = 1, – 1, 1

Fin the area of A if A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\)
Answer:
1

Question 4.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), find 3B – 2A. [Mar, 19 (TS); Mar. 12]
Answer:

If A = \(\left[\begin{array}{ccc}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B
Answer:
\(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
-2 & -2 & -4 \\
-4 & -5 & -5
\end{array}\right],\left[\begin{array}{ccc}
5 & -6 & -7 \\
8 & 7 & 16 \\
16 & 20 & -19
\end{array}\right]\)

If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) find A – B and 4B – 3A
Answer:
\(\left[\begin{array}{ccc}
-1 & 3 & 2 \\
2 & 2 & 5 \\
5 & 5 & 3
\end{array}\right],\left[\begin{array}{ccc}
4 & -11 & -6 \\
-6 & -5 & -16 \\
-16 & -15 & -6
\end{array}\right]\)

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
2 & 3 & -1 \\
-3 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right]\) then examine whether A and B commute with respect to multiplication of matrices. [Nov, 98]
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
2 & 3 & -1 \\
-3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right]\)
Both A and B are square matrices of order 3.
Hence both AB and BA are defined and are matrices of order 3.

which shows that AB ≠ BA
Therefore A and B do not commute with respect to multiplication of matrices.

If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\) do AB and BA exist ? If they exist find them. Do A and B commute with respect to multiplication.
Answer:
\(\left[\begin{array}{cc}
0 & -4 \\
10 & 3
\end{array}\right]\), \(\left[\begin{array}{ccc}
-10 & 2 & 21 \\
-16 & 2 & 37 \\
-2 & -2 & 11
\end{array}\right]\) & AB ≠BA

Question 6.
If A = \(\left[\begin{array}{cc}
\mathbf{i} & 0 \\
0 & -\mathbf{i}
\end{array}\right]\), then show that A² = – 1. [Mar. 16 (AP), 08]
Answer:

Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\).
Answer:
\(\left[\begin{array}{rr}
14 & 10 \\
-5 & -1
\end{array}\right]\)

If A = \(\left[\begin{array}{ll}
\mathbf{i} & \mathbf{0} \\
\mathbf{0} & \mathbf{i}
\end{array}\right]\), find A².
Answer:
\(\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]\)

Question 7.
Find \(\left[\begin{array}{ccc}
\mathbf{0} & \mathbf{c} & -\mathbf{b} \\
-\mathbf{c} & \mathbf{0} & \mathbf{a} \\
\mathbf{b} & -\mathbf{a} & \mathbf{0}
\end{array}\right]\) \(\left[\begin{array}{lll}
\mathbf{a}^2 & \mathbf{a b} & \mathbf{a c} \\
\mathbf{a b} & \mathbf{b}^2 & \mathbf{b c} \\
\mathbf{a c} & \mathbf{b c} & \mathbf{c}^2
\end{array}\right]\) [Mar. 96; May 91]
Answer:
Let A = \(\left[\begin{array}{ccc}
\mathbf{0} & \mathbf{c} & -\mathbf{b} \\
-\mathbf{c} & \mathbf{0} & \mathbf{a} \\
\mathbf{b} & -\mathbf{a} & \mathbf{0}
\end{array}\right]\), B = \(\left[\begin{array}{lll}
\mathbf{a}^2 & \mathbf{a b} & \mathbf{a c} \\
\mathbf{a b} & \mathbf{b}^2 & \mathbf{b c} \\
\mathbf{a c} & \mathbf{b c} & \mathbf{c}^2
\end{array}\right]\)
The order of matrix, A is 3 × 3
The order of matrix, B is 3 × 3
The no. of columns in A = The no. of rows in B.
∴ AB is defined.

Question 8.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & k
\end{array}\right]\) and A² = 0, then find the value of k. [Mar. 17 (AP), 14, 05, May. 11, Mar. 08(TS)]
Answer:

From equality of matrices, – 2 – k = 0 ⇒ k = – 2

Question 9.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\) then find A⁴. [May 01]
Answer:

If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Answer:
\(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
If A = \(\left[\begin{array}{lll}
1 & 4 & 7 \\
2 & 5 & 8
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-3 & 4 & 0 \\
4 & -2 & -1
\end{array}\right]\), then show that (A + B)’ = A’ + B’ [May. 09]
Answer:

Question 11.
If A = \(\left[\begin{array}{rrr}
-2 & 1 & 0 \\
3 & 4 & -5
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & 2 \\
4 & 3 \\
-1 & 5
\end{array}\right]\), then find A + B’. [May. 08]
Answer:

Question 12.
If A = \(\left[\begin{array}{ccc}
2 & -1 & 2 \\
1 & 3 & -4
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -2 \\
-3 & 0 \\
5 & 4
\end{array}\right]\), then verify that (AB)’ = B’A’ [Mar. 13]
Answer:

Question 13.
If A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
-1 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & 1 & -2
\end{array}\right]\), then find (AB)’. [Mar. 19 (TS); May. 12]
Answer:

Question 14.
If A = \(\left[\begin{array}{rr}
-2 & 1 \\
5 & 0 \\
-1 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-2 & 3 & 1 \\
4 & 0 & 2
\end{array}\right]\) then find 2A + B’ and 3B’ – A. [Mar. 10]
Answer:

Question 15.
If A = \(\left[\begin{array}{rr}
2 & -4 \\
-5 & 3
\end{array}\right]\), then find A + A’ and AA’ [May 15 (AP); May 07, 02]
Answer:

If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices ? [Mar. 17(TS)]
Answer:
AA’ = \(\left[\begin{array}{ll}
5 & 2 \\
2 & 1
\end{array}\right]\), A’A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 5
\end{array}\right]\); AA’ ≠ A’A

Question 16.
If A = \(\left[\begin{array}{ccc}
0 & 4 & -2 \\
-4 & 0 & 8 \\
2 & -8 & x
\end{array}\right]\) is a skew symmetric matrix, find the value of x. [Mar. 08]
Answer:

Question 17.
If A = \(\left[\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\) is a symmetric matrix, then find x. [Mar. 16 (AP), 05, 03, May. 15 (TS)]
Answer:

From equality of matrices, x = 6.

Question 18.
If A = \(\left[\begin{array}{rrr}
0 & 2 & 1 \\
-2 & 0 & -2 \\
-1 & x & 0
\end{array}\right]\) is a skew symmetric matrix, then find x. [May. 14, 13, 11]
Answer:
A matrix A is said to be skew symmetric if,

From the equality of matrices, x = 2.
∴ x = 2

Question 19.
Is \(\left[\begin{array}{ccc}
0 & 1 & 4 \\
-1 & 0 & 7 \\
-4 & -7 & 0
\end{array}\right]\) symmetric or skew symmetric? [Mar. 09]
Answer:

∴ A is a skew symmetric matrix since A^T = – A.

Question 20.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that AA’ = A’A = I. [Mar. 07]
Answer:

Question 21.
If ω is complex (non real) cube root of 1, then show that \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\) = 0 [Mar. 14, 11; May. 92]
Answer:
1, ω, ω² are the cube roots of unity.
Then, 1 + ω + ω² = 0, ω³ = 1.

Question 22.
Find the determinant of the matrix. [May. 95]
\(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
det A = \(\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\) = 0(0 – 1) – 1 (0 – 1) + 1(1 – 0)
= 0(- 1) – 1 (- 1) + 1(1) = 0 + 1 + 1 = 2

Find the determinant of the matrix
\(\left[\begin{array}{lll}
\mathbf{a} & \mathbf{h} & \mathbf{g} \\
\mathbf{h} & \mathbf{b} & \mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{c}
\end{array}\right]\).
Answer:
abc + 2fgh – af² – bg² – ch²

Find the determinant of the matrix
\(\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right]\)
Answer:
3abc – a³ – b³ – c³

Question 23.
Find the determinant of the matrix \(\left[\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right]\). [Mar. 10]
Answer:
Let A = \(\left[\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right]=\left[\begin{array}{ccc}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right]\)
det A = 1(225 – 256) – 4 (100 – 144) + 9 (64 – 81)
= – 31 + 176 – 153
= 176 – 184 = – 8.

Question 24.
If A = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) and det A = 45, then find x. [May 09, 03, 99, 96, Mar; 07, 03]
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & \mathrm{x}
\end{array}\right]\) and det A = 45.
det A = 1 (3x + 24) – 0(2x – 20) + 0 (-12 – 15) = 3x + 24 – 0 + 0 = 3x + 24
Given, det A = 45 ⇒ 3x + 24 = 45 ⇒ 3x = 45 – 24 ⇒ 3x = 21 ⇒ x = 7.

Question 25.
Find the adjoint and inverse of the matrix \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\). [Mar. 12]
Answer:
Let A = \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\)
Cofactor of 2 is A₁ = + (6) = 6
Cofactor of 4 is A₂ = – (- 3) = 3
Cofactor of – 3 is B₁ = – (4) = – 4
Cofactor of 6 is B₂ = + (2) = 2
∴ Cofactor matrix of A is B = \(\left[\begin{array}{ll}
\mathrm{A}_1 & \mathrm{~B}_1 \\
\mathrm{~A}_2 & \mathrm{~B}_2
\end{array}\right]=\left[\begin{array}{cc}
6 & -4 \\
3 & 2
\end{array}\right]\)
Adj A = B’ = \(\left[\begin{array}{cc}
6 & 3 \\
-4 & 2
\end{array}\right]\)
det A = ad – bc = 12 – (- 12) = 12 + 12 = 24 ≠ 0
∴ A is invertiable.
A^-1 = \(\frac{{adj} A}{{det} A}=\frac{1}{24}\left[\begin{array}{rr}
6 & 3 \\
-4 & 2
\end{array}\right]\)

Find the adjoint and the Inverse of the matrix A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\). [Mar. 18 (AP); May 06]
Answer:
\(\left[\begin{array}{cc}
-5 & -2 \\
-3 & 1
\end{array}\right],\left[\begin{array}{cc}
\frac{5}{11} & \frac{2}{11} \\
\frac{3}{11} & \frac{-1}{11}
\end{array}\right]\)

Question 26.
Find the adjoint and inverse of the matrix A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) [Mar. 13, 09]
Answer:
Let A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)

Cofactor of cos α is A₁ = + (cos α) = cos α
Cofactor of sin α is A₂ = – (- sin α) = sin α
Cofactor of – sin α is B₁ = – (sin α) = – sin α
Cofactor of cos α is B₂ = + (cos α) = cos

Question 27.
Find the rank of the matrix\(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\). [Mar. 18 (TS); May 10; Mar. 08]
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
det A = 1(1 – 1) – 1(1 – 1) + 1 (1 – 1) = 0 – 0 + 0 = 0
Since det A = 0, Rank [A] ≠ 3
Now, \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) is a submatrix of A, whose determinant is 1 – 1 = 0 ∴ Rank [A] ≠ 2.
Now. [1] is a submatrix of A, whose determinant is 1 ≠ 0. ∴ Rank [A] = 1

Telangana TSBIE TS Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Type Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
Hooke’s law states that within elastic limit stress is proportional to strain.

This constant is known as elastic modulus of the body.

Question 2.
State the units and dimensions of stress.
Answer:

Dimensional formula ML^-1 T^-2

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:

Nm^-2 (or) pascal
Dimensional formula ML^-1 T^-2

Question 4.
State the units and dimensions of Young’s modulus.
Answer:

Dimensional formula ML^-1T²

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity,

(or) pascal
Dimensional formula ML^-1T².

Question 6.
State the units and dimensions of Bulk modulus.
Answer:

unit is Nm^-2 (or) pascal
Dimensional formula ML^-1 T^-2

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:
There is no perfectly elastic body. But behaviour of Quartz fiber is very nearer to perfectly elastic body.

Real bodies are not perfectly plastic, but behaviour of wet clay, butter etc., can be taken as examples for perfectly plastic bodies.

Short Answer Questions

Question 1.
Define Hooke’s law of elasticity, proportionality, permanent set, and breaking stress.
Answer:
Hooke’s Law :
It states that within elastic limit, stress is proportional to strain.

This constant is called elastic constant (E).

Proportionality limit:
When load is increased the elongation of the wire will also increases. The maximum load upto which the elongation is directly proportional to the load is called proportionality limit (A). The graph drawn between load and extension. It is a straight line OA’.

Permanent set :
If the load on the wire is increased beyond elastic limit, the elongation is not pro portional to load. On removal of the load the wire cannot regain its original length. The length of the wire increases permanently. In figure permanent set is given by OP. This is called permanent set.

Breaking stress :
If the load is increased beyond yield point the elongation is very rapid, even for small changes in load and wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
1) Stress:
Restoring force acting on unit area is called stress.

Unit: N/m² (or) pascal;
Dimensional formula: ML^-1T^-2.

2) Strain :
The change in dimension per unit original dimension of a body is called strain.

It is a ratio, so no units and dimensional formula.

3) Modulus of elasticity :
∴ From Hooke’s Law Stress x Strain or

The ratio of stress to strain is called modulus of elasticity.
Unit: N/m².
Dimensional formula: ML^-1T^-2.

4) Poisson’s ratio :
It is defined as the ratio of lateral contraction strain to longitudinal elongation strain.
Poisson’s ratio

It is a ratio, so no units and dimensional formula.

Question 3.
Define Young’s modulus, Bulk modulus, and Shear modulus.
Answer:
1) Young’s Modulus Y:
Within elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young’s modulus

2. Bulk Modulus (B) :
Within elastic limit, the ratio of volumetric stress to volumetric strain is called Bulk modulus

3) Shear Modulus or Rigidity Modulus (G):
Within elastic limit, the ratio of tangential or shearing stress to shearing strain is called Shear modulus or Rigidity modulus.

Question 4.
Define stress and explain the types of stress. [AP War. 19; TS War. 16]
Answer:
Stress:
When a body is subjected to a deforming force, then restoring forces will develop inside the body. These restoring forces will oppose any sort of change in its original shape. The restoring force per unit area of the surface is called stress.

Stress:
Stress is defined as force applied per unit area.

D.F. = ML^-1T^-2 ; Unit: N/m² (or) Pascal.

Types of stress :
It is of three types. They are : 1) Longitudinal stress 2) Tangential stress (or) Shear stress 3) Volumetric stress.

Longitudinal stress:
If the force applied on a body is along its lengthwise direction then it is called longitudinal stress. It produces deformation in length.

Tangential stress (or) Shear stress:
Force applied per unit area parallel to the surface of a body trying to displace the upper layers of the body is called shearing stress.

(Parallel to the surface layers)

Volumetric stress:
If force is applied on all the sides of a body or on the volume of a body then it is called volumetric stress.

Question 5.
Define strain and explain the types of strain.
Answer:
Strain:
Strain is defined as deformation produced per unit dimension. It is a ratio. So no units.

Types of strain :
Strain is of three types. They are: 1) Longitudinal strain 2) Tangential strain or Shear strain 3) Volumetric strain.

Longitudinal strain:
The ratio of elongation to original length along length wise direction is defined as longitudinal strain.

Shearing strain :
If the force applied on a body produces a change in shape only it is called shearing force. The angle through which a plane originally perpendicular to the fixed surface shifts due to the application of shearing stress is called shearing strain or simply shear (θ).

Question 6.
Define strain energy and derive the equation for the same. [TS Mar. ’19; May ’18; AP Mar. ’14; May ’14]
Answer:
Strain energy :
The energy developed in (string) a body when it is strained is called strain energy.

Let a force F be applied on lower end of wire, fixed at the upper end. Let the extension be dl.
∴ Work done = dW = Fdl
Total work done in stretching it from 0

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
For metals Young’s moduli are large. Therefore, these materials require a large force to produce small change in length. To increase the length of a thin steel wire of 0.1 cm² cross-sectional area by 0.1 %, a force of 2000 N is required. The force required to produce the same strain in aluminium, brass, and copper wire having the same cross-sectional area are 690 N, 900N, and 1100 N respectively. It means that steel is more elastic than copper, brass, and aluminium. It is for this reason that steel is preferred in heavy duty machines and in structural designs.

Question 8.
Describe the behaviour of a wire under gradually increasing load. [AP Mar. ’18, ’17, ’16, ’15, ’13, May ’16, ’13; TS Mar. 18, 17, 15; May 17,16; June 15]
Answer:
Behaviour of a wire under increasing load:
Let a wire is suspended at one end and loads are attached to the other end. When loads are gradually increased the following changes are noticed.

1) Proportionality limit (A) :
When load is increased the elongation of the wire gradually increases. The maximum load upto which the elongation is directly pro-portional to the load is called proportionality limit (A). The graph drawn between load and extension is a straight line. So point A is called proportionality limit. In this region Hooke’s Law is obeyed.

2) Elastic limit (B):
If the load is increased above the proportionality limit the elongation is not proportional to the load. Hooke’s law is not obeyed. But it exhibits elasticity which means that it regains the original length if load is removed. The maximum load on the wire upto which it exhibits elasticity is called elastic limit (B in the graph).

3) Permanent set (C) :
If the load on the wire is increased beyond elastic limit say upto C, the elongation is not proportional to load. On removal of the load, the wire does not regain its original length. Length of wire increases perma-nently. In figure permanent set is given by OP. So OP is called permanent set.

4) Point of ultimate tensile strength (D) :
If the load is further increased, upto D’ then strain increases rapidly even though there is no increase in stress.

At this stage the restoring forces seems to be subdued to then deforming forces. Elongation without increase in load is called creeping. This behaviour of metal is called yielding.

5) Fracture point (E) :
If the load is increased beyond Yield point the elongation is very rapid, even for small changes in load the wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Question 9.
Two identical solid balls, one of ivory and the other of wet clay are dropped from the same height on to the floor. Which one will rise to a greater height after striking the floor and why?
Answer:
We know that ivory ball is more elastic than wet-clay ball. Therefore, the ivory ball will tend to regain its original shape in a very short time after the collision. Due to it, there will be large energy and momentum transfer to the ivory ball in comparison to the wet-clay ball. As a result of it, the ivory ball will raise higher after the collision.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why?
Answer:
Use of pillars or columns is very common in buildings and bridges. A pillar with roun-ded-ends supports less load than that with a distributed shape at the ends. Hence, for this reason, while constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km?
Answer:
A mountain base is not under uniform compression and this provides some shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height ‘h’, the force per unit area due to the weight of the mountain is hpg where p is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free. There is a shear component, approximately hpg itself. Now the elastic limit for a typical rock is 3 × 10⁷ Nm^-2. Equating this to hpg with ρ = 3 × 10³ kg m^-3 gives
h = \(\frac{30\times10^{-7}}{3\times10^3\times10}\) = 10 km
Hence, the maximum height of a mountain on earth is approximately 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it. [AP May ’ 18, 17; AP June 15]
Answer:
When a wire is put under a tensile stress, work is done against the inter atomic forces. This work is stored in the form of “Elastic potential energy.”

Expression to elastic potential energy:
To stretch a wire, force is applied. As a result it elongates. So the force applied is useful to do some work. This work is stored in it as potential energy. When the deforming force is removed, this energy is liberated as heat. The energy developed in a body (string) when it is strained is called strain energy.

Let a force F be applied on a wire fixed at the upper end. Let the extension be dl.
∴ Work done = dW = Fdl
Total work done in stretching from

Long Answer Questions

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s Law :
Within elastic limit, stress is directly proportional to strain.
strain ∝ stress

where E constant called modulus of elasticity of the material of a body.

Determination of Young’s modulus of a wire:
The apparatus used to find Young s modulus of a wire consists of two long wires A and B of same length made with same material are used. These two wires are suspended from a rigid support and a vernier scale V’ is attached to them. Wire A is connected to the main scale (M). A fixed load is connected to this cord to keep tension in the wire. This is called reference wire. The second Wire B’ is connected to vernier scale V’. Adjustable load hanger is connected to this wire. This is called experimental wire.

Procedure :
Let a load M₁ is attached to the weight hanger at vernier. Main scale reading (M.S.R) and vernier scale reading (V.S.R) are noted. Weights are gradually increased in the steps of \(\frac{1}{2}\) kg upto a maximum load of say 3 kg. Every time M.S.R and V.S.R are noted. They are placed in tabular form.

Now loads are gradually decreased in steps of \(\frac{1}{2}\) kg. While decreasing M.S.R and V.S.R are noted for every load, values are posted in tabular form.

Let 1^st reading with mass M₁ is e₁ and 2^nd reading with mass M₂ is e₂.

Change in load M = M₂ – M₁
elongation e = e₂ – e₁
‘M’ and ‘e’ values are calculated and a graph is plotted.

Force on the wire = mg
Area of cross section of the wire = πr²
(r = radius of the wire)
Elongation = e
Original length = l

A graph is between load (m) and elongation ‘e’, is straight line passing through the origin. The slope of the graph (tan θ) gives (\(\frac{m}{e}\)). The value is substituted in the above equation to find Young’s modulus of the material of the wire.

Precautions:

1. The load applied should be much smaller than elastic limit.
2. Reading is noted only after the air bubble is brought to centre of spirit level.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
Diameter, d = 1mm
∴ radius, r = 0.5 mm = 0.5 × 10^-3m
Force, F = 10N

Question 2.
A tungsten wire of length 20cm is stretched by 0.1cm. Find the strain on the wire.
Solution:
Length of wire, l = 20cm = 0.2m
elongation, e = 0.1cm = 1 × 10^-3m

Question 3.
If an iron wire is stretched by 1%, what is the strain on the wire?
Solution:

Question 4.
A brass wire of diameter 1mm and of length 2m is stretched by applying a force of 20N. If the increase in length is 0.51mm, find i) the stress, ii) the strain and iii) the Young’s modulus of the wire.
Solution:
Length of wire, l = 2m;
L diameter, d = 1mm = 10^-3 m
Force, F = 20N;
Increase in length, e = 0.51mm

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3: 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_Cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹Nm^-2)
Solution:
Ratio of lengths, l₁ : l₂ = 3 : 2;
Ratio of diameters, d₁ : d₂ = 2 : 3
Ratio of forces, F₁ : F₂ = 4 : 5
Y₁ = Y of copper = 1.1 × 10¹¹
Y₂ = Y of Aluminium = 0.7 × 10¹¹;
Ratio of elongation, e₁ : e₂ = ?

Question 6.
A brass wire of cross-sectional area 2mm² is suspended from a rigid support and a body of volume 100cm³ is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³kgm^-3)
Solution:
Area of cross section, A = 2mm = 2 × 10^-6 m²
Volume of body, V = 100 cc = 100 × 10^-6 m³
Decrease in length, e’ = 0.11mm = 0.11 × 10^-3m
Youngs modulus of brass, Y = 0.91 × 10¹¹ N/m²
Density of water, ρ = 1000 kg / m³
Use e’ = \(\frac{V\rho gl}{AY}\)

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1 : 2. If the extensions produced are equal, what is the ratio of the loads?
Solution:
Ratio of lengths, l₁ : l₂ = 1 : 2
Ratio of radii, r₁ : r₂ = 1 : 2
Extensions produced are equal ⇒ e₁ = e₂;
Made of same material ⇒ Y₁ = Y₂
Ratio of loads m₁ : m₂ = ?

Question 8.
Two wires of different material have same lengths and areas of cross-section. What is the ratio of their increase in length when forces applied are the same? (Y₁ = 0.90 × 10¹¹ Nm^-2, Y₂ = 3.60 × 10¹¹ Nm^-2.)
Solution:
Lengths are same ⇒ l₁ = l₂ ;
Area of cross sections are same, A₁ = A₂
Y₁ = 0.9 × 10¹¹ N/m²
Y₂ = 3.60 × 10¹¹ N/m²

Question 9.
A metal wire of length 2.5m and area of cross-section 1.5 × 10^-6 m² is stretched through 2mm. If its Young’s modulus is 1.25 × 10¹¹ Nm^-2, find the tension in the wire.
Solution:
Length of wire, l = 2.5m
Y = 1.25 × 10¹¹N/m²
Area of cross section, A = 1.5 × 10^-6 m²
Elongation, e = 2 m.m = 2 × 10^-3m
Tension, T = mg = F = ?

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the (i) stress in the two wires and 0Q strain in the two wires. (Y_Al = 0.7 × 10¹¹ Nm^-2, Y_steel = 2 × 10¹¹m^-2)
Solution:
i) Length is same ⇒ l₁ = l₂
Area is same ⇒ A₁ = A₂
In composite wire same load will act on both wires.
∴ Ratio of stress = 1 : 1

ii) Total elongation, e = 1.35mm =e_Al + e_s
Young’s modulus of aluminium = 7 × 10¹⁰ N/m²
Y of steel = 2 × 10¹¹ N/m²
Elongation, e = \(\frac{Fl}{AY}\)
But F, l and A are same

∴ Ratio of strains in the wires is 20 : 7.

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15cm due to a tangential force of 0.3 N while keeping the lower face fixed. Calculate the rigidity modulus of the substance.
Solution:
Side of cube, a = 2.0 cm = 2 × 10^-2 m
Area, A = 4 × 10^-4 m²
Displacement of upper layer = 0.15cm
= 0.15 × 10^-2m
Tangential force, F = 0.30N

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 atmosphere. The change in volume is 10^-8 cm³. If the ball is made of iron, find its bulk modulus. (1 atmosphere = 1 × 10⁵ Nm^-2)
Solution:
Volume of ball, V = 1000 cm³ = 10^-3 m³
(∵ 1M³ = 10⁶ cm³)
Pressure, P = 10 atmospheres
= 10 × 10⁵ pa ( v 1 atm = 10⁵ pascal)
Change in volume, ∆V = 10^-8 cm³

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2 (1 atm = 1 × 10⁵ Nm^-2).
Solution:
Side of cube, a’ = 1cm = 10^-2m
∴ Volume of cube = 10^-6m
Pressure, P = 100 atm = 100 × 10⁵ = 10⁷ pa
Bulk modulus, K = 1.4 × 10¹¹ N/m²;

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2
Solution:

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s’ ratio for steel is 0.19.
Solution:
Length of wire, l = 20, cm = 0.20m,
Poisson’s ratio, σ =0.19
Increase in length, ∆l = 0.2 cm = 2 × 10^-3m
lateral strain = ?
Lateral strain = σ × longitudinal strain ‘e’

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Mathematical Induction Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Mathematical Induction Important Questions

Question 1.
By using mathematical induction show that ∀ n ∈ N, 1² + 2² + 3² + ……………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Answer:
Let S(n) be the statement that 1² + 2² + 3² + …………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
If n = 1, then
LHS = n² = 1² = 1

∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.
∴ 1² + 2² + 3² + ……………… +n² = \(\frac{n(n+1)(2 n+1)}{6}\), ∀ n ∈ N.

Question 2.
By using mathematical Induction show that ∀ n ∈ N, 1³ + 2³ + 3³ + …………. + n³ = \(\frac{n^2(n+1)^2}{4}\). [May 97, 94, 93, 88, Mar. 87]
Answer:
Let S(n) be the statement that 1³ + 2³ + 3³ + ……….. + n³ = \(\frac{n^2(n+1)^2}{4}\)
If n = 1, then
L.H.S = n³ = 1³
R.H.S = \(\frac{n^2(n+1)^2}{4}\) = \(\frac{1^2(1+1)^2}{4}\) = \(\frac{1.4}{4}\) = 1
∴ L.H.S = R.H.S
∴ S(1) is true.
Assume that S(k) is true.

∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.
∴ 1³ + 2³ + 3³ + ………… + n³ = \(\frac{n^2(n+1)^2}{4}\), ∀ n ∈ N

Question 3.
By using mathematical induction show that ∀ n ∈ N, 2.3 + 3.4 + 4.5 + …………. upto n terms = \(\frac{n\left(n^2+6 n+11\right)}{3}\)
Answer:
2, 3, 4 ……………………. are in A.P.
Here a = 2, d = 3 – 2 = 1
∴ t_n = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1.
3, 4, 5 …………………… are in A.P.
Here a = 3, d = 4 – 3 = 1
∴ t_n = a + (n – 1) d = 3 + (n – 1) 1 = 3 + n – 1 = n + 2.
∴ The n^th term of the given series is (n + 1) (n + 2).
Let S(n) be the statement that
2.3 + 3.4 + 4.5 + + (n + 1) (n + 2) = \(\frac{n\left(n^2+6 n+11\right)}{3}\)
If n = 1, then
LHS = (n + 1) (n + 2) = (1 + 1) (1 + 2) = 2.3 = 6
RHS = \(\frac{\mathrm{n}\left(\mathrm{n}^2+6 \mathrm{n}+11\right)}{3}\) = \(\frac{1\left(1^2+6(1)+11\right)}{3}\) = \(\frac{18}{3}\) = 6
∴ LHS = RHS
∴ S(1) is true.
Assume that S(k) is true.

Verification Method:
\(\frac{\mathrm{n}\left(\mathrm{n}^2+6 n+11\right)}{3}\)
Put n = k + 1

∴ S(k + 1) is true.
By the principle of mathematical induction, S(n) is true ∀ n ∈ N.
∴ 2.3 + 3.4 + 4.5 + ……………… + (n + 1) (n + 2) = \(\frac{\mathrm{n}\left(\mathrm{n}^2+6 \mathrm{n}+11\right)}{3}\), ∀ n ∈ N

Question 4.
By using mathematical induction show that ∀ n ∈ N,
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\) [Mar. 18 May 15 (AP): May 14, 97, 92]
Answer:

Question 5.
By using mathematical induction show that ∀ n ∈ N, \(\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\) …………….. upto n terms = \(\frac{n}{3 n+1}\).
Answer:
1, 4, 7, ………………… are in A.P.
Here, a = 1, d = 4 – 1 = 3
t_n = a + (n – 1) d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
4, 7, 10, ………………. are in A.P.
Here, a = 4, d = 7 – 4 = 3
t_n = a + (n – 1) d = 4 + (n – 1)3 = 4 + 3n – 3 = 3n + 1
∴ The n^th term in the given series is \(\frac{1}{(3 n-2)(3 n+1)}\).

Question 6.
By using mathematical induction show that ∀ n ∈ N,
a + (a + d) + (a + 2d) + upto n terms = \(\frac{n}{2}\) [2a + (n – 1) d].
Answer:
a, a + d, a + 2d, ………………….. are in A.P
∴ t_n = a + (n – 1) d
∴ n th term in the given series is a + (n – 1) d.
Let S(n) be the statement that
a + (a + d) + (a + 2d) + ………………. + a + (n – 1) d = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1) d]
If n = 1 then
L.H.S = a + (n – 1)d = a + (1 – 1)d = a
R.H.S = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d] = \(\frac{1}{2}\) [2a + (1 – 1)d] = \(\frac{1}{2}\) [2a] = a
∴ L.H.S = R.H.S
∴ S(1) is true.
[a + (n – 1)d
put n = k + 1
a + (k + 1 – 1)d
= a + kd]
Assume that S(k) is true.
a + (a + d) + (a + 2d) + ………………… + [a + (k- 1) d] = \(\frac{\mathrm{k}}{2}\) [2a + (k – 1)d]
Adding (a + kd) on both sides, we get

∴ S(k + 1) is true.
By the principle of Mathematical Induction, s(n) is true, ∀ n ∈ N
∴ a + (a + d) + (a + 2d) + ….+ a + (n- 1) d = \(\frac{n}{2}\) [2a + (n – 1) d], ∀ n ∈ N

Question 7.
By using mathematical induction show that ∀ n ∈ N,
a + ar + ar² + ………. upto n terms = \(\frac{a\left(r^n-1\right)}{r-1}\), r ≠ 1. [Mar. 19 (AP); Mar. 11, 80; May 87]
Answer:
a + ar + ar² + ………. are in G.P.
∴ t_n = a . r^{n – 1}
The n th term in the given series is a . r ^{n – 1}
Let S(n) be the statement that
a + ar + ar² + …………. + a ∙ r^{n – 1} = \(\frac{a\left(r^n-1\right)}{r-1}\)
If n = 1, then
L.H.S = a ∙ r^{n – 1} = a ∙ r^{1 – 1} = a ∙ r⁰ = a ∙ 1 = a
R.H.S = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{a\left(r^1-1\right)}{r-1}=\frac{a(r-1)}{r-1}\) = a
∴ L.H.S = R.H.S
∴ S(1) is true.
[a.r^{n – 1}
Put n = k + 1
a.r^{k + 1 – 1}
a.r^k]
Assume that S(k) is true.
a + ar + ar² + …….. + a.r^{k – 1} = \(\frac{a\left(r^k-1\right)}{r-1}\)
Adding ar^k on both sides we get,

∴ S(k + 1) is true.
By the principle of mathematical induction, S(n) is true ∀ n ∈ N.
∴ a + ar + ar² + ………………. + a . r^{n – 1} = \(\frac{a\left(r^n-1\right)}{r-1}\)

Question 8.
By using mathematical induction show that ∀ n ∈ N, 1.2.3 + 2.3.4 + 3.4.5 + …………………. upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Answer:
1, 2, 3 …………………. are in A.P.
Here a = 1, d = t₂ – t₁ = 2 – 1 = 1
t_n = a + (n – 1)d = 1 + (n – 1)1 = 1 + n – 1 = n
2, 3, 4 ………………… are in A.P.
Here a = 2, d = 3 – 2 = 1
t_n = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1
3, 4, 5 …………….. are in A.P.
Here a = 3, d = 4 – 3 = 1
t_n = a + (n – 1)d = 3 + (n – 1)1 = 3 + n – 1 = n + 2
∴ The nth term in the given series is n(n + 1) (n + 2).
Let S(n) be the statement that
1.2.3 + 2.3.4 + 3.4.5 + ……………….. + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
If n = 1 then
L.H.S = n(n + 1) (n + 2) = 1(1 + 1) (1 + 2) = 1.2.3 = 6
R.H.S = \(\frac{n(n+1)(n+2)(n+3)}{4}\) = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = \(\frac{1.2 .3 .4}{4}\) = 6
∴ L.H.S = R.H.S
∴ S(1) is true Assume that s(k) is true
1.2.3 + 2.3.4 + 3.4.5 + ……………….. + k (k + 1) (k + 2) = \(\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{4}\)
Adding (k + 1) (k + 2) (k + 3) on both sides we get
1.2.3 + 2.3.4 + 3.4.5 + …………….. + k (k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

∴ S(k + 1) is true.
∴ By the principle of mathematical induction, S(n) is true ∀ n ∈ N.
∴ 1.2.3 + 2.3.4 + 3.4.5 + ……………… + n(n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\) ∀ n ∈ N.

Question 9.
By using mathematical induction show that ∀ n ∈ N,
\(\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}\) + ……………….. upto n terms = \(\frac{n}{24}\) [2n² + 9n + 13]. [Mar. 14, 07, 05]
Answer:
Numerator: n^th = 1³ + 2³ + 3³ + …………. + n³ = Σn³ = \(\frac{n^2(n+1)^2}{4}\)

Denominator: 1 + 3 + 5 + ………………. are in A.P.
Here a = 1, d = 3 – 1 = 2
t_n = a + (n – 1) d = 1 + (n – 1) 2 = 1 + 2n – 2 = 2n – 1
∴ n^th term is

∴ L.H.S = R.H.S
∴ S(1) is true.
Assume that S(k) is true.

Verification Method: \(\frac{\mathrm{n}}{24}\) [2n² + 9n + 13]
Put n = k + 1
= \(\frac{(\mathrm{k}+1)}{24}\) [2(k + 1)² + 9 (k + 1) + 13] = \(\frac{(\mathrm{k}+1)}{24}\) [2k² + 2 + 4k + 9k + 9 + 13]
= \(\frac{1}{24}\) [2k³ + 2k + 4k² + 9k² + 9k + 13k + 2k² + 2 + 4k + 9k + 9 + 13] = \(\frac{1}{24}\) [2k³ + 15k² + 37k + 24]
∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.

Question 10.
By using mathematical induction show that ∀ n ∈ N, 1² + (1² + 2²) + (1² + 2² + 3²) + ……….. upto n terms = \(\frac{n(n+1)^2(n+2)}{12}\).
Answer:

Question 11.
By using mathematical induction show that ∀ n ∈ N, 2 + 3.2 + 4.2² + ……… upto n terms
Answer:
2.1 + 3.2 + 4.2² + ………………… upto n terms = n . 2ⁿ
2, 3, 4 ………….. are in A.P.
Here a = 2, d = 3 – 2 = 1
t_n = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1
1, 2, 2², ……………… are in G.P.
Here a = 1, r = \(\frac{2}{1}\) = 2
t_n = a . r^{n – 1} = 1 . 2^{n – 1} = 2^{n – 1}
∴ The nth term in the given series is (n + 1) (2^{n – 1}).
Let S(n) be the statement that
2.1 + 3.2 + 4.2² + ……….. + (n + 1) 2^{n – 1} = n . 2ⁿ
If n = 1, then
L.H.S. = (n + 1)^{2n – 1} = (1 + 1) 2^{1 – 1} = 2.2⁰ = 2.1 = 2
R.H.S. = n . 2ⁿ = 1.2¹ = 2
∴ LHS = RHS
∴ S(1) is true.
[(n + 1)^{2n – 1}
put n = k + 1
(k + 1 + 1) 2^{k + 1 – 1}
(k + 2) . 2^k]
Assume that S(k) is true.
2.1 + 3.2 + 4.2² + ……….. + (k + 1)^{2k – 1} = k . 2k
Adding (k + 2) . 2^k on both sides, we get
2.1 + 3.2 + 4.2² + ………… + (k + 1) 2^{k – 1} + (k + 2) 2^k = k. 2^k + (k + 2) . 2^k = 2^k (k + k + 2)
= 2^k (2k + 2) = 2^k . 2 (k + 1) = (k + 1) . 2^{k + 1}
∴ S(k + 1) is true.
∴ By the principle of mathematical induction, S(n) is true, ∀ n ∈ N.
2.1 + 3.2 + 4. 2² + ……………… + (n + 1) 2^{n – 1} = n . 2ⁿ, ∀ n ∈ N.

Question 12.
By using mathematical induction show that ∀ n ∈ N, 49ⁿ + 16n – 1 is divisible by 64 for all positive integer n. [Mar. 18 (TS); Mar. 17 (AP); May 13, 05, 98, 93]
Answer:
Let S(n) be the statement that f(n) = 49ⁿ + 16n – 1 is divisible by 64.
If n = 1, then
f(1) = 49¹ + 16.1 – 1 = 49 + 16 – 1 = 49 + 15 = 64 = 64 × 1 is divisible by 64
∴ S(1) is true.
Assume that S(k) is true.
f(k) is divisible by 64 ⇒ 49^k + 16k – 1 is divisible by 64
⇒ 49^k + 16k – 1 = 64 M for some integer M ⇒ 49^k = 64M – 16k + 1
Now
f(k + 1) = 49^{k + 1} + 16 (k + 1) – 1 = 49^k. 49 + 16k + 16 – 1 = (64M – 16k + 1) 49 + 16k + 15
= 64.49M – 784k + 49 + 16k + 15 = 64.49M – 768k + 64
= 64(49M – 12k + 1) is divisible by 64. [ ∵ 49M – 12k + 1 is an integer]
∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true ∀ n ∈ N.
∴ 49ⁿ + 16n – 1 is divisible by 64, ∀ n ∈ N.

Question 15.
By using mathematical induction show that 3 ∙ 5^{2n + 1} + 2^{3n + 1} is divisible by 17 ∀ n ∈ N. [May 12, 10, 08, 01, ’96]
Answer:
Let S(n) be the statement that f(n) = 3 . 5^{2n + 1} + 2^{3n + 1} is divisible by 17.
If n = 1, then
f(1) = 3.5^{2.1 + 1} + 2^{3.1 + 1} = 3.5³ + 24 = 3(125) + 16 = 375 + 16 = 391 = 17 × 23 is divisible by 17
∴ S(1) is true.
Assume that S(k) is true.
∴ f(k) is divisible by 17.
⇒ 3.5^{2k + 1} + 2^{3k + 1} is divisible by 17.
⇒ 3.5^{2k + 1} + 2^{3k + 1} = 17 M for some integer M.
⇒ 3.5^{2k + 1} = 17M – 2^{3k + 1}
Now f (k + 1) = 3 . 5^{2(k + 1) + 1} + 2^{3(k + 1) + 1} = 3.5^{2k + 2 + 1} + 2^{3k + 3 + 1}
= 3.5^{2k + 1} . 5² + 2^{3k + 1} . 2³
= (17M – 2^{3k + 1}) 25 + 8 . 2^{3k + 1} = 17.25 M – 25.2^{3k + 1} + 8 . 2^{3k + 1}
= 17.25 M – 17.2^{3k + 1}
= 17 (25 M – 2^{3k + 1}) is divisible by 17. [∵ 25M – 2^{3k + 1} is an integer]
∴ S(k + 1) is true.
By the principle of mathematical induction, S(n) is true ∀ n ∈ N.
∴ 3 . 5^{2n + 1} + 2^{3n + 1} is divisible by 17, ∀ n ∈ N.

Question 14.
By using mathematical induction show that ∀ n ∈ N, xⁿ – yⁿ is divisible by x – y. [May 04]
Answer:
Let S(n) be the statement that f(n) = xⁿ – yⁿ is divisible by (x – y).
If n = 1 then
f(1) = x¹ – y¹ = x – y = (x – y) × 1 is divisible by (x – y)
∴ S(l) is true.
Assume that S(k) is true.
f(k) is divisible by (x – y).
⇒ x^k – y^k is divisible by (x – y)
⇒ x^k – y^k = (x – y) M, for some integer M.
⇒ x^k = (x – y) M + y^k
Now
f(k + 1) = x^{k + 1} – y^{k + 1} = x^k . x – y^k . y = [(x – y) M + y^k] x – y^k . y = (x – y) . M x + y^k . x – y^k . y
= (x – y) M x + y^k (x – y) = (x – y) [Mx + y^k] is divisible by (x – y).
[Mx + y^k] is an integer.
∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.
∴ xⁿ – yⁿ is divisible by (x – y), ∀ n ∈ N.

Some More Maths 1A Mathematical Induction Important Questions

Question 1.
By using mathematical induction show that ∀ n ∈ N, 4³ + 8³+ 12³ + ……………. upto n terms = 16n² (n + 1)².
Answer:
4, 8, 12, are in A.P.
Here a = 4, d = t₂ – t₁ = 8 – 4 = 4
t_n = a + (n – 1)d = 4 + (n – 1)4 = 4 + 4n – 4 = 4n
∴ The nth term in the given series is (4n)³ = 64n³
Let S(n) be the statement that 4³ + 8³ + 12³ + …………….. + 64n³ = 16n² (n + 1)²
If n = 1, then
LHS = 64n³ = 64(1)³ = 64
RHS = 16n² (n + 1)² = 16.1² (1 +!)² = 16.1.4 = 64
∴ LHS = RHS
∴ S(1) is true.
Assume that S(k) is true.
4³ + 8³ + 12³ + ………………….. + 64k³ = 16k² (k + 1)²
Adding 64(k + 1)³ on both sides, we get
4³ + 8³ + 12³ + …………….. + 64k³ + 64(k + 1)³ = 16k² (k + 1)² + 64 (k + 1)³ = 16(k + 1)² [k² + 4 (k + 1)]
= 16(k + 1)² [k² + 4k + 4] = 16(k + 1)² (k + 2)² = 16(k + 1)² (k + 1 + 1)²
∴ S(k + 1) is true.
By the principle of mathematical induction, S(n) is true ∀ n ∈ N.
∴ 4³ + 8³ + 12³ + ………………. + 64n³ = 16n² (n + 1)², ∀ n ∈ N.

Question 2.
By using mathematical induction show that ∀ n ∈ N. 2.4^{2n + 1} + 3^{3n + 1} is divisible by 11.
Answer:
Let S(n) be the statement that f(n) = 2.4^{2n + 1} + 3^{3n + 1} is divisible by 11.
If n = 1, then
f(1) = 2.4^{2.1 + 1} + 3^{3.1 + 1} = 2.4³ + 34 = 2(64) + 81
= 128 + 81 = 209 = 11 × 19 is divisible by 11.
∴ S(1) is true.
Assume that S(k) is true.
∴ f(k) is divisible by 11.
⇒ 2.4^{2k + 1} + 3^{3k + 1} is divisible by 11
⇒ 2.4^{2k + 1} + 3^{3k + 1} = 11 M, for some integer M
⇒ 2.4^{2k + 1} = 11 M – 3^{3k + 1}
Now f(k + 1) = 2.4^{2(k + 1) + 1} + 3^{3(k + 1) + 1} = 2.4^{2k + 2 + 1} + 3^{3k + 3 + 1} = 2.4^{2k + 1} . 4² + 3^{3k + 1} . 3³
= (11 M – 3^{3k + 1}) 16 + 27. 3^{3k + 1} = 11.16 M – 16. 3^{3k + 1} + 27.3^{3k + 1} = 11.16M + 11.3^{3k + 1}
= 11(16 M + 3^{3k + 1}) is divisible by 11 (∵ 16 M + 3^{3k + 1} is an integer)
∴ S(k + 1) is true.
By the principle of mathematical induction S(n) is true, ∀ n ∈ N.
∴ 2.4^{2n + 1} + 3^{3n + 1} is divisible by 11, ∀ n ∈ N.

Question 3.
Using mathematical induction prove that statement for all ∀ n ∈ N,
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)²
Answer:
Let S(n) be the statement that \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)²
If n = 1, then
LHS = 1 + \(\frac{2 n+1}{n^2}\) = 1 + \(\frac{2.1+1}{(1)^2}\) = 1 + \(\frac{2+1}{1}\) = 1 + 3 = 4
RHS = (n + 1)² = (1 + 1)² = (2)² = 4
∴ LHS = RHS
∴ S(1) = is true
Assume that S(k) is true

Telangana TSBIE TS Inter 1st Year Physics Study Material 9th Lesson Gravitation Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 9th Lesson Gravitation

Very Short Answer Type Questions

Question 1.
State the unit and dimension of universal gravitational constant (G).
Answer:
Units of G = N-m² / kg².
Dimensional formula = M^-1 L³ T^-2.

Question 2.
State the vector form of Newton’s law of gravitation.
Answer:
Vector form of Newton’s Law of gravitation is \(\overline{\mathrm{F}}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2 \overline{\mathrm{r}}}{\overline{\mathbf{r}}^3}\)

Question 3.
If the gravitational force of the Earth on the Moon is F. What is the gravitational force of the moon on the earth? Do these forces form an action-reaction pair?
Answer:
Gravitational force between earth and moon and moon and earth are same
i-e., F_EM = – F_ME

Gravitational force between the bodies are treated as action-reaction pair.

Question 4.
What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant?
Answer:
Acceleration due to gravity, g = \(\frac{GM}{R^2}\)
When mass is kept as constant and radius
is decreased by 2% then \(\frac{\Delta \mathrm{R}}{\mathrm{R}}\) × 100 = 2

From distribution of errors in multiplications and divisions \(\frac{\Delta \mathrm{g}}{\mathrm{g}}\) × 100 = -2\(\frac{\Delta \mathrm{R}}{\mathrm{R}}\) × 100

% Change in g = – 2 × 2 = – 4% – ve sign indicates that when R decreases ‘g’ increases.

Question 5.
As we go from one planet to another, how will a) the mass and b) the weight of a body change?
Answer:

1. As we go from one planet to another planet mass of the body does not change. Mass of a body is always constant.
2. As we move from one planet to another planet weight of the body gradually decreases. It become weightless. When we approaches the other planet the weight will gradually increases.

Question 6.
Keeping the length of a simple pendulum constant, will the time period to be the same on all planets? Support your answer with reason.
Answer:
Even though length of pendulum / is same the time period of oscillation T value changes from planet to planet.
Time period of pendulum, T = 2π\(\sqrt{\frac{l}{g}}\)
i.e., T depends on l and g.

Acceleration due to gravity, (g = \(\frac{GM}{R^2}\)) changes from planet to planet.

Hence Time period of pendulum changes even though length ‘l’ is same.

Question 7.
Give the equation for the value of g at a depth ‘d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth?
Answer:
Acceleration due to gravity at a depth ‘d’ below the ground is, gd = g(1 – \(\frac{D}{R}\))
Acceleration due to gravity at centre of earth is zero. (Since D = R)

Question 8.
What are the factors that make ‘g’ the least at the equator and maximum at the poles?
Answer:
‘g’ is least at equator due to
1) The equatorial radius of earth is maximum ∵ g = \(\frac{GM}{R^2}\) (∵ R = maximum)

2) Due to rotation of earth centrifugal force will act on the bodies. It opposes gravitational pull of earth on the bodies. At equator centrifugal force is maximum. So ‘g’ value is least at equator.
The g’ ⇒ maximum at poles due to

1) The polar radius of earth is minimum
(∵ g = \(\frac{GM}{R^2}\))

2) Centrifugal force due to rotation of earth is zero at poles. This centrifugal force reduces earth’s gravitational pull.

Since Centrifugal force is zero, ‘g’ value is maximum at poles.

Question 9.
“Hydrogen is in abundance around the sun but not around earth”. Explain.
Answer:
The escape velocity on the sun is very high compared to that on the earth. The gravitational pull of the sun is very large because of its larger mass compared to that of the earth. So it is very difficult for hydrogen to escape from the Sun’s atmosphere. Hence hydrogen is abundant on sun.

Question 10.
What is the time period of revolution of a geostationary satellite? Does it rotate from West to East or from East to West?
Answer:
Time period of geostationary satellite is equal to time period of rotation of earth.

∴ Time period of geostationary orbit T = 24 hours. Satellites in geostationary orbit will revolve round the earth in west to east direction in an equatorial plane.

Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).

Question 11.
What are polar satellites?
Answer:
Polar satellite :
Polar satellites are low altitude satellites. They will revolve around the poles of the earth in a north-south direction. Time period of polar satellites is nearly 100 minutes.

Short Answer Questions

Question 1.
State Kepler’s Laws of planetary motion. [TS Mar. ’17]
Answer:
Kepler’s Laws :
Law of orbits (1st law):
All planets will move in elliptical orbits with the sun lies at one of its foci.

Law of areas (2nd law) :
The line that joins any planet to the sun sweeps equal areas in equal intervals of time, i.e., \(\frac{\Delta \mathrm{A}}{\Delta \mathrm{T}}\)= constant. i.e., planets will appear to move slowly when they are away from sun, and they will move fast when they are nearer to sun.

Law of periods (3rd law) :
The square of time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet.
i.e., T² ∝ R³ or \(\frac{T^2}{R^3}\) = constant

Question 2.
Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:
Relation between g and G :
Each and everybody was attracted towards centre of earth with some force. This is called weight of the body,
W = mg …………. (1)

This force is due to gravitational pull on the body by the earth.

For small distances above earth from centre of earth is equal to radius of earth ‘R’.

According to Newton’s law of gravitation.

Question 3.
How does the acceleration due to gravity (g) change for the same values of height (h) and depth(d)?
Answer:
Variation of ‘g’ with altitude :
When we go to a height ‘h’ above the ground ‘g’ value decreases.
On surface of earth (g) = \(\frac{GM}{R^2}\)
At an altitude ‘h’ g(h) = \(\frac{GM}{(R+h)^{2}}\) because

R + h is the distance from centre of earth to the given point at ‘h’.
h << R ⇒ g(h) = g(1 – \(\frac{2h}{R}\))
So acceleration due to gravity decreases with height above the ground.

Variation of ‘g’ with depth :
When we go deep into the ground ‘g’ value decreases.

At a depth ‘d’ inside the ground mass of earth upto the point d from centre will exhibit force of attraction on the body. The remaining mass does not exhibit any influence. So effective radius is (R – d) only. Acceleration due to gravity ‘g’ at a depth ‘d’ is given by

So ‘g’ value decreases with depth below the ground.

Question 4.
What is orbital velocity? Obtain an expression for it. [AP Mar. 17, 14; May 18. 14]
Answer:
Orbital velocity (V₀) :
Velocity of a satellite moving in the orbit is called orbital velocityog.

Let a satellite of mass m is revolving round the earth in a circular orbit at a height ‘h’ above the ground.

Radius of the orbit = R + h where R is radius of earth.

In orbital motion is “The centrifugal and centripetal forces acting on the satellite”.

Centrifugal force = \(\frac{mV^2}{r}=\frac{mV^{2}_{0}}{R+h}\) ……… (1)
(In this case V = V₀ and r = R + h)

Centripetal force is the force acting towards the centre of the circle it is provided by gravitational force between the planet and satellite.

V₀ = \(\sqrt{gR}\) is called orbital velocity. Its value is 7.92 km/sec.

Question 5.
What is escape velocity? Obtain an expression for it. [TS Mar. ’19, ’16; AP Mar. ’19. ’18, ’15, ’13. May ’17, ’16]
Answer:
Escape speed (v₁)_min :
It is defined as the minimum velocity required by a body to overcome gravitational field of earth is known as escape velocity.

For a body of mass ‘m’ gravitational potential energy on surface of earth PE = – \(\frac{G.m.M_E}{R_E}\)

For a body to escape from gravitational field of earth its kinetic energy must be equal or more than gravitational potential energy.

Question 6.
What is a geostationary satellite? State [AP Mar. ’16, June ’15; May ’13; TS Mar. ’18, ’15, May ’18, ’16, June ’15]
Answer:
A geostationary satellite will always appears to be stationary relative to earth.

The time period of geostationary satellite is equals to time perfod of rotation of earth.

∴ Time period of geostationary orbit t = 24 hours. Satellites in geostationary orbit will revolve round the earth in west to east direction in an equatorial plane.

Uses of geostationary satellites:

1. For study of the upper layers of the atmosphere.
2. For forecasting the changes in atmosphere and weather.
3. For finding the size and shape of earth.
4. For investigating minerals and ores present in the earth’s crust.
5. For transmission of T.V. signals.
6. For study of transmission of radio waves.
7. For space research.

Question 7.
If two places are at the same height from the mean sea level; One is a mountain and other is in air. At which place will ‘g’ be greater? State the reason for your answer.
Answer:
‘g’ value on the mountain is greater than g’ value in air even though both are at same height.

For a point on mountain while deciding the ‘g’ value, mass of mountain is also considered which leads to change in ‘g’ value depending on local condition such as concentration of huge mass at a particular place. Whereas for a point in air no such effect is considered. Hence ‘g’ on the top of mountain is more.

Question 8.
The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight? State the reason for your answer.
Answer:
If we are using common balance to measure sugar we will get some quantity of sugar both at equator and at poles.

Whereas if we are using spring balance to weigh sugar then weight of sugar at poles is more. So we will get less quantity.

Weight of sugar at equator is less. So we will get more quantity of sugar at equator.

Question 9.
If a nut becomes loose and gets detached from a satellite revolving around the earth, will it fall down to earth or will it revolve around earth? Give reasons for your answer.
Answer:
If a nut is detached from a satellite revolving in the orbit then its velocity is equals to orbital velocity. So it continues to revolve in the same orbit. It does not fall to earth.

Question 10.
An object projected with a velocity greater than or equal to 11.2 km.s^-1 will not return to earth. Explain the reason.
Answer:
Escape velocity of earth is 11.2 km/sec. If any body acquires a velocity of 11.2 km/ sec. or more its kinetic energy is more than gravitational potential energy. So earth is not able to stop the motion of that body. So any body with a velocity 11.2 km/s or more will escape from gravitational field of earth and never comes back to earth.

Long Answer Questions

Question 1.
Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:
Gravitational potential energy of a body at a point in a gravitational field of another. It is defined as the amount of work done in brining the given body from infinity to that point in the field is called Gravitational potential energy.

Expression for gravitational potential energy :
Consider two particles of masses m, and m2 are placed at the points O’ and p respectively. Let the distance between the two particles is r’ i.e., OP = r.

Let us calculate the gravitational potential energy of the particle of mass m₂ placed at point p in the gravitational field of m₁. Join OP and extended it in forward direction. Consider two points A and B on this line such that OA = x and AB = dx.

The gravitational force of attraction on the particle at A is, F = \(\frac{\mathrm{Gm_1m_2}}{\mathrm{x^2}}\)

Small amount of work done in bringing the particle without acceleration through a very small distance AB is, dW = F dx
= \(\frac{\mathrm{Gm_1m_2}}{\mathrm{x^2}}\)

Total workdone in bringing the particle from infinity to the point P is,

Since, this work done is stored in the particle as its gravitational potential energy (U). Therefore, gravitational potential energy of the particle of mass m₂ placed at point p’ in the gravitational field of particle of mass
m₁ at distance r is, U = \(\frac{\mathrm{-Gm_1m_2}}{\mathrm{r}}\)

Here, negative sign shows that the potential energy is due to attractive gravitational force between two particles.

Question 2.
Derive an expression for the variation of acceleration due to gravity (a) above and (h) below the surface of the Earth.
Answer:
Variation of acceleration due to gravity above the surface of earth :
We know ‘g’ on planet, g = \(\frac{GM}{R^2}\). But on earth g’ value changes with height above the ground h’.

Variation of ‘g’ with altitude :
For a point h’ above the earth total mass of earth seems to be concentrated at centre of earth. Now distance from centre of earth is (R + h).
Acceleration due to gravity at ‘h’ = g(h)
= \(\frac{GM_E}{(R_E+h)^2}\)

For small values of ‘h’ i.e., h << R than

Variation of acceleration due to gravity below the surface of earth :
At a depth d’ inside the ground mass of earth upto the point d from centre will exhibit force of attraction on the body. The remaining mass does not exhibit any influence.
Mass of spherical body M ∝ R³

∴ M_S/M_E = (R_E – d)³/ R³^E where M_s is mass of earth’s shell upto a depth ‘d’ from centre. Gravitational force at depth ‘d’ is

So ‘g’ value decreases with depth below the ground.

Question 3.
State Newton’s Universal Law of Gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.
Answer:
Newton’s law of gravitation :
Every body in the universe attracts every other body with a force which is directly proportional to theproduct of their masses and inversely proportional to the square of the distance between them.

This is always a force of attraction and acts along the line joining the two bodies.

Cavendish experiment to find gravitational constant ‘G’ :
Cavendish experiment consists of a long metalic rod AB to which two small lead spheres of mass ‘m’ are attached.

This rod is suspended from a rigid support with the help of a thin wire. Two heavy spheres of mass M are brought near to these small spheres in opposite direction. Then gravitational force will act between the spheres.

Force between the spheres, r = \(\frac{GMm}{d^2}\)

Two equal and opposite forces acting at the two ends of the rod AB will develop force couple and the rod will rotate through an angle ‘θ’.
∴ Torque on the rod F × L = \(\frac{GM.m}{d^2}\) → (1)
Restoring force couple = τθ → (2)
Where τ = Restoring couple per unit twist

By measuring 0 we can calculate ‘G’ value when other parameters are known. Practical value of G is 6.67 × 10^-11 Nm²/kg².

Problems

(Gravitational Constant ‘G’ = 6.67 × 10^-11 Nm²kg^-2; Radius of earth ‘R’ = 6400 km; Mass of earth ‘M_E‘ = 6 × 10²⁴ kg)

Question 1.
Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.
Solution:
Mass of each ball, m = 1 kg;
Separation, r = 1 cm = 10^-2 m
Gravitational force of attraction,

Question 2.
The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the gravitational force between them is 6.67 × 10^-7 N. Find the masses of the two balls.
Solution:
Mass of 1st ball = m;
Mass of 2nd ball = 4m.
Separation, r = 10 cm = 0.1 m;
Mass of the 1st ball = m = ?
Gravitational force, F = 6.67 × 10^-7 N
∴ Mass of the balls are 5 kg, 20 kg.

Question 3.
Three spherical balls of masses 1 kg, 2 kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of the gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass.
Solution:
Side of equilateral triangle, a = lm.
Masses at corners = 1 kg, 2 kg, 3 kg.
Force between 1 kg, 2g = F₁ = G.\(\frac{2\times1}{1^2}\) = 2 G
Force between I kg, 3kg = F₂ = G.\(\frac{3\times1}{1^2}\)= 3G.
Now F₁ & F₂ act with an angle of 60°
∴ Resultant force

Question 4.
At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of the earth. Determine the height.
Solution:
Acceleration due.to gravity at a height, h = 4% of g.
Radius of earth, R = 6400K.M. = 6.4 × 10⁶m.

Question 5.
A satellite orbits the earth at a height of 1000 km. Find its orbital speed.
Solution:
Radius of earth, R = 6,400 km = 6.4 × 10⁶ m;
Mass of earth, M = 6 × 10²⁴
Height of satellite h = 1000 km;
G = 6.67 × 10¹¹ N – m² /kg²

Question 6.
A satellite orbits the earth at a height equal to the radius of earth. Find it’s (i) orbital speed and (ii) Period of revolution.
Solution:
Radius of earth, R = 6400 k.m.;
-height above earth, h = R.
Mass of earth, M = 6 × 10²⁴
G = 6.67 × 10^-11 N – m²/kg²

Question 7.
The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:
Let force between the objects = F;
Distance between them = r.
For Case II distance, r₁ = (r + 4);
New force, F₁ = 36% less than F

⇒ 100 r² = 64 (r + 4)² Take square roots on both sides.
10r = 8 (r + 4) ⇒ 10 r = 8r + 32
⇒ (10 – 8) r = 2r = 32
∴ r = 16 m.

Question 8.
Four identical masses m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:
Given all masses are equal
∴ m₁ = m₂ = m₃ = m₄
Force between m₁, m₄ = F₁ = \(\frac{G.m^2}{a^2}\) ……… (1)
Force between m₄, m₃ = F₂ = \(\frac{G.m^2}{a^2}\) ……… (2)
Forces F₁ and F₂ act perpendicularly.

Their magnitudes are equal.

Now forces F_R and F₃ are like parallel. So resultant is sum of these forces.
Total force at m₄ due to other masses

Question 9.
Two spherical balls of 1 kg and 4 kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:
Mass, m₁ = 1 kg ; Mass, m₂ = 4 kg ;
Separation, d = 12 cm
Mass of 3rd body m₃ = ?

For m₃ not to experience any force the condition is
Force between m₁, m₃ = Force between m₂, m₃.

Take square roots on both sides,
d – x = 2x ⇒ d = 3x or x = \(\frac{12}{3}\) = 4 cm
∴ Distance from 1 kg mass = 4 cm

Question 10.
Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:
Mass m and radius R are same for all spheres.
Force between 1, 3 spheres = F₁ = \(\frac{G.m^2}{(2R)^2}\)
Force between 1, 2 spheres = F₂ = \(\frac{G.m^2}{(2R)^2}\)

Now F₁ and F2 will act with an angle θ = 60° between them so from Parallelogram law

Question 11.
Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds is 2 : 1. If one of them is at a height of 100 km what is the height of the other satellite?
Solution:
Mass of earth, m = 6 × 10²⁰ kg ;
G = 6.67 × 10^-11 N-m² / kg²
Ratio of orbital velocities V₀₁ : V₀₂ = 2 : 1;
Height of one satellite, h = 100 k.m

4R + 4h₂ = R + h₁ ⇒ h₁ = 3R + 4h₂. ;
Put h₂ = 100 km
∴ h₁ = 3 × 6400 + 400 = 19600 km.

Question 12.
A satellite is revolving round in a circular orbit with a speed of 8 km/ s^-1 at a height where the value of acceleration due to gravity is 8 m/s^-2. How high is the satellite from the Earth’s surface? (Radius of planet 6000 km.).
Solution:
Orbital velocity of satellite, V₀ = 8 km/s.
= 8 × 10³ m/s.
Acceleration due to gravity in the orbit = g
= 8 m/s²

Orbital velocity, V = \(\sqrt{gR}\)
where R is radius of the orbit and g is acceleration due to gravity in the orbit.

Height of satellite = 8000 – radius of earth ;
Radius of earth = 6000 km.
∴ Height above earth = 8000 – 6000
= 2000 km.

Question 13.
(a) Calculate the escape velocity of a body from the Earth’s surface, (b) If the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood?
Solution:
Radius of earth, R = 6400 km = 6.4 × 10⁶ m.
Mass of earth, M = 6 × 10²⁴ kg; g = 9.8 ms^-2.
a) Escape velocity, V_e = \(\sqrt{2gR}\)
∴ V_e = \(\sqrt{2\times9.8\times6.4\times10^6}\) = 11.2 km/s

b) If earth is made of wood mass,
M₁ = 10% of M = 6 × 10²³

Additional Problems

Question 1.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
A comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 2.
A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.5 × 10⁸ km away from the sun? Solution:
Here, T_s = 29.5 T_e; R_e = 1.5 × 10⁸ km; R_s = ?

Question 3.
A body weighs 63 N on the surface of Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:
Weight of body = mg = 63 N
At height h, the value of g’ is given by,

Question 4.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of earth if it weighed 250 N on the surface?
Solution:
Weight of body at a depth, d = mg’

Telangana TSBIE TS Inter 1st Year Physics Study Material 8th Lesson Oscillations Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 8th Lesson Oscillations

Very Short Answer Type Questions

Question 1.
Give two examples of periodic motion which are not oscillatory.
Answer:

1. Motion of seconds hand of a watch.
2. Motion of fan blades which are rotating with constant angular velocity ‘w’.

For these two cases, they have constant centrifugal acceleration which does not change with rotation so it is not considered

Question 2.
The displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by2π/ω?
Answer:
Displacement :
Displacement remains constant ; \(\frac{2 \pi}{\omega}\) = time period T. After a time (0
period T, there is no change in equation of S.H.M.
i.e. Y = A sin (20t + 4) = Y = A sin (201 + 4 + T)
∴ There is no change in change in displacement.

Question 3.
A girl is swinging seated in a swing. What is the effect on the frequency of oscillation if she stands?
Answer:
The frequency of oscillation (n) will increase because in the standing position, the location of centre of mass of the girl shift upwards. Due to it, the effective length of the swing decreases. As n ∝ \(\frac{1}{\sqrt{l}}\), therefore, n increases.

Question 4.
The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere?
Answer:
When water begins to drain out of the sphere, the centre of mass of the system will first move down and then will come up to the initial position. Due to this the equivalent length of the pendulum and hence time period first increases, reaches a maximum value and then decreases till it becomes equal to its initial value.

Question 5.
The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum?
Answer:
The time period of a simple pendulum does not change, if the wooden bob is replaced by an identical bob of aluminium because the time period of a simple pendulum is independent of the material of the bob.

Question 6.
Will a pendulum clock gain or lose time when taken to the top of a mountain?
Answer:
At higher altitudes i.e., on mountains the acceleration due to gravity is less as compared on the surface of earth. Since time period is inversely proportional to the square root of the acceleration due to gravity, the time period increases. The pendulum clock loses time on the top of a mountain.

Question 7.
What is the length of a simple pendulum which ticks seconds? (g = 9.8 ms^-2) [AP Mar. ’18: TS Mar. ’15]
Answer:
In simple pendulum T = 2π\(\sqrt{\frac{l}{g}}\) or l = \(\sqrt{\frac{gt^2}{4\pi^2}}\)
For seconds pendulum T = 2s ⇒ t² = 4
∴ = \(\frac{9.8\times4}{4\pi^2}\) = 1 m (∴ π² nearly 9.8)

Question 8.
What happens to the time period of a simple pendulum if its length is increased upto four times?
Answer:
In simple pendulum Time period T ∝ √l

From the above equation time period is doubled.

Question 9.
A pendulum clock gives correct time at the equator. Will it gain or lose time if it is taken to the poles? If so, why?
Answer:
When a pendulum clock showing correct time at equator is taken to poles then it will gain time.

Acceleration due to gravity at poles is high. Time period of pendulum T = 2π\(\sqrt{\frac{l}{g}}\).

When g increases T decreases. So number of oscillations made in the given time increases hence clock gains time.

Question 10.
What fraction of the total energy is K.E when the displacement is one half of a amplitude of a particle executing S.H.M?
Answer:
Kinetic energy is equal to three fourth (i.e.,\(\frac{3}{4}\)) of the total energy, when the displacement is one-half of its amplitude.

Question 11.
What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?
Answer:
Energy of a simple harmonic oscillator,

From the above equation, energy increases by four times.

Question 12.
Can a simple pendulum be used in an artificial satellite? Give the reason
Answer:
No, this is because inside the satellite, there is no gravity, i.e., g = 0. As T = 2π\(\sqrt{\frac{l}{g}}\) where T = ∞ for g = 0. Thus, the simple pendulum will not oscillate.

Short Answer Questions

Question 1.
Define simple harmonic motion? Give two examples.
Answer:
Simple Harmonic Motion :
A body is said to be in S.H.M, if its acceleration is directly proportional to its displacement, acts opposite in direction towards a fixed point.

Examples:

1. Projection of uniform circular motion on a diameter.
2. Oscillations of simple pendulum with small amplitude.
3. Oscillations of a loaded spring.
4. Vibrations of a liquid column in U – tube.

Question 2.
Present graphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M.
Answer:
The variations of displacement, velocity and acceleration with time for a particle in S.H.M can be represented graphically as shown in the figure.

From the graph

1. All quantities vary sinusoidally with time.
2. only their maxima differ and the different plots differ in phase.
3. Displacement x varies between – A to A; v(t) varies from – ωA to ωA and a (t) varies from – ω²A to ω²A.
4. With respect to displacement plot, velocity plot has a phase difference of \(\frac{\pi}{2}\) and acceleration plot has a phase difference of π.

Question 3.
What is phase? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
Answer:
Phase (θ) :
Phase is defined as its state or condition as regards its position and direction of motion at that instant.
In S.H.M phase angle, θ = ωt = 2π(\(\frac{t}{T}\))

a) Phase between velocity and displacement :
In S.H.M, displacement,
y = A sin (ωt – Φ)
Velocity, V = Aω cos (ωt – Φ)
So phase difference between displacement and velocity is 90°.

b) Phase between displacement and acceleration :
In S.H.M, acceleration ‘a’ = – ω²y
or y = A sin ωt and a = – ω² A sin ωt
– ve sign indicates that acceleration and displacement are opposite.

So phase difference between displacement and acceleration is 180°.

Question 4.
Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached.
Answer:
Let a spring of negligible mass is suspended from a fixed point and mass m is attached as shown. It is pulled down by a small distance ‘x’ and allowed free it will execute simple harmonic oscillations.

Displacement from mean position = x.
The restoring forces developed are opposite to displacement and proportional to ‘x’. ∴ F ∝ – x or F = – kx where k is constant of spring, (-ve sign for opposite direction)

Question 5.
Derive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
Answer:
Expression for K.E of a simple harmonic oscillator :
The displacement of the body in S.H.M., X = A sin ωt
where A = amplitude, ωt = Angular displacement.

Velocity at any instant, v = \(\frac{dx}{dt}\) = Aω cos ωt
∴ K.E = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\)mA²ω² cos² ωt

At mean position velocity is maximum and displacment x = 0
∴ K.E_max = \(\frac{1}{2}\)mA²ω²

Expression for P.E of a simple harmonic oscillator:
Let a body of mass m’ is in S.H.M with an amplitude A.

Let O is the mean position.
Equation of a body in S.H.M is given by, x = A sin ωt
For a body in S.H.M acceleration, a = – ω²Y
Force, F = ma = – mω²x
∴ Restoring force, F = mω²x

Potential energy of the body at any point say ‘x’ :
Let the body is displaced through a small distance dx
Work done, dW = F . dx = P.E.
This work done.
∴ P.E = mω²x. dx(where x is its displacement)
Total work done, W = ∫dW = \(\int_0^x m \omega^2 x\).dx
⇒ Work done, W = \(\frac{m\omega^2x^2}{2}\).
This work is stored as potential energy.
∴ P.E at any point = \(\frac{1}{2}\)mω²x²

Question 6.
How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:
The total energy of a simple pendulum is,
E = \(\frac{1}{2}\)mA² (or) E = \(\frac{1}{2}\frac{mg}{l}\)A²

The above equation, shows that the total energy of a simple pendulum remains constant irrespective of the position at any time during the oscillation i.e., the law of conservation of energy is valid in the case of a simple pendulum. At the extreme positions P and Q the energy is completely in the form of potential energy and at the mean position 0 it is totally converted as kinetic energy.

At any other point the sum of the potential and kinetic energies is equal to the maximum kinetic energy at the mean position or maximum potential energy at the extreme position. As the bob of the pendulum moves from P to O, the potential energy decreases but appears in the same magnitude as kinetic energy. Similarly as the bob of the pendulum moves from 0 to P or Q, the kinetic energy decreases to the extent it is converted into potential energy, as shown in figure.

Question 7.
Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.
Answer:
Displacement of a body in S.H.M.
X = A cos (ωt + Φ).

i) Displacement (x) :
At t = 0 displacement x = A i.e., at extreme position when ωt + Φ = 90° displacement x = 0 at mean position at any point x = A cos (ωt + Φ).

ii) Velocity (V): Velocity of a body in S.H.M.

When (ωt + Φ) = 0 then velocity v = 0. For points where (ωt + Φ) = 90°
Velocity V = – Aω i.e., velocity is maximum,

iii) Acceleration (a): Acceleration of a body in S.H.M. is a = \(\frac{dv}{dx}\)
= \(\frac{d}{dt}\)(-Aω sin(ωt + Φ) = -Aω²cos(ωt + Φ) = -ω²x)
a_max = -ω²A

Long Answer Questions

Question 1.
Define simple harmonic motion. Show that the motion of projection of a particle performing uniform circular motion, on any diameter is simple harmonic. [TS May 18, Mar. 16, June 15; AP Mar. ;19, 18, AP May 16, 14]
Answer:
Simple harmonic motion :
A body is said to be in S.H.M, if its acceleration is directly proportional to its displacement, acts opposite in direction towards a fixed point.

Relation between uniform circular motion and S.H.M.:
Let a particle ‘P’ is rotating in a circular path of radius ‘ω’ with a uniform angular velocity ‘P’. After time ‘t’ it goes to a new position ‘P’. Draw normals from ‘P’ on to the X – axis and on to the Y – axis. Let ON and OM are the projections on X and Y axis respectively.

As the particle is in motion it will subtend an angle θ = ωt at the centre.

From triangle OPN
ON = OP cos θ
But OP = r and θ = ωt
∴ Displacement of particle P on X – axis at any time t is
X = r cos ωt ………… (1)
From triangle OPM
OM = Y = OP sin θ
But OP = r and θ = ωt
∴ Displacement of particle P on Y- axis is
Y = r sin ωt ………… (2)

As the particle rotates in a circular path the foot of the perpendiculars OM and ON will oscillate with in the limits X to X¹ and Y to Y¹.

At any point the displacement of particle P is given by OP² = OM² = ON²
Since OM = X = r cos ωt and ON = Y = r sin ωt.

So a uniform circular motion can be treated as a combination of two mutually perpendicular simple harmonic motions.

Question 2.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is a seconds pendulum? [TS Mar. 18, 17, 15, May 17, 16; AP Mar. 17, 16. 15, 14, 13; AP May 18. 17. 13; June 15]
Answer:
Simple pendulum :
Massive metallic bob is suspended from a rigid support with the help of inextensable thread. This arrangement is known as simple pendulum.

So length of simple pendulum is ‘l’. Let the pendulum is pulled to a side by a small angle ‘θ’ and released it oscillate about the mean position.

Let the bob is at one extreme position B. The weight (W = mg) of body acts vertically downwards.

By resolving the weight into two perpendicular components :

1. One component mg sin θ is responsible for the to and fro motion of pendulum.
2. Other component mg cos θ will balance the tension in the string.

Force useful for motion F = mg sin θ = ma (From Newton’s 2nd Law)
From the above equations
∴ a = g sin θ

Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of simple pendulum is “simple harmonic”.

Time period of simple pendulum :

Seconds pendulum :
A pendulum whose time period is 2 seconds is called “seconds pendulum.”

Question 3.
Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
Answer:
Expression for K.E of a simple harmonic oscillator :
The displacement of the body in S.H.M, X = A sin ωt
where A = amplitude and ωt = Angular displacement.

Velocity at any instant, v = \(\frac{dx}{dt}\) = Aω cos ωt

At mean position velocity is maximum and displacement x = 0
∴ K.E_max = \(\frac{1}{2}\)mA²ω²

Expression for P.E of a simple harmonic oscillator :
Let a body of mass’m’ is in S.H.M with an amplitude A.

Let O is the mean position.
Equation of a body in S.H.M is given by, x = A sin ωt
For a body in S.H.M acceleration, a = – ω²Y
Force, F = ma = – mω²x
∴ Restoring force, F = mω²x

Potential energy of the body at any point say ‘x’:
Let the body is displaced through
a small distance dx
⇒ Work done, dW = F . dx
This work done = RE. in the body
∴ P.E = mω²x. dx(where x is its displacement)
Total work done, W = ∫dW = \(\int_0^x m \omega^2 x\).dx
work done, W = \(\frac{m\omega^2 x^2}{2}\)
This work is stored as potential energy.
∴ P.E at any point = \(\frac{1}{2}\)ω²x²
For conservative force total Mechanical Energy at any point = E= P.E + K.E
∴ Total energy,
E = \(\frac{1}{2}\)mω²(A² – x²) + \(\frac{1}{2}\)mω²x²
E = \(\frac{1}{2}\)mω²{A² – x² + x²} = \(\frac{1}{2}\)mω²A²

So for a body in S.H.M total energy at any point of its motion is constant and equals to \(\frac{1}{2}\)mω²A²

Problems

Question 1.
The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely? Why?
Solution:
If the hollow brass sphere is completely filled with water, then time period of simple pendulum does not change. This is because time period of a pendulum is independent of mass of the bob.

Question 2.
Two identical springs of force constant “k” are joined one at the end of the other On series). Find the effective force constant of the combination.
Solution:
When two springs of constant k each are joined together with end to end in series then effective spring constant k = \(\frac{k_1k_2}{k_1+k_2}\) in this case k_eq = \(\frac{k.k}{k+k}=\frac{k}{2}\)

In series combination, force constant of springs decreases.

Question 3.
What are the physical quantities having maximum value at the mean position in SHM?
Solution:
In S.H.M at mean position velocity and kinetic energy will have maximum values.

Question 4.
A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period? [TS June ’15]
Solution:
Given maximum velocity, V_max = \(\frac{1}{2}\) maximum acceleration (a_max)
But V_max = Aw and a_max = ω² A
∴ Aω = \(\frac{1}{2}\) . Aω² ⇒ ω = 2
Time period of the body, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{2}\)

Question 5.
A mass of 2 kg attached to a spring of force constant 260 Nm^-1 makes 100 oscillations. What is the time taken?
Solution:
Mass attached, m = 2 kg ; Force constant, k = 260 N/m
∴ Time period of loaded spring, T = 2π\(\sqrt{\frac{m}{k}}\)
= 2π\(\sqrt{\frac{2}{260}}\) = 0.5509 sec
∴ Time for 100 oscillations = 100 × 0.551
= 55.1 sec

Question 6.
A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the lift (i) moves up with uniform velocity (ii) moves down with uniform velocity (iii) moves up with uniform acceleration ‘a’ (iv) moves down with uniform acceleration ‘a’ (v) begins to fall freely under gravity?
Solution:
i) When the lift moves up with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

ii) When the lift moves down with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

iii) When lift is moving up with acceleration ‘a’ then relative acceleration = g + a
∴ Time period, T = 2 π\(\sqrt{\frac{l}{g+a}}\) so when lift is moving up with uniform acceleration time period of pendulum in it decreases.

iv) When lift is moving down with acceleration ‘a’ time period, T = 2π\(\sqrt{\frac{l}{g-a}}\)
(g – a = relative acceleration of pendulum)
So time period of pendulum in the lift decreases.

v) If the lift falls freely, a = g then the time period of a simple pendulum becomes infinite.

Question 7.
A particle executing SHM has amplitude of 4cm, and its acceleration at a distance of 1cm from the mean position is 3cms^-2. What will its velocity be when it is at a distance of 2cm from its mean position?
Solution:
Amplitude, A = 4cm = 4 × 10^-2m
Acceleration, a = 3cm/s² = 3 × 10^-2 m/s²;
Displacement, y = 1cm = 10^-2 m
∴ Angular velocity, ω = \(\sqrt{\frac{a}{y}}=\sqrt{\frac{3}{1}}=\sqrt{3}\)

To find velocity at a displacement of 2cm

Question 8.
A simple harmonic oscillator has a time period of 2s. What will be the change in the phase after 0.25 s after leaving the mean position?
Solution:
Time period, T = 2 sec; time, t = 0.25 sec
Phase difference after t sec = Φ = \(\frac{t}{T}\) × 2π
= \(\frac{0.25}{2}\) × 2π = \(\frac{2 \pi}{4}\) = 90°
For a phase of \(\frac{2 \pi}{4}\) starting from mean position the body will be at extreme position. (Phase difference between mean position and extreme position is \(\frac{2 \pi}{4}\) Rad or 90°)

Question 9.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.
Solution:
Given that, A = 5 cm = 5 × 10^-2m and T = 0.2 s
Angular velocity, ω = \(\frac{2 \pi}{T}=\frac{2 \pi}{0.2}\) = 10π rad s^-1

a) Displacement, y = 5 cm = 5 × 10^-2 m
i) Acceleration of the body, a = – ω²y
= -(10π)² × 5 × 10^-2 = -5π²ms^-2
ii) Velocity of the body,

b) Displacement, y = 3 cm = 3 × 10^-2 m
i) Acceleration of the body, a = – ω²y
= -(10π)² × 3 × 10^-2 = -3π²ms^-2
ii) Velocity of the body, v = ω\(\sqrt{A^2 – y^2}\)

= 10π × 4 × 10^-2 = 0.4π ms^-1

c) Displacement, y = 0 cm
i) Acceleration of the body, a = – ω²y = 0
ii) Velocity of the body, v = ω\(\sqrt{A^2 – y^2}\)
= 10π^(5xl0’2)2-(0)2\(\sqrt{(5\times10^{-2})^2-(0)^2}\)
= 10π × 5 × 10^-2 =0.5π ms^-1

Question 10.
The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
Solution:
Mass of planet, M_P = 2 M_e ;
Radius of planet, R_P = 2R_e
Time period of pendulum on earth = T ;
Time period on planet = T’

Question 11.
Calculate the change in the length of a simple pendulum of length lm, when its period of oscillation changes from 2 s to 1.5 s. [TS Mar. ’18]
Solution:
For seconds pendulum T₁ = 2 sec ;
Length l₁ = 1 m.
New time period T₂ = 1.5 sec; Length l₂ = ?

Question 12.
A freely falling body takes 2 seconds to reach the ground on a plane, when it is dropped from a height of 8m. If the period of a simple pendulum is seconds on the planet. Calculate the length of the pendulum.
Solution:
Height, h = 8m;
Time taken to reach the ground, t = 2 sec
But for a body dropped, t = \(\sqrt{\frac{2h}{g}}\)
⇒ 2 = \(\sqrt{\frac{16}{g}}\) ⇒ g = \(\frac{16}{4}\) = 4m/s² on that planet
Time period of pendulum, T = 2π\(\sqrt{\frac{l}{g}}\) = π
∴ 2\(\sqrt{\frac{l}{g}}\) = 1 or \(\frac{l}{g}=\frac{1}{4}\) ⇒ l = \(\frac{g}{4}\)
Length of pendulum = \(\frac{4}{4}\) = 1m = 100cm on that planet

Question 13.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period.
Find the length of a simple pendulum which ticks seconds, (g = 9.8 ms^-2) [AP Mar. ’18. ’16, ’15, May ’17, June ’15; TS Mar.’17, 15, May 17]
Solution:
Simple pendulum :
In a laboratory a heavy metallic bob is suspended from a rigid support with the help of a spunless thread. This arrangement is known as “simple pendulum”.

Let the length of simple pendulum is ‘l’ and the point of suspension is ‘S’. Let the pendulum is drawn to a side by a small angle ‘θ’ and allowed free to oscillate in the vertical plane. Then it will oscillate between the extreme positions A and B with a displacement say ‘x’ at any given time.

Let the bob is at one extreme position say B. The force vertically acting downwards is Weight W = mg.

By resolving the weight into two per-pendicular components:

1. The component mg sin θ is responsible for the to and fro motion of the bob.
2. The component mg cos θ will balance the tension in the string.

Force useful for motion F = mg sin θ
= ma (From Newton’s 2nd Law)
∴ a = g sin θ

Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of simple pendulum is “simple harmonic”.

Time period of simple pendulum :

Problem:
In simple pendulum T = 2π\(\sqrt{\frac{l}{g}}\) or l = \(\frac{gt^2}{4\pi^2}\)
For seconds pendulum T = 2s ⇒ t² = 4
∴ l = \(\sqrt{\frac{9.8\times4}{4\pi^2}}\) = 1 m (∴ π² nearly 9.8)

Question 14.
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m/s².
Solution:
a) Increase in length of pendulum = 0.6m ;
Increase in time period = 50% = 1.5T
Let original length of pendulum = 1
Original time period = T; g = 9.8 m/s².
For 1st case 9.8 = π² \(\frac{1}{T^2}\) → 1 ;
For 2nd case l₁ = (l + 0.6), T₁ = 1.5 T

But l₁ = l + 0.6 ;
∴ l + 0.6 = 2.25l ⇒ 0.6 = 1.25l
∴ Length of pendulum l = \(\frac{0.6}{1.25}\) = 0.48 m
b) Time period T = 2π\(\sqrt{\frac{l}{g}}\)
= 2 × 3.142\(\sqrt{\frac{0.48}{9.8}}\) = 6.284 × 0.2213
= 1.391 sec.

Question 15.
A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.02m. How much will the clock gain or lose in one day?
Solution:
Time period of seconds pendulum,
T = 2 sec
Length of seconds pendulum,
L = gT² / 4π² = 0. 9927 m
Length of seconds pendulum during summer = 1.02 m
∴ Error in length, ∆l = 1.02 – 1 = 0.0273
In pendulum T × √l. From principles of error

Question 16.
The time period of a body suspended from a spring is T. What will be the new time period, if the spring is cut into two equal parts and (i) the mass is suspended from one part? (ii) the mass is suspended simultaneously from both the parts?
Solution:
Time period of spring, T = 2π\(\sqrt{\frac{m}{K}}\)
When a spring is cut into two equal parts force constant of each part K₁ = 2K

ii) When mass is suspended simultaneously from two parts ⇒ they are connected in parallel. For springs in parallel K_p = K₁ – K₂ = 4K

Question 16.
What is the length of a seconds pendulum on the earth? [AP Mar. ’17, ’16; June ’15; TS Mar. ’17]
Solution: