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Telangana TSBIE TS Inter 2nd Year Physics Study Material 16th Lesson Communication Systems Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 16th Lesson Communication Systems

Very Short Answer Type Questions

Question 1.
What are the basic blocks of a communication system? [TS June ’15]
Answer:
The basic blocks of a communication system are

1. Transmitter
2. Communication channel
3. Receiver.

Question 2.
What is “world wide web” (www)? [IMP]
Answer:
It is an encyclopedia of knowledge accessible to everyone round the clock throughout year with the help of computer connectivity.

Question 3.
Mention the frequency range of speech signals.
Answer:
Frequency range of speech signals is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation? (IMP)[AP May ’16; June ’15]
Answer:
In the frequency range from a few MHz upto about 30 MHz, long distance communication can be achieved by the ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave pro-pagation and it is used by short wave broadcast services.

Question 5.
Mention the various parts of the ionosphere? [TS May ’16]
Answer:
Parts of ionosphere are

1. D – Layer at a height of 65 to 75 km from ground.
2. E – Layer at a height of nearly 100 km.
3. F₁ – Layer at a height of 170 to 190 km.
4. F₂ – Layer at a height of 250 to 400 km.
   Note : F₁ and F₂ will merge during night.

Question 6.
Define modulation. Why is it necessary? [AP Mar. 18, 17, 16, 14; May 17, 14; TS Mar. 19, 16, 15, May 18, 17]
Answer:
Modulation :
The process of combining audio frequency signal with high frequency signal is called modulation. Modulation is necessary for the following reasons.

1. to reduce the size of antenna.
2. to increase the effective power radiated by antenna.
3. to avoid mixing up of signals from different transmitters.

Question 7.
Mention the basic methods of modulation [AP Mar. 19. 16; TS Mar.’ 18, ’17, ’15]
Answer:
The basic methods of modulation are :
a) Amplitude modulation
b) Frequency modulation and
c) Phase modulation.

Question 8.
Which type of communication is employed in mobile phones? [AP May ’18, Mar. ’15]
Answer:
Space wave communication is employed in mobile phones.

Short Answer Questions

Question 1.
Draw the block diagram of a generalized communication system and explain it briefly.
Answer:
The block diagram of a generalized communication system are as shown. It consists of the following three parts.

1. Transmitter
2. Communication channel and
3. Receiver.

In this system, the transmitter and receiver are located at two different places and the channel is the physical medium that connects them depending upon the type of communication system. A channel may be in the form of wires (or) cables connecting the transmitter and the receiver, or it may be wireless.

Using the transmitter, the original message is converted into a form so that it can be transmitted over the communication channel. Hence it is called as a-message signal.

The uses of these three parts are :
1) Transmitter :
Purpose of transmitter is to convert the message signal produced by source into a form suitable for transmission through channel.

2) Channel :
Channel is used for transmission of signals. It may consist a medium like coaxial cable, optical fibre, (or) even space is used as channel in wireless communication where E.M waves are used.

3) Receiver :
Receiver will convert the signals received through channel into a recognisable form of the original message signal.

Question 2.
What is a ground wave? When is it used for communication?
Answer:
Ground wave :
In A.M. broadcasting ground based vertical metallic towers called antennas are used for transmission. For such antennas ground has strong influence on transmission. Near transmitting antennas energy of E.M wave is high. They have strong influence on propagation of signals along the ground surface. A strong E.M. wave near antenna will induce current on surface of ground over which it passes. So E.M waves slides along surface of ground.

For ground wave propagation attenuation on surface of earth is high. The magnitude of attenuation is proportional to frequency. Due to very high energy absorption of ground.

The ground wave propagation is limited to few kilometers from transmitting antenna.

Question 3.
What are sky waves? Explain sky wave propagation, briefly.
Answer:
Sky waves :
The electromagnetic waves which suffers reflection of ionosphere and reaches earth are called sky waves.

Sky wave propagation :
Electromagnetic waves with in the range of 30 to 40 MHz are reflected by ionosphere. The degree of ioni-sation varies with height above ground. Due to ionospheric reflection long range com-munication is possible with sky waves. Because these reflected waves reach earth at a longer distance. Short wave broadcast for long distance is possible with sky waves. Generally E.M waves with in the frequency range of 3 to 30 MHz are highly suitable for sky wave propagation. E.M waves of higher frequencies i.e., frequency > 30 MHz will penetrate ionosphere. Hence sky wave propagation is not possible with high frequency electro magnetic waves.

Question 4.
What is space wave communication? Explain.
Answer:
Space wave :
A space wave will travel from transmitting antenna to receiver in the form of straight lines. Space waves will obey line of sight properties.

Space wave propagation :
This type of propagation of E.M waves is called space wave propagation.

Space wave propagation is mainly used in television transmission. Due to curvature of earth range of space wave propagation is limited to certain distance only. Range depends on height of the antenna. Let height of transmitting antenna is ‘h_T’ and radius of earth is ‘R’ the distance upto which signals will reach with LOS (hne-of-sight) properties is d_r = \(\sqrt{2Rh_T}\). If receiving antenna is at a height h_R then also distance d_r upto which signals can be reached with LOS properties will increase. In this case d_r = \(\sqrt{2Rh_T}+\sqrt{2Rh_R}\).

Question 5.
What do you understand by modulation? Explain the need for modulation.
Answer:
Modulation :
The process of super imposing a low frequency signal onto a high frequency carrier wave is known as modulation.

Modulation is necessary for long range transmission of signals.

Modulation is a very essential part of communication. Transmission of low frequency of voice signals (frequency range 20 Hz to 20,000 Hz) to longer distances is not possible because energy associated with low frequency signals is less.

So our voice signals after converting into electrical signals must be superposed on to a high frequency signal called carrier wave. Generally frequency of carrier wave is high. This will help for efficient transmis¬sion and reception. Modulation is necessary for the following reasons,

1. To reduce the size of antenna for E.M. waves to be transmitted as antenna is necessary.
2. To increase effective power radiation by transmitter.
3. To avoid mixing up of signals from different transmitters.

To obtain the above advantages we are using modulation techniques where a low frequency signal is superposed on a high requency carrier wave.

Question 6.
What should be the size of the antenna or aerial? How the power radiated is related to length of the antenna and wavelength?
Answer:
Antenna and its size :
It is a must for every transmitter to have an antenna. Without antenna, it is not possible to radiate electrical energy output of transmitter into space in the form of electromagnetic radiation.

Antenna size must be comparable to wavelength of the signal to be transmitted.

To transmit all information contained in a signal without any loss in quality the minimum size of antenna must be at least \(\frac{\lambda}{4}\).

For good transmission output power of transmitter must be high.

Relation between length of antenna ‘l’ and wavelength ‘ λ’ and power radiated is
Power radiated ∝ (\(\frac{1}{\lambda}\))²

Question 7.
Explain amplitude modulation.
Answer:
Amplitude modulation (A.M.) :
In amplitude modulation (A.M.) the amplitude of carrier wave is varied in accordance with the voice signal superimposing on it.

Theory :
Let time varying carrier wave is c(t) = A_c sin ω_ct

Time varying modulating wave is m(t) = A_m sin ω_mt

when these two waves are superposed modulated wave is given by
c_m(t) = (A_c + A_m sin ωt) sin ω_ct

Because modulating wave changes only amplitude of carrier wave but not its frequencies

is the modulated wave. Where (ω_c – ω_m) and (ω_c + ω_m) are lower and upper side frequencies. Production of amplitude modulated wave.

Question 8.
How can an amplitude modulated wave be generated?
Answer:
The block diagram to produce amplitude modulated wave is as shown. Here modulating signal A_m sin ω_mt is mixed with a carrier signal A_c sin ω_ct in a square law device.

The output of square law device y(t) = B x (t) + Cx²t

Where B and C are constants.
∴ y(t) = BA_m sin ω_mt + BA_c sin ω_ct + C (A_m sin ω_mt + A_c sin ω_ct)² …………… (1)

By using trigonometric relations (1)
sin²A = (1 – cos 2A) / 2

and (2) sin A sin B = \(\frac{1}{2}\) [cos (A – B) – cos (A + B)] equation (1) can be written as
y(t) = BA_msin ω_mt + BA_c sin ω_ct
+ C\(\frac{\mathrm{A}_{\mathrm{m}}^2}{2}\) + A²_c – \(\frac{c\mathrm{A}_{\mathrm{m}}^2}{2}\) cos 2ω_mt – cos 2ω_mt
\(\frac{C\mathrm{A}_{\mathrm{c}}^2}{2}\) cosω_ct + cA_mA_ccos(ω_c – ω_m
CA_mA_c cos(ω_c

This signal is passed through a b filter which rejects the D.C compc sinusoidal frequencies ω_m , 2ω_m a, ailows the frequencies ω_c, ω_c ω_m + ω_m at the out put side. This me signal is transmitted through ante

In this way amplitude modulat is produced.

Question 9.
How can an amplitude modulated wave be detected?
Answer:
The block diagram of amplitude modulated wave detector is as shown.

The receiving antenna will collect the amplitude modulated waves from space. These are passed through an amplifier to increase strength of signals.

These signals are passed through intermediate frequency stage (I.F. stage) to frequency of carrier.

For recovery of original massage signal m(t) with angular frequency ω_m the modulated signal is passed detector which consists (1) a rectifier where lower half (- ve half) p modulated signals is eliminated. (2) Later it was sent through envelope detector to recive the original signal.

This signal is passed through a amplifier and finally signal with sufficient strength is going to a microphone to convert electrical signals into audio signals.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Type Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:
When intrinsic Germanium / Silicon crystal is doped with a pentavalent impurity then “n-type semiconductor” will be formed.

The majority charge carriers are electrons and minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? [AP May ’18, Mar, ’15; June ’15]
Answer:
Intrinsic semiconductor :
Ultra high pure semiconductor are called “intrinsic semi-conductor”.

Extrinsic semiconductor :
The doped semi-conductor is called extrinsic semiconductor.

Question 3.
What is a p-type semiconductor? What are the majority and minority charge carriers in it? [TS May ’18, Mar. ’17; AP Mar. ’17]
Answer:
When third group impurities like boron, aluminium, galium, indium etc., are added to intrinsic semiconductor then it is called “p-type semiconductor.”

In p-type semiconductor majority, charge carriers are holes and minority charge carriers are electrons.

Question 4.
Give examples of “photosensitive substances”. Why are they called so? [AP May ’16]
Answer:

1. Metals like zinc, cadmium and magnesium will respond to ultraviolet rays.
2. Alkalimetals such as sodium, potassium, caesium and rubidium will respond to visible light.

These substances are called photo sensitive surfaces because they will emit electrons when light falls on them.

Question 5.
What is p-n junction diode? Define depletion layer. [TS Mar. ’19 May ’16]
Answer:
p-n semiconductor diode :
When a semiconductor material such as silicon or germanium crystal is doped in such a way that one side of it becomes a p-type and the other side becomes n-type then a p-n semi-conductor diode is formed.

Depletion layer :
The formation of a narrow region on either side of the junction which becomes free from mobile charge carriers is called “depletion layer”.

Question 6.
How is a battery connected to a Junction diode in i) forward and ii) reverse bias?
Answer:
In Forward Bias :
In a p-n junction diode if p-side is connected to positive terminal of a battery and n – side to negative terminal it is called “forward biased”.

In Reverse Bias :
If p-side is connected to negative terminal of the battery and n-side to positive terminal of the cell it is called “reverse biased”.

Question 7.
What is die maximum percentage of rectification in half wave and full wave rectifiers?
Answer:
The maximum percentage of rectification in a half wave rectifier is 40.6% and in a full wave rectifier maximum percentage of rectification is 81.2%.

Question 8.
What is zener voltage (V_Z) and how will a zener diode be connected in circuits generally?
Answer:
When a zener diode is reverse biased at a particular voltage the current increases suddenly. The voltage at which the current increases is called “breakdown voltage” or “zener voltage”, so zener is always connected in “reverse bias”.

Question 9.
Write the expressions for the efficiency of a half wave rectifier and a full wave rectifier.
Answer:
Efficiency of half wave rectifier

Efficiency of full wave rectifier

Question 10.
What happens to the width of the depletion layer in a p – n junction diode when it is i) forward biased and ii) reverse biased? [AP Mar. ’19]
Answe:
When a p – n diode is forward biased thickness of depletion layer decreases and in reverse bias condition the thickness of depletion layer increases.

Question 11.
Draw the circuit symbols for p – n – p and n – p – n transistors. [TS & AP Mar. ’18, May ’17; AP May ’16, ’14; Mar. ’14; TS Mar. ’16]
Answer:

Question 12.
Draw the symbol of NOT gate and explain its operation. Give its truth table. [TS June ’15]
Answer:
NOT gate :
It has one input terminal and one output terminal. The output of NOT gate is the opposite of input i.e., if input is ‘0’ then output is ‘1’. If input is ‘1’ then output is ‘O’.

Implementation of NOT gate using a transistor :
NOT gate can be implemented with transistor. If A = 0 the emitter base junction is open and there is no current through the transistor. The current through the resistor. R_L = 0 and Q becomes equal to a potential of 5V i.e., Q when A = 1 then Q = 1 in the emitter base junction. So large current flows and Q approximately 0 volt i.e., Q =0. Thus output is same as that of a NOT gate.

Truth tables of NOT gate :
The truth tables of NOT gate interms of low and high (0 and 1) are as given below.

Question 13.
Define amplifier and amplification factor.
Answer:
Amplifier :
Raising the strength of a weak signal is known as amplification and the device used for this purpose is called amplifier.

Amplification factor :
It is the ratio between output voltage to the input,
Voltage, (A) = \(\frac{V_0}{V_i}\)

Question 14.
In which bias, can a zener diode be used as voltage regulator? [AP Mar. 16; TS June 15]
Answer:
In reverse bias, zener diode can be used as voltage regulator.

Question 15.
Which gates are called universal gates? [TS Mar. ’15]
Answer:
NAND and NOR gates are known as the basic building blocks of logic gates or universal gates.

Question 16.
Write the truth tables of NAND gate. How does it differ from AND gate?
Answer:

Truth Table
Input		Output
A	B	Q
0	0	1
1	0	1
0	1	1
1	1	1

The output of NAND gate is opposite to output of AND gate.

Short Answer Questions

Question 1.
What are n-type and p-type semiconductors? How is a semiconductor junction formed?
Answer:
n-type semiconductors :
When pentavalent impurities such as phosphorous (P), arsenic (As), antimony (Sb) are added to intrinsic semiconductors then they are called n-type semiconductors.

p-type semiconductors :
When trivalent. impurities such as Boron (B), Aluminium (AI), Galium (Ga), Indium (In) etc. are added to intrinsic semiconductor then it is called p-type semiconductor.

p-n junction :
A p-n junction is formed by adding a small quantity of pentayalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

• During the formation of p-n junction diffusion and drift of charge carriers takes place.
• In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side.

Due to the concentration gradient between p-type and n-type region holes diffuse to n- region and electrons diffuse to p-region.

Due to diffusion of changes a chargeless region is formed near junction called depleted layer.

Question 2.
Discuss the behaviour of p-n junction. How does a potential barrier develop at the junction?
Answer:
p-n junction :
A p-n junction is formed by adding a small quantity of pentavalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

• During the formation of p-n junction diffusion and drift of charge carriers takes place.
• In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side.

Due to the concentration gradient between p-type and n-type region holes diffuse to n- region and electrons diffuse to p-reglon. This leads to diffusion current.

Due to diffusion of electron an ionised donor is developed at n-region and due to diffusion of holes to n- region an ionised acceptor. These ions are immobile. So some – ve charge is developed in p – region and positive charge is developed in n-region. This space charge prevents further motion of electrons and holes near junction.

Depletion layer :
Both the negative and positive space charge regions near junc-tion are called depletion region.

Question 3.
Draw and explain the current-voltage (I-V) characteristic curves of a junction diode in forward and reverse bias.
Answer:
When a graph is plotted between junction potential ‘V’ and junction current T of a p-n junction then it is called V-I characteristics.

In forward bias p-side of p-n junction is connected to + ve terminal and n – side is connected to – ve terminal of external voltage ‘V’. The external voltage ‘V’ is gradually increased and junction current T is measured.

Initially junction potential is slowly increased in steps of 0.1 V. Until junction potential reaches a minimum value called threshold potential junction current is almost zero. ;

Once applied potential crosses threshold potential then junction current increases exponentially with applied voltage. On the reverse bias side nearly a steady current of few micro amperes was observed with applied voltage. When reverse bias potential reaches a high valued suddenly the diode is thrown into conduction. This is called breakdown potential.

Question 4.
Describe how a semiconductor diode is used as a half wave rectifier. [TS May. 18. Mar. 16; AP Mar. 16, 14]
Answer:
A junction diode allows current through it in forward bias only. In a half wave rectifier an a.c. source, p-n junction and load resistance (R_L) are connected in series as shown.

For ‘+ve’ half cycle p-n junction is for-ward biased so current flows through diode and we will get output current across load resistance.

For ‘+’ ve half cycle the p-n junction is reverse biased so current does not flow through p-n junction. So we are not able to get current through load resistance.

In half wave rectifier the out put voltage changes sinusoidally. But still it is flowing in only one direction through load resistance (R_L) so input a.c. voltage is rectified.

Question 5.
What is rectification? Explain the work¬ing of a full wave rectifier. [AP Mar. ’18, ’15; May ’17, ’14; TS Mar. ’19. ’15, May ’17]
Answer:
Rectification :
The process of converting alternating current (a.c) into direct current (d.c) is called rectification. Instruments used for rectification is called rectifier.

In a full wave rectifier, two p-n diodes are connected at the output side of a center-tapped transformer through a load resistance as shown.

The centre tap will divide the output a.c. wave exactly into two equal halves say + ve half cycle and – ve half cycle.

Let diode D₁ is connected to ‘+ve’ half cycle then it is forward biased. So current flows through D₁ at the same time ‘-ve’ half cycle is applied to diode D₂ so it is reverse biased and current does not flow through it. Hence we will get output through diode D₁.

As the applied a.c. wave is progressing we will get ‘+ve’ half cycle to diode D₂ at that time diode D₂ is forward biased so current flows through D₂. But now ‘-ve’ half cycle is applied to diode D₁. So it is in reverse bias hence current does not flow through D₁.

In full wave rectifier diodes D₁ and D₂ will conduct current alternately. Even though out put current oscillates between a minimum and maximum value it always passes through same direction through load resistance (R_L). So input a.c. is rectified.

Question 6.
Distinguish between half-wave and full-wave rectifiers. [AP Mar. ’19. ’17; May. ’16; TS Mar. ’18, May, ’18]
Answer:

Half wave rectifier	Full wave rectifier
i) In half wave rectifier only one diode is used.	i) In full wave rectifier two diodes are used.
ii) Every positive half cycle is rectified.	ii) Both positive and negative half cycles are rectified.
iii) Electric current is not continuous.	iii) Electrical current is continuous.
iv) In the negative half cycle of a.c. rectification will not take place.	iv) In the negative half cycle of a.c. also rectification will take place.
v) Efficiency is less.	v) Efficiency is high.
vi) Ripple is less.	vi) Ripple is high.

Question 7.
Distinguish between zener breakdown and avalanche breakdown.
Answer:
Avalanche breakdown:

1. In avalanche breakdown, the thermally generated electrons and holes acquire sufficient energy from the applied potential to produce new carries by removing valence electrons from their bonds.
2. These new carries, in turn produce additional carriers again through the process of disrupting bonds.
3. This cumulative process is referred to as avalanche multiplication. It results in the large flow of current, and the diode finds itself in avalanche breakdown.
4. This occurs in lightly doped diodes at high reverse bias voltages.

Zener breakdown:

1. If a diode is heavily doped, direct rupture of covalent bonds takes place because of strong electric field at the junction.
2. As a result of heavy doping of p and n regions, the depletion region width becomes very small and an applied voltage causes an electric field of 10⁷ V/m of the junction making condition suitable for zener breakdown. This occurs in heavily doped diodes at low reverse bias voltages.

Question 8.
Explain hole conduction in intrinsic semiconductors.
Answer:
Intrinsic semiconductors :
Semiconductors with ultra high pure state are called “intrinsic semiconductors”.

In pure germanium (Ge) or silicon(Si) crystal every germanium or Silicon atom forms four covalent bonds with neighbouring Ge/Si crystal.

At very low temperatures intrinsic semiconductors are insulators. When temperature increases electrons absorbs more thermal energy and it may become a free electron and that atom will become positive.

In intrinsic semiconductors number of free electrons (n_e) is equal to number of holes (n_h)

Due to thermal energy some of the electrons escapes from the bonds and an empty spaces left behind in the valence band. This vacancy in the valence band is called a “hole”.

Due to applied electric field the holes drift in opposite direction to the electrons with lesser speed and behave like positive charge carriers and current is produced due to the both electrons and holes.

Current contribution by electrons (l_e) and holes (I_h) is same.
∴ In an intrinsic semiconductor n_e = n_h = n_i.
Total current I = I_e + I_h

Question 9.
What is a photodiode? Explain its working with a circuit diagram and draw its I-V characteristics.
Answer:
Photodiode :
A photodiode consists of a p-n junction diode with a transparent window to allow light to fall on to the junction.

When light photons falls on this p-n junction it will produce electron – hole pair. These pairs are separated by the applied potential ‘V’ before they recombine. The magnitude of photo current depends on intensity of incident light.

i.e., current, i ∝ intensity of light I. So a photodiode can convert light intensity variations into current variations. This property is used to detect optical signals.

The I-V characteristics of photodiode can easily studied with reverse bias potential on it. The junction voltage ‘V’ and current ‘I’ characteristics are as shown in figure.

Question 10.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) :
It is a highly doped p-n junction. Under forward biased condition it emits spontaneous radiation in visible region. This p-n junction is coated with a transparent cover so that emitted light can come out.

Working :
When LED is forward biased holes are driven to n-region and electrons are driven to p-region due to electric potential of battery. As a result charge concentration near junction region increases. The electrons and holes while recombining with them they will release the recombination energy in the form of light photons.

Generally LED breakdown voltages are very low such as 5V. For the fabrication of visible LED the energy band gap of 1.8 eV to 3 eV is necessary. The minimum energy gap must be 1.8 eV. It LED materials are selected with in this range then we can get visible light in the wavelength range of 0.7 pm to 0.4 pm or in wavelength range of 7000Å to 4000Å.

Generally LEDs are biased to emit light with maximum efficiency. Intensity of light emitted depends on strength of current. Colour of light depends on energy band width.

For gallium arsenide-phosphide energy gap is ≅ 1.9 eV, it will emit red light.

For galium Arsenide Ga, As energy gap is ≅ 1.4 eV, it will emit infrared light. These LEDs are widely used in T.V remote control and in burglar alarm systems.

Advantages of LED:

1. They require low operational voltage and consumes less power.
2. They are surged and life period is very high.
3. They will respond very quickly to current changes.

Question 11.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cells :
A solar cell is also a p-n junction which generates emf when solar radiation falls on it. A p-type silicon wafer of nearly 300 pm is taken. A n-type impurity layer of nearly 0.3 µm is developed on it through diffusion process, junction surface area Is kept large. A metallic grid is deposited on n-region and the other p- side is also coated with metal. These two will provide electrical contact.

Working :
Generally semiconductors width a band width of nearly 1.5 eV or less are selected in solar cells when solar radiation falls on p-n junction electron hole pair is produced.

Separation of electrons and holes will takes place (before they recombine) due to the electric field produced across depleted region.

Electrons on reaching n – side are collected by front metallic grill. Holes reaching p – side are collected by back contact.

V.I characteristics of solar cells are drawn in fourth quadrant of the coordinate axis. It is as shown in figure.

Solar cells are widely used to tame solar energy from which electricity is produced.

These solar cell plays an important role in supplying electrical energy to satellites and in remote forest areas.

Question 12.
Explain the different transistor configurations with diagrams.
Answer:
Transistors are connected in three ways. They are :

1. Common base configuration
   Common emitter configuration
2. Common collector configuration

1) Common base configuration :
In this configuration, base is common to both input and output. Base terminal is earthed and input is given across base emitter and output is taken across base collector.

2) Common emitter configuration :
In this configuration emitter is common to both input and output. The emitter is earthed and input is given across base emitter and output is taken across collector emitter.

3) Common collector configuration :
In this configuration, collector is common to both input and output. The collector is earthed and input is given across base collector and output is taken across emitter-collector.

Question 13.
Explain how transistor can be used as a switch?
Answer:
In a transistor D.C. input voltage V_i = I_BR_B + V_BE …….. (1)

Sum of D.C potential between emitter base i.e., and product of base current and base resistance (V_BE) / I_BR_B.

D.C. output voltage V₀ = V_CC – I_CR_C ……….. (2)

As far as V_i is less than active region minimum voltage of 0.6 V (nearly), output voltage V₀ is high because Collector current (I_C) is zero.

When input voltage V_i increases collector current I_C increases and output voltage V₀ decreases because V₀ = V_i – I_CR_C. i.e., when V_i is less V₀ is high and whenV_i is high output voltage V₀ is less. By changing the input potential a transistor can be made to move between high and low states or ON and OFF states. Hence by selecting proper input voltage V_i a transistor can be used as a switch.

Question 14.
Explain how transistor can be used as an oscillator.
Answer:
In an oscillator, we will get out put without any external input. The oscillator circuit is as shown in figure.

When switch ‘S₁‘ is closed a surge current will flow in transistor and it produces output in collector circuit. Current in coil T₂ will gradually increase from minimum value ‘x’ to maximum value ‘y’ with changing time. Current in T₂ is connected to oscillator LC in collector circuit. This induces current in coil T₁ due to inductive coupling between T₁ and T₂. Coil T₁ is connected to emitter base circuit. So a part of output is given as feedback to input.

When collector current reaches maximum value rate of change in collector current is zero. So induced current in T₁ is zero i.e., input feedback is zero. So collector current begin to decrease while decreasing again emf is induced in T₁ and feedback is given to emitter base circuit. Like this the signal is self sustained
Frequency of oscillator υ = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

In this way a transistor can be used as an oscillator.

Question 15.
Define NAND and NOR gates. Give their truth tables. [TS Mar. ’17; AP June ’15]
Answer:
NOR Gate :
It is the combination of OR gate and NOT gate OR + NOT = NOR.

In this logic gate the output of OR gate is given to the input of NOT gate as shown in figure.

Truth Table
Input		Output
A	B	Q
0	0	1
1	0	0
0	1	0
1	1	0

NAND gate :
It is the combination of AND gate and NOT gate.
AND + NOT = NAND.
In this logic gate the output of AND gate is given to the input of NOT gate.

Truth Table
Input		Output
A	B	Q
0	0	1
1	0	1
0	1	1
1	1	0

The NAND and NOR gates are basic build¬ing blocks of logic gates; because any logic gates can be constructed by using only NAND or only NOR gates.

Question 16.
Explain the operation of a NOT gate and give its truth table. [TS June ’15]
Answer:
NOT gate :
It has one input terminal and one output terminal. When the input is low, the output is high and when the input is high, the output is low.

Implementation of NOT gate using a transistor :
NOT gate can be implemented with transistor. If A = 0 the emitter base junction is open and there is no current through the transistor. The current through the resistor. R_L = 0 and Q becomes equal to a potential of 5V i.e., Q when A = 1 then Q = 1 in the emitter base junction. So large current flows and Q approximately 0 volt i.e., Q =0. Thus output is same as that of a NOT gate.

Truth tables of NOT gate :
The truth tables of NOT gate interms of low and high (0 and 1) are as given below.

Long Answer Questions

Question 1.
What is a junction diode? Explain the for-mation of depletion region at the junction. Explain the variation of depletion region in forward and reverse biased condition.
Answer:
p-n junction :
A p-n junction is formed by adding a small quantity of pentavalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

During the formation of p-n junction diffusion and drift of charge, carriers takes place.

In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side. Due to the concentration gradient between p-type and n-type regions holes diffuse to n-region and electrons diffuse to p-region. This leads to diffusion current.

Due to diffusion of electron an ionised donor is developed at n-region and due to diffusion of holes to n- region an ionised acceptor. These ions are immobile. So some – ve charge is developed in p – region and positive charge is developed in n-region. This space charge prevents further motion of electrons and holes near junction.

Depletion layer :
Both the negative and positive space charge regions near junction are called depletion region.

Variation of depletion region :
In forward bias due to applied voltage electrons from n – region crosses junction layer and goes to p-region. Similarly holes from p-region crosses junction and goes to n-region. Since charge carriers are freely crossing the junction depleted region vanishes in forward bias condition.

In reverse bias condition electrons in n- region are attracted by + ve terminal of bat-tery connected to it. Similarly holes in p – region are attracted by ’-ve’ terminal con-nected to it. As a result charges will travel towards battery terminals. We can not find charge carriers near p-n junction.

So in reverse bias condition width of depleted region increases.

Question 2.
What is a rectifier? Explain the working of half wave and full wave rectifiers with diagrams.
Answer:
The process of converting alternating current (a.c) into direct current (d.c) is called “rectification”. Instruments used for rectification is called “rectifier”.

A junction diode allows current through it in forward bias only. In a half wave rectifier an a.c. source, p-n junction and load resistance (R_L) are connected in series as shown.

For ‘+ve’ half cycle p-n junction is forward biased so current flows through diode and we will get output current across load resistance.

For + ve half cycle the p-n junction is reverse biased so current does not flow through p-n junction. So we are not able to get current through load resistance.

In half wave rectifier the out put voltage changes sinusoidally. But still it is flowing in only one direction so input a.c. voltage rectified.

In a full wave rectifier two p-n diodes are connected at the output side of a center tapped transformer through a load resistance as shown.

The centre tap will divide the output a.c. wave exactly into two equal halves say + ve half cycle and – ve half cycle.

Let diode D₁ is connencted to ’+ve’ half cycle then it is forward biased. So current flows through D₁ at the same time ‘-ve’ half cycle is applied to diode D₂. So it is reverse biased and current does not flow through it. Hence we will get output through diode D₁.

As the applied a.c. wave is progressing we will get ‘+ve’ half cycle to diode D₂ at that time diode D₂ is forward biased so current flows through D₂. But now -ve’ half cycle is applied to diode D₁. So it is in reverse bias hence current does not flow through D₁.

In full wave rectifier diodes D₁ and D₂ will conduct current alternately. Even though out put current oscillates between a minimum and maximum value it always passes through same direction. So input a.c. is rectified.

Question 3.
What is a zener diode? Explain how it is used as a voltage regulator.
Answer:
Zener diode :
A zener diode is a highly doped p-n junction with sharp breakdown voltage. Generally zener is operated in “reverse bias condition”.

Note:
In forward bias condition zener diode will also act as ordinary p-n junction.

Zener diode as voltage regulator :
Principle :
Zener diode has sharp breakdown voltage in “reverse bias condition”.

In zener diode release of large scale of charge carriers at breakdown voltage is due to field emission of electrons from host atoms. For field ionisation nearly an electric field of 10⁶ V/m is necessary.

The speciality of zener diode is even though current through zener increases largely its potential (zener potential) remains almost constant.

Working :
Let a zener diode is connected to unregulated supply through a series resistance R_s. The value of zener voltage is selected such that zener voltage V_z is less than the minimum value of unregulated supply. A load resistance R_L is connected parallel to zener. Output is taken across load resistance R_L.

Case – I :
Let voltage of unregulated supply is at Its minimum value say V_{i min}. Zener potential V_z is constant. So V_{i min} – V_z is voltage drop across R_s. Current through R_s = I_min
= \(\frac{V_{i min}-V_z}{R_s}\)
Load current, I_L = \(\frac{V_z}{R_L}\)
∴ Current through zener I_{z min} = I_min – \(\frac{V_z}{R_L}\)

Case – II :
When applied voltage is maximum say, V_{i max} then voltage drop across
R_s = V_{i max} – V_z

Maximum current through R_s is
I_max = \(\frac{V_{i max}-V_z}{R_s}\)
Current through zener is also max.
I_{z max} = I_max = \(\frac{V_z}{R_L}\)

In zener diode voltage regulator voltage changes in unregulated supply are converted into current changes in series resistance R_s. These current changes are absorbed by zener.

As a result we will get a constant voltage and current at output. In this way zener diode acts as a voltage regulator.

Question 4.
Describe a transistor and explain its working.
Answer:
Transistor :
A transistor is a three layered electronic device. These layers are called emitter, base and collector.

Transistors are two types 1) p-n-p 2) n-p-n.

Emitter :
Emitter region is of moderate size. It is heavily doped. It supplies large number of majority charge carriers for the current flow through transistor.

Base :
Width of base region is very less. It is lightly doped nearly with 3 to 5% Impurity concentration of emitter.

Collector :
Size of collector region is larger than emitter. It is moderately doped (i.e., impurity concentration is less than emitter).

Biasing of transistor :
In a transistor for transistor action to takes place (i) Emitter base region must be forward biased, (ii) Base collector region must be reverse biased, (iii) Forward bias emitter base potential V_EB must be less than reverse bias collector base potential V_CB.

If a transistor is biased as above, then the transistor is said to be in active state.

Working principle :
Electrons/holes in emitter region are injected into base region due to forward bias potential in emitter base region.

In base region nearly 3 to 5% of electrons/ holes are recombined so a small amount of current (I_B) will flow in emitter base circuit. Due to less impurity concentration the remaining charges injected into base will cross base collector region.

The high reverse bias potential in base collector region will act as forward bias for the charges left unrecombined in base region.

So these charges are attracted by collector region. These charges will recombine at collector region and some current (I_C) will flow in base collector region.

In a transistor emitter current (I_E) = Base current (I_B) + Collector current (I_C)
∴ I_E = I_B + I_C

Hence transistor is a current controlled device.

Question 5.
What is amplification? Explain the working of a common emitter amplifier with necessary diagram.
Answer:
Amplifier :
An amplifier is an electronic device used to strengthen weak signals.

Transistor as an amplifier (C.E. configuration) :
A transistor can be used as an amplifier in its active region. In this region output voltage V₀ increases drastically even for a small change in input voltage’ V_i‘.

In transistor amplifier the mid point of active region is taken as operating point (also called input potential) on which varying signal voltage is superposed. This variation is magnified at output side by a factor equals to amplification factor p.

In a transistor output voltage of collector, V_OC = V_CE + I_LR_L

Input voltage V_BB = V_BE + I_BR_B when input signal voltage V_i ≠ 0 the,

V_BE + V_i + V_BE + I_BR_B + ∆I_B(R_B + r_i)

Where r_i is input resistance.

Input signal voltage
V_i = ∆I_B(R_B + r_i) = r∆I_B.

Current amplification factor β is defined as the ratio of change in collector current ∆I_C to change In base current ∆I_B when V_CC Is constant.

Voltage amplification factor A_v :
It is defined as the ratio of output signal voltage (V₀) to input signal voltage (V_i).

Question 6.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
Implementation of OR gate using diodes :
Let ‘D₁‘ and ‘D₂‘ represent two diodes. A potential of 5V represent the logical value 1 and a potential of 0V represents the logical value zero.

When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at ‘Q’ is zero, i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential behaves like a closed switch. The output potential then becomes 5V. i.e., Q=l. When both A and B are 1, both the diodes are forward biased and the potential at Q’ is same as that at A and B which is 5V i.e., Q =1. The output is same as that of the OR gate.

Truth table of OR gate :
The truth table of OR gate interms of low, high; 0 and 1 are below.

Truth Table
Input		Output
A	B	Q
0	0	0
1	0	1
0	1	1
1	1	1

Logical symbol of OR gate :
The logical function OR is represented by the symbol plus. So that the output, Q = A + B

Questuion 7.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors?
Answer:
AND gate :

1. It has two input terminals and one output terminal.
2. When both the inputs low or one of the input is low the output is low in an AND gate.
3. The output of the gate is high only when both the inputs are high.
4. The output of the gate is high only when both the inputs are high.
5. If the input of the gate are A and B and the outputs in Q than Q is logical function of A and B. The value of Q for different combinations of A and B is shown by means of a table called truth table.

Truth Table :
It is defined as the table that shows the values of the output of all possible combinations of the value of the input variables.

Truth Table
Input		Output
A	B	Q
0	0	1
1	0	0
0	1	0
1	1	0

Implementation of AND gate using diodes:

1. Let D₁ and D2 represent two diodes. A potential of 5V represents the logical value 1 and potential of OV represents the logical value zero (0).
2. When A = 0, B = 0, both the diodes D₁ and D₂ are forward biased and they behave like open switches. There is no current through the resistance R’ making the potential of Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping – Conductivity of semiconductors:
When a p – type impurity is doped it will develop acceptor energy levels near valance band in Forbidden region. So Forbidden gap width decreases and electrons can easily sent to conduction band.

When n-type impurities are doped they will develop donor energy levels near conduction band in Forbidden gap. As a result Forbidden gap decreases and electrons in valance band will be sent to conduction band with very little energy (< 0.01 eV).

As a result due to doping conductivity of semiconductors will increase.

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Answer:
Internal resistance of diode r_f = 20 Ω;
Load resistance R_L = 2kΩ = 2000 Ω
Efficiency of half wave rectifier =

\(\frac{812}{2020}\) = 0.4019
% of η = 0.4019 × 100 = 40.19% = 40.2%

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Answer:
Load resistance R_L = 1300 Ω;
Internal resistance of diode r_f = 9 Ω
Efficiency of full wave rectifier

% of efficiency η = 0.8064 × 100 = 80.64%

Question 3.
Calculate the current amplification factor b (beta) when change in collector current is 1 mA and change in base current is 20 mA.
Answer:
Change in collector current ∆I_C = 1mA = 1 × 10^-3 A
Change in base current ∆I_b =20µA = 20 × 10^-6 A
Current amplification

Question 4.
For a transistor amplifier, the collector load resistance R_L = 2k ohm and the input resistance R_i = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Answer:
Load resistance RL = 2 kD = 2000 Ω;
Input resistance Rj = 1 kQ = 1000 Ω
Current gain b = 50 ; Voltage gain A_v = ?

Telangana TSBIE TS Inter 2nd Year Physics Study Material 14th Lesson Nuclei Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 14th Lesson Nuclei

Very Short Answer Type Questions

Question 1.
What are isotopes and isobars?
Answer:
Isotopes :
The nuclei having the same atomic number (Z) but different mass number (A) are called isotopes.
Ex: ₈O¹⁶ , ₈O¹⁷, ₈O¹⁸.

Isobars :
The nuclei having the same mass number (A) but different atomic numbers (Z) are called isobars.
Ex: 14 ¹⁴₆C, ¹⁴₇N.

Question 2.
What are isotones and isomers?
Answer:
Isotones :
The nuclei having same neutron number (N) but different atomic number (Z) are called isotones.
Ex: ₈₀Hg¹⁹⁸, ¹⁹⁷₇₉Au.

Isomers :
Nuclei having the same atomic number (Z) and mass number (A) but different nuclear properties such as radioactive decay and magnetic moments are called isomers.
Ex: ⁸⁰₃₅Br^m, ⁸⁰₃₅Br^g. Here’m’ denotes metastable state and ‘g’ denotes ground state.

Question 3.
What is a.m.u.,? What is its equivalent energy?
Answer:
Atomic mass unit (lu) :
1/12th mass of ¹²₆C atom is taken as atomic mass unit.

Energy equivalent of 1 u = 931.5 MeV.

Question 4.
What will be the ratio of the radii of two nuclei of mass numbers A₁ and A₂?
Answer:
We know the radius of the nucleus
R = R₀ A^1/3

Question 5.
Natural radioactive nuclei are mostly nuclei of high mass number, why?
Answer:
As the atomic number increases coulombian repulsive force increases in the nucleus and hence the stability of the nucleus decreases.

That is why the nuclei after lead are unstable and they exhibit natural radioactivity.

Question 6.
Does the ratio of neutrons to protons in a nucleus increase, decrease or remain the same after the emission of an a – particle?
Answer:
α – particle means Helium nucleus (₂He⁴).

If the nucleus emits α -particle, it loses 2 protons and 2 neutrons. But in the nuclei of radio active elements number of neutrons is greater than the number of protons.

When neutrons number and protons number decreases equally, their ratio increases.

Question 7.
A nucleus contains no electrons but can emit them. How?
Answer:
Natural radioactive elements undergo β – decay. Then a neutron loses one electron and converts into proton.

This electron is ejected out with high velocity called β – ray. The proton remains inside the nucleus.

Question 8.
What are the units and dimensions of the disintegration constant?
Answer:
Since \(\frac{N}{N_0}\) = e^λt . λt has no unit and no
dimensions. So unit of λ = s^-1
Dimensional formula λ = T^-1

Question 9.
Why do all electrons emitted during β^– – decay not have the same energy?
Answer:
In negative beta decay (β^–) emission of electron is accompanied by a neutrino. These neutrinos have very small mass cempared to electron and some kinetic energy. Due to this neutrino energy of electrons liberated in β^– decay is not constant.

Question 10.
Neutrons are the best projectiles to produce nuclear reactions. Why?
Answer:
Neutron is an uncharged particle so it is not deflected by electric and magnetic fields and has high penetrating power. So the neutron required lesser energy than a positive charged particle for producing nuclear reactions. Hence neutron is the best projectile for producing nuclear reactions.

Question 11.
Neutrons cannot produce ionization. Why?
Answer:
Since neutron does not posses any charge, its ionising power is very less when compared to α and β rays.

Question 12.
What are delayed neutrons?
Answer:
The neutrons which are liberated over a period of time after the fission process has taken place are called delayed neutrons.

They play an important role in the nuclear reactor.

Question 13.
What are thermal neutrons? What is their importance?
Answer:
Slow neutrons are called thermal neutrons. Their energy is 0.025 eV. They produce nuclear fission.

Question 14.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction?
Answer:
In controlled chain reaction K = 1.
In uncontrolled chain reaction K > 1.

Question 15.
What is the role of controlling rods in a nuclear reactor?
Answer:
They can control chain reaction by absorbing neutrons in nuclear reactor.

Question 16.
Why are nuclear fusion reactions called thermo nuclear reactions?
Answer:
As nuclear fusion reactions occur at very high temperature of the order of 10⁷K, the fusion reactions are known as thermonuclear reactions.

Question 17.
Define Becquerel and Curie.
Answer:
Becquerel (Bq) :
It is a unit to measure radioactivity of a substance. If a radioactive substance undergoes one disintegration or decay per second then it is called Becquerel.

Curie :
It is a unit to measure radioactivity of a substance. If a radioactive substance undergoes 3.7 × 10¹⁰ decays per second the radioactivity of that substance is called curie.
1 Curie = 3.7 × 10¹⁰ Bq (Becquerel)

Question 18.
What is a chain reaction?
Answer:

1. Chain reaction : If a fission reaction is self-maintained due to the neutrons released in that reaction then it is called chain reaction.
2. For chain reaction to takes place (1) At- least one external neutron is necessary 2) neutron multiplication factor K ≥ 1.
3. Initial mass of uranium must be greater than critical mass.

Question 19.
What is the function of moderator in a nuclear reactor?
Answer:
The function of moderator in a nuclear reactor is, to slow down the neutrons, thus neutrons will participate actively in fission reaction. It decreases the speed of neutrons from 2 MeV to 0.025 eV in the nuclear reactor. A good moderator does not absorb the neutrons.

Question 20.
What is the energy released in the fusion of four protons to form a helium nucleus?
Answer:
Energy released due to fusion of four protons is 26.7 MeV.

Short Answer Questions

Question 1.
Why is the density of the nucleus more than that of the atom? Show that the density of nuclear matter is same for all nuclei.
Answer:
a) In an atom nearly 99.9% of mass is concentrated in a very small volume called nucleus. Volume of nucleus is 10^-12 times less than volume of atom. From above explanation density of nucleus is clearly more than density of atom.

b) Density = ρ = mass / volume. But mass of nucleus m = A.m_p.

The above equation is independent of mass number of nucleus A. So density of nuclear matter is same for all nuclei.

Question 2.
Write a short note on the discovery of neutron.
Answer:
Chadwick observed emission of neutral radiation when beryllium nuclei were bombarded with alpha particles.

This neutral radiation when passed through lighter elements like helium, CO₂ and nitrogen knock out a proton.

If the particle is a photon, then its energy must be for higher than the α – particle participating in the reaction.

The energy of neutral particle and also that of a proton released when that neutral particle passes through the lighter element gave some energy discrepancy.

To explain this energy discrepancy Chadwick proposed that the neutral particle is not photon. But a new particle with its mass equal to that of proton and he called it neutron.

Mass of neutron is m_n = 1.00866 u.

In this way Chadwick discovered Neutron.

Question 3.
What are the properties of a neutron?
Answer:
Properties of a neutron :

1. Neutron is a chargeless particle and its mass is almost equal to mass of proton.
2. A free neutron is unstable when it is out side the nucleus. Its mean life period is 1000 sec.
3. Inside nucleus neutron is stable.
4. Number of neutrons in an atom is (A – Z) where A is mass number and Z is atomic number.

Question 4.
What are nuclear forces? Write their pro-perties. .
Answer:
Nuclear forces :
Forces between nucleons present inside the nucleus are called nuclear forces. Properties of nuclear forces are

1. A nuclear force is mych stronger than the coulomb force and the gravitational force.
2. Nuclear force between two nucleons are distance dependent.
3. From potential energy graph of a pair of nucleons these forces are found to be attractive forces when separation between nucleons is 0.8 Fermi or more. These forces are found to be repulsive forces when separation between nucleons is less than 0.8 Fermi.
4. Nuclear forces are saturated forces.
5. Nuclear forces are independent on charge. So nuclear force between proton-proton, proton – neutron and neutron – neutron are equal.

Question 5.
For greater stability, a nucleus should have greater value of binding energy per nucleon. Why?
Answer:
When a graph is plotted between binding energy per nucleon (E_bn) and mass number (A) for different elements it is called
E_bn – A graPh

Salient features of the graphs :
i) The binding energy per nucleon (E_bn) is constant in mass number range of 30 to 170 these elements are found to be more stable. E_bn value is maximum at 8.75 MeV / nucleon for Iron A = 56 which is highly stable.

ii) For heavy nuclei (A > 170) Binding energy per nucleon gradually decreases with increasing mass number.
Ex : Uranium has low binding energy per nucleon of 7.6 MeV. To obtain greater stability under suitable conditions it always tries to break up into two intermediate masses.

For E_bn Vs mass number A graph for region A > 30 to A < 170 is almost same we cannot break easily in two separate nuclei.

Hence atoms with high E_bn are more stable.

Question 6.
Explain α – decay.
Answer:
Alpha decay :
In α – decay ⁴₂He nuclie is emitted from given radioactive substance. Mass number of product nucleons (called daughter nucleus) is decreased by Four units and Atomic number is decreased by two units. Equation of α – decay is

Disintegration energy (OR) Q – value :
The Q – value of a nuclear reaction is the difference between the initial mass energy and total mass-energy of decay products. For α – decay Q value is given by Q = (m_x – m_y – m_He)c².

Question 7.
Explain β – decay.
Answer:
Beta decay (β) :
In Beta decay an electron (or) a positron is liberated from given radioactive substance.

Positive beta decay (β⁺) : In this decay a positron (e⁺) and a neutrino (v) are liberated from radioactive substance.
Ex: ²²₁₁Na → ²²₁₀Ne + e⁺ + v

In + Ve beta decay a proton looses positron and converts into neutron.
⇒ p → n + e⁺ + ν

Negative beta decay (β^–) :
In this decay an electron (e^–) and an antineutrino (\(\overline{\mathrm{ν}}\))are liberated.

In – Ve beta decay a neutron loses electron and converted into proton i.e.,
n → p + e“ + \(\overline{\mathrm{ν}}\)

Question 8.
Explain γ – decay.
Answer:
Nucleus also has discrete energy levels like that of an atom. The energy difference bet-ween these energy levels is in the order of MeV. They are called ground state and exi-ted states.

When a nucleon is in an excited state it will spontaneously return ground state.

While returning to ground state the nucleons will emit energy photons whose energy is equal to difference of those two energy levels.

Since energy difference is in the order of MeV. The photons emitted are highly energetic. This highly energetic radiation is called gamma(γ) radiation.

Question 9.
Define balf-life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half-life period (T_½) : It is the time taken for the number of nuclei (N) to become half of initial nuclei (N₀) i.e., N = \(\frac{N_0}{2}\).

Decay constant (λ) :
Let N is the number of nuclie in a sample. The number of nuclei (∆N) undergoing radioactive decay during the time ‘∆t’ is given by \(\frac{\triangle N}{\triangle t}\) ∝ N or \(\frac{\triangle N}{\triangle T}\) = λN.

where λ = disintegration constant (or) decay constant.

Relation between half-life peirod T_½ and decay constant λ :
From definition of halflife period N = \(\frac{N_0}{2}\) ; dt = T_½ …………… (1)
But N (t) = N₀ e^-λt ………….. (2) i:e., Number of nuclei after a time ‘t’
From above equation
T_(½) = \(\frac{log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 10.
Define average life of a radioactive substance. Obtain the relation between decay constant and average life.
Answer:
Average life time :
In a radioactive substance, some nuclei may live for a long time and some nuclei may live for a short time. So we are using average life time τ.

Relation between decay constant and average life time:

When integrated with in the limits o to ∞
⇒ τ = \(\frac{1}{\lambda}\).

Question 11.
Deduce the relation between half-life and average life of a radioactive substance.
Answer:
1) Half-life period (T_½) :
The half-life period of a radioactive nuclide is the time taken for the number of nuclei (N) to become half of initial nuclei (N₀) i.e., N = \(\frac{N_0}{2}\).

2) Average life time :
In a radioactive substance, some nuclei may live for a long time and some nuclei may live for a short time. So we are using average life time τ.

Relation between half-life period and Average life period :
half-life period τ_½ =\(\frac{log 2}{\lambda}\) …………. (1)
But from average life period τ = \(\frac{1}{\lambda}\) ….. (2)
From equations (1) & (2)
∴ τ_½ = t log²_e

Question 12.
What is nuclear fission? Give an example to illustrate it.
Answer:
Nuclear fission:
It is a nuclear reaction in which a heavy nucleus is divided into smaller nuclei along with some energy.

Explanation:
When uranium isotope ²³⁵₉₂U is bombarded with a neutron it will break up into two intermediate mass nuclei.

The atomic number of products is not constant it may vary from 35 to 57.

The fragments produced are radioactive nuclei. They will emit β – particles to achieve stable end products.

Energy liberated in this process is in the form of kinetic energy of fragments and neutrons. Later it is transferred into surroundings in the form of heat energy.

Question 13.
What is nuclear fusion? Write the Conditions for nuclear fusion to occur.
Answer:
Fusion :
The process of combining lighter nuclei to form a heavy nucleus is called “fusion reaction” (or) “nuclear fission”.

Conditions for fusion to occur :

1. For fusion to take place the two nuclie must come close enough so that nuclear attractive short range force is able to effect them.
2. Since protons have positive charge on them coulomb repulsive forces will come into account. These forces will develop nearly a potential barrier of 400 keV.
3. Fusion will be achieved if temperature is raised to nearly 108 K. So that protons will overcome coulomb repulsive forces. At this temperature the particles have enough kinetic energy to overcome coulomb repulsive force.

Question 14.
Distinguish between nuclear fission and nuclear fusion. [TS June ’15]
Answer:

NUCLEAR FISSION	NUCLEAR FUSION
1) When a heavy nucleus like U-235 splits up into nearly two equal parts by the bombardment of slow moving energy is released.	1) When two or more lighter atoms of hydrogen or protons are fused into heavier neutrons considerable amount of atom, large amount of energy is released.
2) The principle behind atom bomb is Nuclear fission.	2) The principle behind hydrogen bomb is nuclear fusion.
3) In this reaction 200 MeV of energy is released per fission.	3) In this reaction 26.70 MeV of energy is released.
4) Energy released per nucleon is less.	4) Energy released per nucleon is more.
5) Particles involved are neutrons.	5) Particles involved are protons.
6) Fission takes place at room temperature.	6) Fusion takes place only at very high temperature and high pressure.
7) Fission produces radioactive elements like Barium and Krypton which are harmful (radioactive).	7) The products of fusion are harmless (not radioactive).

Question 15.
Explain the terms ‘chain reaction’ and ‘multiplication factor’. How is a chain reaction sustained?
Answer:
1) Chain reaction :
If a fission reaction is self maintained due to the neutrons released in that reaction, then it is called “chain reaction”.

2) Multiplication factor K :
The ratio of number of fissions produced by neutrons of present generation to the number of fissions produced in preceding generation is called “multiplication factor”.

3) Condition to sustain chain reaction :
For chain reaction to takes place

1. Atleast one external neutron is necessary.
2. Neutron multiplication factor K ≥ 1.
3. Initial mass of uranium must be greater than critical mass.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number? What is its significance?
Answer:
Mass defect :
The difference in mass of nucleus and its constituents is known as mass defect. In every nucleus the theoretical mass (Mt) is always less than practical mass (M).

Mass defect ∆m = [Zm_p + (A – Z) m_n] – M

Binding energy :
The energy released while forming a nucleus is called Binding energy E_b. Binding energy E_b = ∆mc².

When a certain number of protons and neutrons are brought together to form a nucleus then certain amount of energy E_b is released.

To divide a nucleus into its constituents, we have to supply an amount of energy equals to E_b from outside.

Binding energy per nucleon :
The ratio of Binding energy E_b to mass number ‘A’ of nucleus is cated’Binding energy per nucleon”.

Binding energy per nucleon E_bn = \(\frac{E_b}{A}\).

Binding energy per nucleon (E_bn) – Mass number (A) graphs :
When a graph is plotted between binding energy per nucleon (E_bn) and mass number (A) for different elements it is called “E_bn -A graph”.

Salient features of the graphs:

1. The binding energy per nucleon (E_bn) is almost constant within mass number range of 30 to 170.
2. For mass number A < 30 Binding energy per nucleon is less it gradually increases with increasing of mass number.
   This region suggested that when lighter nuclei are fused to form large nucleus then they release energy (Fusion process).
3. For heavy nuclei(where A > 170) Binding energy per nucleon gradually decreases with increasing mass number (A).

This region suggested that when larger nuclei are divided into small nuclei then energy is released (Fission process).

Question 2.
What is radioactivity? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [TS May 18. 16; Mar. 16]
Answer:
Radioactive decay :
The spontaneous disintegration of unstable nucleus is referred as “radioactivity or radioactive decay”.

When a nucleus undergoes radioactive decay three types of radioactive decay takes place.

1. α – decay : In this process ⁴₂He nuclei are emitted.
2. β – deay : In this process electrons or positrons are emitted.
3. γ – decay : In this process high energy photons (E.M. Waves) are liberated.

Law of radioactive decay :
Let N is the number of nuclie in a sample. The number of nuclie (∆N) undergoing radioactive decay during the time ‘∆t’ is given by
\(\frac{\triangle N}{\triangle t}\) ∝ N or \(\frac{\triangle N}{\triangle t}\) = λN

Where λ is disintegration constant or decay constant.

Decay rate (R) (OR) activity :
The total decay rate of a sample is the number of nuclei disintegrating per unit time.
∴ Total decay rate R = – \(\frac{dN}{dt}\) (OR)
R = R₀ e^-λt (or) R = λN (activity)

Total decay rate is also called activity. Radioactive decay is exponential because the total decay rate R = R₀ e^-λt Where R0 is a constant.
∴ R ∝ e^-λt Here e-W is exponential function whose value decreases with time.

Hence total radioactive decay of a substance decreases exponentially.

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [TS Mar. 17, 15, AP Mar. ’17, ’16, ’15, ’14; May 18, 16, 14; AP & TS Mar. 19, 18; May 17; AP June 15]
Answer:
Principle :
A nuclear reactor works on the principle of “sustained and controlled chain reaction”.

The labelled diagram of nuclear reactor is as shown. The important parts of nuclear reactor are

1) Core :
The core of the reactor is the site of nuclear fission. Generally it is made with graphite bricks. Core contains fuel elements in suitably fabricated form.

2) Fuel :
Generally used fuel is enriched uranium which contain ²³⁵₉₂U at high conentration than in natural uranium. Fuel is responsible for fission reaction in reactor.

3) Reflector :
Generally inner part of core is provided with some reflecting material which prevents leakage of neutrons from core.

4) Coolant :
The purpose of coolant is to remove heat energy produced in reactor, generally water is used as a coolant.

5) Moderator :
Core is filled with Heavy water. It is a good moderator. Purpose of moderator is to slow down the neutrons. Slow neutrons will effectively participate in fission reaction.

6) Control rods :
Purpose of control rods is to reduce number of neutrons inside the core. Generally control rods are made with neutron absorbing material like boron or cadmium.

Radiation shield :
The whole assembly of core is shielded with suitable material which prevents leakage of radioactive radiation from core.

Working :
Due to fission reaction energy is released in the core. It is transferred to an outside tank by means of coolant. As a result water in the tank is heated and steam is produced. This steam is used to run a turbine and power is produced. Power produced in this method is called atomic power.

Question 4.
Explain the source of stellar energy. Explain the carbon-nitrogen cycle, proton-proton cycle occurring in stars. [TS June ’15]
Answer:
The source of stellar energy is nuclear fusion reaction, for which hydrogen is the fuel.

Energy from Sun and Stars :
Sun and Stars have been radiating huge amounts of energy by nuclear fusion reactions taking place in their core, where the temperature is of the order 10⁷ K. More scientists proposed two types of cyclic processes for the source of energy in the Sun & Stars. They are :

Carbon – Nitrogen Cycle :
Carbon – Nitrogen cycle is one of the most important nuclear reactions for the production of solar energy by fusion.

The entire cycle can be summed up as,
4₁H¹ → ₂He⁴ + 2 _{+ 1}e⁰ + 3γ + 2υ + 26.70MeV

The energy released during this process is 26.70 MeV.

Proton-Proton Cycle : [TS June ’15]
At higher temperature, the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form higher nuclei of helium ³₂He. Two such helium nuclei combine with another proton releasing a total amount of energy 25.71 MeV. The nuclear fusion reactions are given below.
₁H¹ + ₁H¹ → ₁H² + ₊₁e⁰ + υ + 0.42MeV ………. (1)
e⁺ + e^– → γ + γ + 1.02MeV …………. (2)
₁H² + ₁H² → ₂He³ + γ + 5.49MeV ………. (3)
₂He³ + ₂He³ → ₂He⁴ + 2 ₁H¹ + energy 12.86 MeV ………….. (4)

For this equation to takes place first three equations must occur twice.

The nef result of the above reactions is 4 ₁H¹ → ₂He⁴ + 2 ₊₁e⁰ + 2γ + 2υ + 26.70 MeV

The energy released during this process is 26.70 MeV.

Problems

Question 1.
Show that the density of a nucleus does not depend upon its mass number (density is independent of mass).
Answer:
Volume of nucleus

But volume V ac A (mass number )
Density of nucleus matter

The above expression is independent of mass number ‘A’. Hence density of nuclear matter is independent of mass number of nucleus.

Question 2.
Compare the radii of the nuclei of mass numbers 27 and 64.
Answer:
Ratio of radii of nuclie is

Question 3.
The radius of the oxygen nucleus ¹⁶₈O is 2.8 × 10^-15m. Find the radius of lead nucleus ²⁰⁵₈₂Pb.
Answer:
Given, Radius of the oxygen nucleus R_o = 2.8 × 10^-15 m ;
Mass number of oxygen, A₀ = 16 ;
Mass number of lead = A_pb = 205 ; Radius of the lead nucleus R_pb = ?
Radius of nucleus

Question 4.
Find the binding energy of ⁵⁶₂₆Fe. Atomic mass of Fe is 55.9349 u and that of Hydrogen is 1.00783u and mass of neutrons is 1.00876 u.
Answer:
Mass of Hydrogen atom M_H = 1.00783u;
Mass of neutron m_n = 1.00867u
Mass of Iron atom M₁ = 55.9349u
Atomic number Z = 26;
Mass number A = 56;
Mass defect ∆m = [Zm_H + (A – Z)m_n – M_I]
∆m = [26 × 1.00783 + (56 – 26) 1.00867] – 55.9349 u
∆m = 0.52878 u
Binding energy = ∆mc² = 0.52878 × (3 × 10⁸)²
∴ B.E. = 0.52878 × 9 × 10¹⁶ or 0.52878 × 931.5 MeV
∵ [lu × C²] = 931.5 MeV energy
∴ B.E. =492.55
Binding energy per nucleon BE/ n = \(\frac{492.55}{56}\)
= 8:79 MeV

Question 5.
How much energy is required to separate the typical middle mass nucleus ¹²⁰₅₀Sn k its constituent nucleons? (Mass of ¹²⁰₅₀Sn = 119.902199 u, mass of proton = 1.007825 u and mass of neutron = 1.008665 u)
Answer:
Mass of proton m_p = 1.007825 amu ;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of Sn, Z = 50 ;
Mass number of Sn, A = 120 ;
Mass of nucleus of Sn atom = 119.902199 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z) m_n – M_N]
= [ (50) (1.007825) + (120 – 50) (1.008665) – 119.902199]
= (50 × 1.007825 + 70 × 1.008665 -119.902199)
= (50.39125 + 70.60655 – 119.902199)
∆m = [120.9978 – 119.902199] = 1.095601 amu

(ii) Energy required to separate the nucleons = Binding energy of the nucleus
B.E = ∆mc² (∵ lamu × c² = 931.5MeV)
= ∆m × 931.5MeV = 1.095601 × 931.5 MeV
= 1021 MeV

Question 6.
Calculate the binding energy of an α – particle. Given that mass of proton – 1.0073u, mass of neutron = 1.0087u, and mass of α – particle = 4.0015u.
Answer:
An α – particle is nothing but helium nucleus ⁴₂He.
It contains 2 protons, 2 neutrons with a mass number A = 4.
Mass of proton m_p = 1.0073 amu ;
Mass of neutron m_n = 1.0087 amu ;
Atomic number of helium Z = 2 ;
Mass number of helium A = 4 ;
Mass of helium atom M_N = 4.0015 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [(2) (1.0073) + (4 – 2) (1.0087) – 4.00260]
= (2 × 1.0073 + 2 × 1.0087-4.00260)
= (2.0146 + 2.0174)-4.0015 Am = [4.032 – 4.0015] = 0.0305 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5MeV)
= ∆m × 931.5 MeV = 0.0305 × 931.5 MeV
= 28.41 MeV

Question 7.
Find the energy required to split ¹⁶₈O nucleus into four α – particles. The mass of an α – particle is 4.002603u and that of oxygen is 15.994915u.
Answer:
The energy required to split
Q = [ Total mass of the products
– Total mass of the reactants] c²
= [Mass of four ⁴₂He – Mass of ¹⁶₈O] × c²
= [(4 × 4.002603) -15.994915] amu × c²
= [16.010412 – 15.994915] amu × c²
= (0.015497) 931.5 MeV = 14.44 MeV.

Question 8.
Calculate the binding energy per nucleon of ³⁵₁₇Cl nucleus. Given that mass of ³⁵₁₇Cl nucleus = 34.9800u, mass of proton = 1.007825u, mass of neutron = 1.008665 u and lu is equivalent to 931 MeV.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ³⁵₁₇Cl, Z = 17 ;
Mass number of ³⁵₁₇Cl, A = 35 ;
Mass of nucleus of ³⁵₁₇Cl atom =34.98 amu.

(i) Mass defect ∆m = [Zm_p + (A – Z)m_n – M_N]
= [(17) (1.007825) + (35 – 17) (1.008665) – 34.98]
= (17 × 1.007825 + 18 × 1.008665 – 34.98)
= (17.13303 + 18.15597 -34.98)
∆m = [35.289 – 34.98] = 0.3089

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ 1amu × c² = 931.5 MeV)
B.E = ∆m × 931.5 MeV = 0.3089×931.5 MeV
= 287.83 MeV

(iii) Binding energy per nucleon of Cl = \(\frac{B.E}{A}\)
= \(\frac{287.83}{35}\) = 8.22 MeV

Question 9.
Calculate the binding energy per nucleon of ⁴⁰₂₀Ca. Given that mass of ⁴⁰₂₀Ca nucleus = 39.962589 u, mass of a proton = 1.007825 u ; mass of neutron = 1.008665 u and 1 u is equivalent to 931 MeV.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ⁴⁰₂₀Ca Z = 20 ;
Mass number of ⁴⁰₂₀Ca A = 40 ;
Mass of nucleus of ⁴⁰₂₀Ca atom = 39.962589 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [ (20) (1.007825) + (40 – 20) (1.008665) – 39.962589]
= (20 × 1.007825 + 20 × 1.008665 – 39.962589)
= (20.1565 + 20.1733 – 39.962589)
∆m = [ 40.3298 – 39.962589 ] = 0.3672 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5 MeV)
∴ B E = ∆m × 931.5MeV
= 0.3672 × 931.5 MeV = 342.06 MeV

(iii) Binding energy per nucleon of Ca = \(\frac{B.E}{A}\)
= \(\frac{342.06}{40}\) = 8.55 MeV

Question 10.
Calculate(i) maw defect, (ii) binding energy and (iii) the binding energy per nucleon of ¹²₆C nucleus. Nuclear mass of ¹²₆C 12.000000 u; maw of proton = 1.007825 u and maw of neutron = 1.008665 u.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ¹²₆C, Z = 6 ;
Mass number of ¹²₆C, A = 12;
Mass of nucleus of ¹²₆C atom = 12.00amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [6(1.007825) + (12 – 6) (1.008665) – 12.00]
= (6 × 1.007825 + 6 × 1.008665 – 12.00)
= (6.04695 + 6.05199 – 12.00)
∆m = [12.09894 – 12.00] = 0.09894 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5 MeV)
∴ B.E = ∆m × 931.5 MeV
= 0.09895 × 931.5 MeV = 92.16 MeV

(iii) Binding energy per nucleon of carbon =
\(\frac{B.E}{A}=\frac{92.16}{12}\) = 7.86 MeV

Question 11.
The binding energies per nucleon for deuterium and helium are 1.1 MeV and 7.0 MeV respectively. What energy in joules will be liberated when 10v deuterons take part in the reaction.
Answer:
Given, Binding energies per nucleon for deuterium (\(\frac{B.E}{A}\))_D = 1.1 MeV
Mass number of deuterium A = 2
Binding energy of deuterium
B.E = (\(\frac{B.E}{A}\))_D × A = 1.1 × 2 = 2.2 MeV
Binding energies per nucleon for helium (\(\frac{B.E}{A}\))_He = 7.0 MeV
Mass number of helium A = 4
Binding energy of helium
B.E = (\(\frac{B.E}{A}\))_D × A = 7 × 4 = 28MeV
We know, ₁H² + ₁H² → ₂He⁴
Energy released = B.E. of 10⁶ deuterons
– B.E. of \(\frac{1}{2}\) × 10⁶ Helium atoms
Binding energy = 2.2 × 10⁶ – \(\frac{1}{2}\) × 10⁶ × 28
= 10⁶ (2.2 – 14) = – 11.8 × 10⁶ MeV
= -11.8 × 10⁶ × 1.6 × 10^-31J
= -18.88 × 10^-7J
(- ve sign indicates that energy is released)
∴ Energy released = 18.88 × 10^-7J

Question 12.
Bombardment of lithium with protons gives rise to the following reaction :

Find the Q – value of the reaction. The atomic masses of lithium, proton, and helium are 7.016u, 1.008u, and 4.004u respectively.
Answer:
Given, Mass of Lithium = 7.016 amu ;
Mass of Proton = 1.008 amu ;
Mass of Helium = 4.004 amu ;
1 a.m.u = 931.5 MeV energy
Q = [Total mass of the reactants – total mass of the products] c²
= [mass of Lithium + mass of Proton – ( 2 × mass of Helium)] 931.5 MeV
= [ 7.016 + 1.008 – 2 ( 4.004 ) ] 931.5 MeV
= [8.024 – 8.008] 931.5. MeV
∴ Energy Q. = 0.016 × 931.5 = 14.904 MeV

Question 13.
The half-life of radium is 1600 years. How much time does 1 g of radium take to reduce to 0.125 g. [TS May, ’18, ’16, Mar. ’16]
Answer:
1g become \(\frac{1}{2}\) g after one half-life period and it become \(\frac{1}{4}\) g after 2 half-life periods and \(\frac{1}{8}\) after 3 half-life periods.
But \(\frac{1}{8}\) g = 0.125g.
So number of half-life periods, n = 3.
Time taken = n x half-life period
= 3 × 1600 = 4800 years.

Question 14.
Plutonium decays with a half-life of 24.000 years. If plutonium is stored for 72,000 years, what fraction of it remains?
Answer:
Given, Half-life of Plutonium = 24,000 years;
The duration of the time = 72,000 years
Initial mass = x g ; Final mass = mx g

Fraction of Plutonium that remains

Question 15.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Answer:
Given, Fraction of substance decays

Question 16.
The half-life period of a radioactive sub¬stance is 20 days. What is the time taken for 7 / 8th of its original mass to disintegrate?
Answer:
Given, Half-life period = 20 days

∴ Time taken to disintegrate
= n × Half-life time = 3 × 20 = 60 days

Question 17.
How many disintegration per second will occur in one gram of ²³⁸₉₂U if its half – life against ∝ – decay is 1.42 × 10¹⁷ s?
Answer:
Given, Half-life period T = 1.42 × 10¹⁷s.;
mass of uranium m = 238

Number of disintegrations / sec = nλ

Question 18.
The half – life of a radioactive substance is 100 years. Calculate in how many years die activity will decay to 1/10th of its initial value.
Answer:
Given, Half – life period = 100 years

∴ Time taken to disintegrate
= n × Half-life time = 3 × 20 = 60 days

Question 19.
One gram of a radium is reduced by 2 milligram in 5 years by β – decay. Calculate the half-life of radium.
Answer:
Initial Mass of radium M₀ = 1 gram ;
∴ Mass reduced = 2mg
Final mass of radium M = 1 – 0.002
= 0.998 mg;
Time taken to reduce the mass t = 5 years
But number of atoms N α mass of the sample

Question 20.
The half-life of a radioactive substance is 5000 years. In how many years, its activity will decay to 0.2 times of its initial value? Give log₁₀ 5 = 0.6990.
Answer:
Half-life period T = 5000 years ;
Activity A = Nλ = 0.2 times initial value
Initial activity A₀ = N₀λ
In radioactivity N = N₀e^-λt

Question 21.
An explosion of atomic bomb releases an energy of 7.6 x 1013 J. If 200 MeV energy is released on fission of one atom calculate CD the number of uranium atoms undergoing fission. 00 the mass of uranium used in the bomb.
Answer:
i) Energy released E = 7.6 × 10¹³J
Energy released per fission = 200 MeV
= 200 × 10⁶ × 1.6 × 10^-19J
= 200 × 1.6 × 10^-13J = 3.2 × 10^-11 J

ii) Number of Uranium atoms participated

Question 22.
If one microgram of ²³⁸₉₂U is completely destroyed in an atom bomb, how much energy will be released? [AP Mar. ’19]
Answer:
Given, Mass of Uranium destroyed = 1
Micro gram = 1 × 10^-6 gr = 10^-9 Kg.
According to Einstein mass energy relation Energy released E = me² = 10^-9 × [ 3 × 10⁸]² (∵ Velocity of light c = 3 x 108 m/s)
∴ Energy released = 9 × 10^-9 × 10¹⁶ = 9 × 10⁷J

Question 23.
Calculate the energy released by fission from 2g of ²³⁸₉₂U in kWh. Given that the energy released per fission is 200 MeV.
Answer:
Given, Mass of Uranium (m) = 2g ;
Energy per fission = 200 MeV
Number of atoms in 2 grams of Uranium

Energy released per fission = 200 MeV
= 200 × 10⁶ × 1.6 × 10^-19J
= 200 × 1.6 × 10^-13 J = 3.2 × 10^-11J

Total Energy released ’Q’ = 5.1256 × 3.2 × 10^-11 = 1640.2 × 10⁸J
But, 1 kWh = 1000 × 60 × 60 = 36 × 10⁵J

Question 24.
200 MeV energy is released when one nucleus of ²³⁸U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Answer:
Given, Total power produced = 1 Mega Watt = 10⁶ Watt
Energy per fission = 200 MeV
= 200 × 1.6 × 10^-13 J = 3.2 × 10^-11 J
∴ Number of fissions per second

Question 25.
How much ²³⁸U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass ²³⁸U is converted into energy?
Answer:
Power P = 400 MW = 400 × 10⁶ watt;
Time T = 1 day = 86,400 sec.
∴ Energy produced per day
= 400 × 10⁶ × 86400 = 3.456 × 10¹³J
But Energy E = me² × m = E/c²
Mass of uranium consumed

Telangana TSBIE TS Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves

Very Short Answer Type Questions

Question 1.
What is the average wavelength of X-rays?
Answer:
Wavelength range of X-rays = 10^-8 m to 10^-13 m i.e., 1 nm to 10^-4 nm.
∴ Average wavelength of X-rays is \(\frac{1+10^{-4}}{2}\) nm = 0.5 nanometers (nearly)

Question 2.
Give any one use of infrared rays. [TS Mar. ’19; AP May ’18. ’17]
Answer:

1. Infrared detectors are used in satellites both for military purpose and to observe growth of crops.
2. Infrared radiation is responsible to keen the atmosphere warm through green house effect.

Question 3.
If the wavelength of electromagnetic radiation is doubled, what happens to die energy of photon? [TS Mar. ’16; June ’15]
Answer:
Energy of electromagnetic photon E = hυ = \(\frac{hc}{\lambda}\)

So when wavelength λ is doubled energy of photon is reduced to half.

Question 4.
What is the principle of production of electromagnetic waves?
Answer:
Electromagnetic waves are produced by accelerating charges through conductors.

Question 5.
How are Microwaves produced? [AP Mar. ’15]
Answer:
Microwaves can be produced by special type of vacuum tubes. Namely Klystrons, magnetrons and yun diodes.

Question 6.
What is sky wave propagation? [AP June ’15]
Answer:
In the frequency range from a few MHz upto about 30 MHz, long distance communication can be achieved by the ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation and it is used by short wave broadcast services.

Question 7.
What is the ratio of speed of infrared rays and ultraviolet rays in vacuum?
Answer:
Ratio =1:1
In vacuum speed of electromagnetic waves v = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
It is applicable to whole range of electromagnetic spectrum.
∴ In vacuum velocity of UV rays = Velocity of I.R rays.

Question 8.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave?
Answer:
Amplitude of Magnetic field

Relation between amplitude of Magnetic field B0 and Electric field E0 in vacuum is B₀ = \(\frac{E_0}{c}\)

Question 9.
What are the applications of microwaves? [AP Mar. 18, 17, 15; May 18, 16 , 15; June 15; TS Mar. 18, 17,15, May. 18]
Answer:

1. Microwaves are widely used in radar because of their small wavelength.
2. Microwaves are used in microwave ovens due to the frequency of microwave region is nearly equals to resonant frequency of water molecule.

Question 10.
Microwaves are used in Radars, why?
Answer:
Due to short wavelengths microwaves are suitable for the radar systems used in air craft navigation.

Question 11.
Give two uses of infrared rays. [TS Mar. ’17; May ’16; AP Mar. ’19, ’16; May ’14]
Answer:

1. Infrared detectors are used in satellites both for military purpose and to observe growth of crops.
2. Infrared radiation is responsible to keen the atmosphere warm through green house effect.

Question 12.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates?
Answer:
Charging current i = conduction current

∴ Displacement current i_d = 0.6 A across the plates.

Short Answer Questions

Question 1.
What does an electromagnetic wave consists of? On what factors does its velocity in vacuum depend?
Answer:
According to Maxwell’s theory, accelerated charges radiate electromagnetic waves.

Suppose a charge oscillates with some frequency. It produces an oscillating electric field in space. This oscillating electric field produces oscillating magnetic field. It again regenerates oscillating electric field. These electric and magnetic fields are mutually perpendicular and also perpendicular to direction of propagation. Electric field component E_x = E₀ sin (kz – ωt) Magnetic field component By = B0 sin (kz – ωt)

where k = \(\frac{2 \pi}{\lambda}\) and speed of wave v = \(\frac{\omega}{k}\)

According to Maxwell’s equations, relation between E₀ and B₀ \(\frac{E_0}{B_0}\) = c or B₀ = \(\frac{E_0}{c}\)

In vaccum velocity of electromagnetic wave c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
∴ Velocity of electromagnetic wave in vacuum depends on permeability µ₀ and permittivity ∈₀ of vacuum.

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth?
Answer:
Greenhouse effect :
Infrared rays plays an important role in maintaining average temperature of earth’s atmosphere.

During daytime, earth absorbs the energy of incoming Infrared & Visible radiation from Sun. As a result earth’s surface get heated. Hot earth will reradiate energy in the form of infrared radiation.

Many molecules such as Carbon dioxide (CO₂), Ammonia (NH₃), Chloro Fluoro Carbons, Methane (CH₄), etc. will absorb the long range infrared radiation because the resonant frequency of these gases matches with infrared region. So energy reradiated by earth is trapped by these molecules as a result earth’s atmosphere is heated. This is known as “greenhouse effect”.

Effect of greenhouse gases :
When concentration of greenhouse gases such as CO₂, NH₃, CH₄ etc., increases average temperature of .earth is also gradually increase. This effect is called global warming.

Long Answer Questions

Question 1.
Give the brief history of discovery of knowledge of electromagnetic waves.
Answer:
Maxwell conducted an experiment to apply the ampere’s law to study the magnetic field at a point outside the capacitor by applying a time varying current.

Consider a point ‘P’ outside the parallel plate capacitor [See Fig (a)]. Consider a loop of radius ‘r’. Its direction is perpendicular to the conductor.
From Amperes Law B(2πr) = µ₀ i (t) → (1)

Consider another different surface [See Fig (b)]. It is a pot like dielectric surface without electrical contact with capacitor plate. But it encloses the capacitor plate.

By applying ampere’s Law LHS of eq. (1) is not changed but RHS = 0. Because there is no current flowing in the plate.
∴ B(2πr) = 0 → (2)

Consider another surface like a tiffin box [See Fig (c)] apply ampere’s circuital law with same parameters LHS of eq (1) is same. But RHS is zero
i.e. B(2πr) = 0 → (3)

From eq. (1) there is a magnetic field at ‘P’ But for the figures (b) & (c) field at ‘P’ is zero. So there is some contradiction in calculating magnetic field.

Maxwell suggested that there is some electric field E = \(\frac{Q}{Ae_0}\) between the plates
and it is perpendicular to the plates. E has some value in between the plates and vanishes outside the plate.

Electric flux between the plates Φ_E = |E|A\(\frac{Q}{\epsilon_0}\)

Charge Q on plates of capacitors changes with time as a result Φ_E changes with time.

Changing electric flux will produce the current called displacement current i_d = ∈₀\(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\).

So Maxwell suggested that source of magnetic field is not only due to conduction current i_c

but due to a total current i = i_c + i_d and

So source of magnetic field is not just conduction current, it is also produced by displacement current.

Displacement current will have same physical properties just like conduction current.

Basing on this he formulated that a time varying electric field will produce a time varying magnetic field. Similarly a time varying magnetic field will produce a time varying electric field. In this way energy oscillates between perpendicular electric and magnetic fields in an electromagnetic wave.

Question 2.
State six characteristics of electromagnetic waves. What is Greenhouse effect?
Answer:
From Maxwell’s theory

1. accelerated charges radiate electromag¬netic waves.
2. Frequency of electromagnetic wave is equal to frequency of oscillator.
3. Energy associated with the propagation of wave is obtained from oscillating source.
4. From Maxwell’s equations relation between E₀ and B₀ is \(\frac{E_0}{B_0}\) = c or B₀ = \(\frac{E_0}{c}\)
5. In vacuum velocity of electromagnetic wave c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
6. Hertz experiments on electromagnetic waves showed that electromagnetic waves of wavelength 10 million times more than light waves could be diffracted, reflected and polarised.
   Electromagnetic waves carry energy and momentum like other waves.
7. In vacuum speed of electromagnetic waves v = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

It is applicable to whole range of electromagnetic spectrum.

Greenhouse effect :
Infrared rays plays an important role in maintaining average temperature of earth’s atmosphere.

During daytime earth absorbs the energy of incoming Infrared & Visible radiation comming from Sun. As a result earth’s surface get heated. Hot earth will reradiate energy in the form of infrared radiation.

Many molecules such as CO₂ Ammonia (NH₃), Chloro Fluro Carbons, Methane CH₄ etc., will absorb the long range infrared radiation because the resonant frequency of these gases matches with infrared region. So energy reradiated by earth is trapped by these molecules as a result earth’s atmosphere is heated. This is known as “greenhouse effect”.

Intext Question and Answers

Question 1.
What physical quantity is the same for X- rays of wavelength 10^-10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 in?
Answer:
The speed of light (3 × 10⁸ m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

Question 2.
A plane electromagnetic wave travels in ‘ vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave, o = 30 MHz
= 30 × 10⁶ s^-1
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Wavelength of a wave is given as, λ = \(\frac{c}{v}\)
\(\frac{3\times10^8}{30\times10^6}\) = 10 m

Question 3.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
A radio can tune to minimum frequency, υ₁ = 7.5 MHz = 7.5 × 10⁶ Hz
Maximum frequency, υ₂ = 12 MHz
= 12 × 10⁶ Hz ;
Speed of light, c = 3 × 10⁸m/s
Relation between λ and υ is λ₁ = \(\frac{c}{υ_1}\)

Thus the wavelength band of the ratio is 40 m to 25 m.

Question 4.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 10⁹ Hz.

Question 5.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Magnetic field B₀ = 510 nT = 510 × 10^-9 T ;
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀ = 3 × 10⁸ × 510 × 10^-9 = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 13th Lesson Atoms Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 13th Lesson Atoms

Very Short Answer Type Questions

Question 1.
What is the angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom?
Answer:
Angular momentum L = \(\frac{nh}{2 \pi}\)
For 2nd orbit ⇒ n = 2.
∴ Angular momentum of 2nd orbit
L₂ = \(\frac{2h}{2 \pi}=\frac{h}{\pi}\)

Question 2.
What is the expression for fine structure constant and what is its value?
Answer:
Fine structure constant a = \(\frac{2 \pi e^2}{ch}=\frac{1}{137}\)
= 7.30 × 10^-3.

Question 3.
What is the physical meaning of ‘negative energy of an electron’?
Answer:
Energy is always +ve. There is no negative energy. But negative energy of electron means it is the force of attraction with which an electron is bounded to the nucleus.

Question 4.
Sharp lines are present in the spectrum of a gas. What does this indicate?
Answer:
When gases are excited they will absorb the exact quanta of energy equals to energy level difference of the orbits.
i.e., E = hv = E_f – E_i

So electron goes to higher orbit. As a result we will see sharp lines in the spectrum of gases corresponding to that energy.

Question 5.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Angular momentum (L) = mvr,
Dimensions = ML²T^-1
Planck’s constant h has same dimensions of L.
∵ L = \(\frac{nh}{2 \pi}\), where \(\frac{n}{\pi}\) is dimensionless.

Question 6.
What is the difference between α-particle and helium atom?
Answer:
α – particle is helium nucleus. But not helium atom.

α – particle contains 2 protons and two neutrons where as helium atom contains 2 protons, two neutrons and two electrons.

Question 7.
How is impact parameter related to angle of scattering?
Answer:
When impact parameter is minimum head on collision between α – particle and nucleus takes place and α- particle will be deviated by θ = π radians, when impact parameter is high or large the α – particle goes nearly undeviated.

Question 8.
Among alpha, beta and gamma radiations, which get affected by the electric field?
Answer:
α and β particles are affected by electric fields because both of them carries charge on them.

γ – ray is radiation so it is not affected by electric field.

Question 9.
What do you understand by the phrase ‘ground state atom’?
Answer:
The lowest state of the atom which is the lowest energy, with the electron revolving in the orbit of smallest radius is called the ground state.
For ground state hydrogen atom E = – 13.6 eV,

Question 10.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment?
Answer:
In Rutherford experiment gold foil is used. For gold Z = 79 and its atomic weight is nearly 50 times more than α – particle. So mass of nucleus used to bombard gold atom has no significant effect.

Question 11.
The Lyman series of hydrogen spectrum line in the ultraviolet region Why?
Answer:
For Lyman series \(\frac{1}{\lambda}\) = R(\(\frac{1}{l^2}-\frac{1}{n^2}\))
n = 2, 3, 4 ….
Wavelength of 1st member of Lyman series is 1216A°. It is in ultraviolet region.

Question 12.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:

Question 13.
The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 1216Å, 6463 Å and 9546Å. Which one of these wavelengths belongs to the Paschen series?
Answer:
λ = 9546Å belongs to Paschen series. Paschen series is in infrared region. So wavelength λ = 9546Å is in infrared region. So it belongs to Paschen series.

Question 14.
Give two drawbacks of Rutherford’s atomic model. [TS Mar. ’19]
Answer:

1. Electron revolving around the nucleus must be continuously accelerated. An accelerating electron must lose energy continuously due to radiation and finally, atom must destroy.
2. When electron radiates energy it will spiral around nucleus. As a result its angular velocity and frequency of spectral lines must change continuously. But these two things are not taking place.

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering? How are they related to each other?
Answer:
Impact parameter :
It is the perpendicular distance of the initial velocity vector of α-particle from centre of nucleus.

In case of head-on collision, impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small. It, in turn, suggested that mass of atom is much concentrated in a small volume.

Angle of scattering ‘0’ :
It is the angle between the direction of incident α – particle and scattered α – particle.

Relation between impact parameter and angle of scattering, when Impact parameter is less ⇒ angle of scattering is high and vice-versa.

Question 2.
Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr’s atomic model. How does P.E. change with increasing n? [TS June 15; Mar. 15]
Answer:
According to Bohr model, electrons are revolving around the nucleus in certain permitted orbits.

For electron to revolve in orbit, the electrostatic force and centrifugal force must be equal i.e., F_c = F_e

∴ Kinetic energy of electron (K) = \(\frac{1}{2}\) mv²

Potential energy between electron and nucleus is

– ve sign indicates force of attraction. Relation between potential energy and radius of orbit.
From the equation (2) U ∝ \(\frac{1}{r}\) …………… (3)

From eq. (3) & (4) potential energy
U ∝ \(\frac{1}{n^2}\)
∴ Potential energy of orbit is inversely proportional to square of number of orbit (n²).

Question 3.
What are the limitations of Bohr’s theory of hydrogen atom? [TS May 18, 17; AP Mar. 17, 14, May 17]
Answer:
Limitations of Bohr model:
i) Bohr model is applicable to hydrogen atom only. It ean not be extended even to a two electron system such as helium.

Because it involves force between + vely charged nucleus and electron.

Electrical forces between electrons are not taken into account.

ii) It is not able to explain the intensity variation in spectral lines of different frequencies and why some transitions are more preferred than others.

Question 4.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :
In α- particle scattering experiment the α – particle will move near to gold nucleus until it is just stopped.

Kinetic energy of α – particle just before stopped is equal to electrostatic potential.

where = Z atomic number of gold, ’2e’ charge on α – particle.
distance of closest approach d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{k}}\)

Impact parameter :
It is the perpendicular distance of the initial velocity vector of α – particle from centre of nucleus.

In case of head-on collision impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small. It in turn suggested that mass of atom is much concentrated in a small volume.

Question 5.
Give a brief account of Thomson model of atom. What are its limitations?
Answer:
J.J. Thomson Model :
J.J.Thomson thought that the positive charge of the atom is uniformly distributed through out the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.

Limitations:

1. Rutherford α – particle scattering experiment showed that in an atom positive charge is concentrated at nucleus of atom.
2. In Thomson model the radiation emitted by solids and gases is due to oscillations of atoms and molecules are governed by the interactions between them.

But experiments on rarefied gases showed that hydrogen always gives rise to a set of lines with fixed wavelength.

Balmer experiments and his formula for Balmer series of wavelength of a group of lines are emitted by atomic hydrogen only. Not by interaction of atoms or molecules.

Question 6.
Describe Rutherford atom model. What are the drawbacks of this model? [AP & TS Mar. ’16]
Answer:
Rutherford’s nuclear model:

1. According to Rutherford the entire positive charge and most of the mass of the atom is concentrated at nucleus. Electrons are revolving around the nucleus as planets revolve around the sun.
2. Rutherford experiments on α – particle suggested that size of atom is about 10^-15 to 10^-14 and size of nucleus is about 10^-10 m.

Drawbacks :

1. When electron is revolving round the nucleus it is in continuously accelerated state. As per classical mechanics, it must emit energy continuously. So electron must follow spiral path instead of circular path. Finally electron must fall on nucleus i.e., + ve charge and atom will destroy which is not happening.
2. This model is not able to explain why electron in an orbit is not radiating any energy.

Question 7.
Distinguish between excitation potential and ionization potential.
Answer:
Excitation potential :

1. Generally gases or vapours are excited at low pressure by passing high current through them. For this we have to apply a very high voltage.
2. When excited atoms or molecules will absorb certain amount of energy from applied potential and gives rise to series of spectral lines called emission spectrum.

Ionisation potential :
It is the amount of minimum energy required to release an electron from the outer most orbit of the nucleus.

From Bohr’s model, energy of the orbit is the ionisation energy of electron in that orbit.
Ex: Energy of 1st orbit in hydrogen is 13.6 eV.

Practically ionisation potential of hydrogen is 13.6 eV.

Difference:
→ Excitation potential will permit electrons to transit between various energy levels whereas ionisation potential will liberate an electron from the influence of nucleus of that atom.

Question 8.
Explain the different types of spectral series of hydrogen atom.
(OR)
Write the different types of Hydrogen Spectral series. The Lyman series of Hydrogen spectrum lies in the ultraviolet region. [TS May 16, Mar. 16, AP Mar. 19, 15, May 18, 16, June 15]
Answer:
Lyman series :
When electrons are jumping from higher energy levels to the first orbit then that series of spectral lines emitted are called “Lyman series”.
In Lyman series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{l^2}-\frac{1}{n^2}\))
where n = 2, 3, …………… etc.

These spectral lines are in ultraviolet region.

Balmer series :
When electrons are jumping the from higher levels to 2nd orbit then that series of spectral lines are called “Balmer series”.

For Balmer series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\))
where n = 3, 4, ………… etc.

Spectral lines of Balmer series are in visible region.

Paschen series :
When electrons are jumping on to the 3rd orbit from higher energy levels then that series of spectral lines are called “Paschen series”.
For Paschen series \(\frac{1}{\lambda}\) = R(\(\frac{1}{3^2}-\frac{1}{n^2}\))
where n = 4, 5, …………..

These spectral lines are in near infrared region.

Brackett series :
When electrons are jumping on to the 4th orbit from higher levels then that series of spectral lines are called “Brackett series”.
For Brackett series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{4^2}-\frac{1}{n^2}\))
where n = 5, 6, ………… etc.

Brackett series are in middle infrared region.

Pfund series :
When electrons are jumping on to the 5th orbit from higher energy levels then that series of spectral lines are called “Pfund series”.
For Pfund series \(\frac{1}{\lambda}\) = R(\(\frac{1}{5^2}-\frac{1}{n^2}\))
n = 6, 7, ……….. etc.

These spectral lines are in far infrared region.

Reason for Lyman series in ultraviolet region :
The wavelength of Lyman series is nearly 1200 A° and less. So Lyman series is ultraviolet region.

Question 9.
Write a short note on de Broglie’s explanation of Bohr’s second postulate of quantization. [TS Mar. ’17]
Answer:
De – Broglie’s explanation for Bohr 2nd postulate :
According to de Broglie, the electron in an orbit must be seen as a particle wave. Particle waves can lead to standing waves under resonance condition.

A standing wave can be formed in a wire when length of the wire is equal to wave-length λ or integral multiple of wavelength λ (i.e., nλ). In the same way for an electron moving in a circular orbit, if its circum-ference 2πr_n is equal to nλ then standing waves will be formed in that orbit.
∴ From de-Broglie’s explanation 2πr_n = nλ
But λ = h/p = \(\frac{h}{mv_n}\)
∴ de Broglie wavelength 2πr_n = nh / mv_n
mv_nr_n = \(\frac{nh}{2 \pi}\)
Where mv_nr_n is angular momentum of electron in nth orbit.
In this way de – Broglie hypothesis provided an explanation to Bohr’s 2nd postulate.

Long Answer Questions

Question 1.
Describe Geiger – Marsden Experiment on scattering of α – particles. How is the size of the nucleus estimated in this experiment?
Answer:
Geiger – Marsden experiment :
Description :
In this experiment α – particles from ²¹⁴₈₃Bi are passed through lead blocks containing coaxial holes to emit narrow beam of α – particles. This α – particle beam falls on gold foil and scatters. The scattered α – particles are detected. By rotating the microscope at different angles (θ) number of scattered particles are measured.

Explanation :
In this experiment a beam of α – particles of energy 5.5 MeV are made to fall on gold foil.

α – particle carries 2 units of positive charge and mass is equal to that of helium nucleus.

→ Assume that the gold foil is very thin and α – particle will suffer not more than one scattering during its passage through gold foil.

For gold Z = 79. Its mass is nearly 50 times more than that of α – particle. So during collision it remains almost stationary.

Magnitude of force between α – particle and gold nuclei (F) = \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\), ‘r’ = the distance between them.

The magnitude and direction of force changes continuously as it approaches the nucleus.

Impact parameter :
It is the perpendicular distance of the initial velocity vector of α – particle from centre of nucleus.

In case of head – on collision impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small.

Estimation of size of nucleus :
In this experiment a graph is plotted between number of scattered particles (n) and angle of scattering ‘θ’. It suggested that size of nucleus is about 10^-15 to 10^-14 m. From kinetic theory size of atom is about 10^-10 m. So size of atom is 10,000 to 100,000 times more than nucleus.

Question 2.
Discuss Bohr’s theory of the spectrum of hydrogen atom. [AP Mar. ’16]
Answer:
From Bohr’s model of hydrogen atom

1. Electrons are revolving in certain per-mitted non-radiating orbits around nucleus.
2. For non-radiating orbits, orbital angular momentum L = mvr = nh / 2π.
3. Electrons may jump from one orbit to another orbit. While doing so they will emit or absorb the energy equal to diffe-rence of those orbital energies.
   E = hν = E_i – E_f

When electrons are revolving in the orbit centrifugal force and centripetal force on it are equal F_c = F_e.

For hydrogen atom n = 1
∴ r₁ = h²ε₀/π me² ……………. (5)
This is also called Bohr radius
a₀ = 5.29 × 10^-11 m
Energy of the orbit E = K + U = Kinetic energy + Potential Energy of electron.
∴ E_n = –\(\frac{e^2}{8\pi \epsilon_0 r}\)
By using value of r1 from eq. 4.

From 3rd postulate :
when electrons are jumping from higher to lower orbit.

Where R = Rydberg’s constant. (∵ Electron is initially at orbit n₂ and finally at n₁)

In this way Bohr atom model successfully explained energy of the orbits and origin of spectral lines.

Question 3.
State the basic postulates of Bohr’s theory of atomic spectra. [AP Mar. ’16]
Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom.
Answer:
Bohr’s postulates :
Bohr model of hydrogen atom consists of three main postulates.
i) Electrons in an atom could revolve in certain permitted stable orbits. Electrons revolving in these stable orbits do not emit or radiate any energy.

ii) The stable orbits are those whose orbital angular momentum is an integral multiple of h/2π.
i.e., L = nh / 2π where n = 1, 2, 3 etc. (an integer.)
These stable orbits are also called as non – radiating orbits.

iii) An electron may take a transition between non-radiating orbit.

when electron transition takes place a photon of energy equals to the energy difference between initial and final states will be radiated.
E = hν = E_i – E_f.

Bohr Model of Hydrogen atom :
When electrons are revolving in the orbit electrostatic force and centripetal force on it are equal F_c = F_e.

This is also called Bohr radius
a₀ = 5.29 × 10^-11 m
Energy of the orbit E = K + U = Kinetic energy + Potential Energy of electron.

Intext Question and Answer

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 × 10^-11m. What is the radius of the second orbit?
Answer:
Radius of first orbit r₁ = 5.3 × 10^-11m.
Radius of Bohr orbit r =
For 2nd orbit n = 2,
∴ r = n₂ a₀ =4 × 5.3 10^-11 = 2.12 × 10^-11 m.

Question 2.
Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ? Given: h = 6.62 × 10^-34J s, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, k = 9 × 10⁹ m²C.
Answer:
Given h = 6.62 × 10^-34 J,
Mass m = 9.1 × 10^-31 kg
Charge e = 1.6 × 10^-19 C;

Question 3.
The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the potential energy of the electron in this state?
Answer:
Total energy, T.E. = – 3.4 eV.
Potential energy U = 2 x T.E. = – 3.4 × 2
= – 6.8 eV in same orbit.

Question 4.
The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the kinetic energy of the electron in this state?
Answer:
In 1st excited state
Total energy T.E. = – 3.4 eV.
Kinetic energy in 1st excited state = T.E. – U
Where U = 2 × T.E.
∴ K = U – 2U = – U = – (-3.4) = 3.4 eV.

Question 5.
Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in n = 1 orbit. Given h = 6.63 × 10^-34 Js, m = 9.1 × 10^-31kg, e = 1.6 × 10^-19 C, k = 9 × 10⁹ N m²C^-2.
Answer:
Planck’s constant h = 6.63 × 10^-34 Js.;
Mass of electron m = 9.1 × 10^-31 Kg.
Charge on electron e = 1.6 × 10^-19 C;
K = 9 × 10⁹ m²C²
a) Radius of orbit, r \(\frac{n^2h^2\epsilon_0}{\pi me^2}\) where \(\frac{h^2\epsilon_0}{\pi me^2}\) = a₀ = 5.3 × 10^-11
For 1st orbit n = 1 ⇒ r = 1 × 5.3 × 10^-11
= 5.3 × 10^-11m

Question 6.
Prove that the ionization energy of hydrogen atom is 13.6 eV.
Answer:
From Bohr atom model energy radiated when electrons are jumping from one orbit
to another orbit is E = hυ = \(\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}\) = E₁ – E₂

By definition, ionization potential is the minimum amount of energy required to remove the electron from the orbit. Now for an electron of nearest orbit n₁ = 1 and n₂ = ∞ (because electron is free from influence of nucleus).

Question 7.
Calculate the Ionization energy fora lithium atom.
Answer:

For lithium Z = 3 (But outer most orbit n = 2 in lithium)
∴ lonization energy = Energy of the orbit = \(\frac{13.6\times9}{4}\) = 30.6 eV.

Question 8.
The wavelength of the first member of Lyman series is 1216 Å. Calculate the wavelength of second member of Balmer series.
Answer:
Wavelength of 1st member of Lyman series = 1216 A°.

2nd member of Balmer series ⇒ \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{4^2}\))
∵ For 2nd member of Balmer series n = 4.

Question 9.
The wavelength of first member of Balmer series is 6563 A. Calculate the wavelength of second member of Lyman series.
Answer:
Wavelength of 1 st member of Balmer series = 6563.

Question 10.
The second member of Lyman series in hydrogen spectrum has wavelength 5400 Å. Find the wavelength of first member.
Answer:
Wavelength of 2nd member of Lyman series λ = 5400

Question 11.
Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer series limit.
Given : R = 10970000m^-1.
Answer:
Given Rydberg constant R = 10970000 m^-1
= 1.097 × 10⁷ m^-1
For Balmer series \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\))
Short wavelength ⇒ n₂ = ∞

Question 12.
Using the Rydberg formula, calculate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum.
Answer:
For Balmer series \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\)) where
n = 2, 3, 4, 5 etc.
Rydberg constant R = 1.097 × 10⁷ m^-1


= 4102 A°

Telangana TSBIE TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Type Questions

Question 1.
What are “Cathode rays”? [TS Mar. 19; AP mar. 1 7, may 18, 14]
Answer:
Cathode rays consists of a steam of fast moving negatively charged particles.

Speed of cathode rays ranges about 0.1 to 0.2 times light velocity (3 × 10⁸m/s)

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan proved the validity of Einstein’s photo electric equation and his successful explanation of photo electric effect using light quanta. He experimentally found the value of Planck’s constant ‘h’ and work function on ‘Φ’ of photo electric surface.

Question 3.
What is “work function”? [AP Mar. 19, May 16; TS Mar. 18. 17, 15]
Answer:
Work function (Φ) :
The mininum energy required by an electron to escape from metal surface is called “work function”.

Work function depends on nature of metal.

Question 4.
What is “photoelectric effect”? [AP Mar. ’18, ’17, ’15, ’14, May ’14; TS May ’17, Mar. ’18. ’16]
Answer:
Photoelectric effect :
The process of liberating an electron from the metal surface due to light energy falling on it is called “photoelectric effect”.

Question 5.
Give examples of “photosensitive substances”. Why are they called so?
Answer:

1. Metals like zinc, cadmium and magnesium will respond to ultraviolet rays.
2. Alkalimetals such as sodium, potassium, caesium and rubidium will respond to visible light.

These substances are called “photo sensitive surfaces” because they will emit electrons when light falls on them.

Question 6.
An electron, an a particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength? [TS May, ’18, Mar. ’15]
Answer:
De-Broglie wavelength λ = \(\frac{h}{p}\); but KE = \(\frac{P^2}{2m}\)

For a particle m is more them given other particles.
So De-Broglie wavelength of a – particle is less.

Question 7.
What is Photo-electric effect? How did Einstein’s photo-electric equation explain the effects of intensity (of light) and potential on photo-electric current? [AP May 18, TS June 15]
Answer:
Photoelectric effect :
The process of emitting electron from the metal surface when light energy falling on it is called “photo-electrie effect”.

According to Einstein radiation consists of discrete units of energy called quanta of energy radiation.

Energy of quanta is called photon in light E = hυ

Maximum kinetic energy of photoelectron (K_max) is the difference of energy of incident radiation (hυ) and work function (Φ)
∴ K_max = hυ – Φ (when υ > υ₀)

Photoelectric equation can also written as

Effect of intensity :
As per Einstein’s photoelectric equation energy of photon E = hυ decides weather a photon will come out of metal surface or not. If frequency of incident light υ > υ₀ then electron will come out of that surface.

Number of electrons liberated depends on the number of photons striking the surface i.e., on intensity of light. So as per Einstein’s equation photocurrent liberated must be linearly proportional to intensity of light which is a practically proved fact.

Effect of voltage on photocurrent :
When positive potential on collector is gradually increased then photocurrent i.e., also gradually increased upto certain limit called saturation current all the photoelectrons liberated from. Photosurface reached the collector.

When υ >υ₀ photoelectron is released. The positive potential on collector will accelerate the electron. So it reaches the collector quickly. So photocurrent increases. But collector potential deals nothing with liberation of electron from photosurface.

In this way Einstein’s photoelectric equation explained the saturation of photocurrent with increasing collector +ve potential. Effect of frequency and stopping potential.

From Einstein’s photoelectric equation
K_max = hυ – Φ

i.e., kinetic energy of photoelectron is directly proportional to frequency of incident light.

Question 8.
Write down Einstein’s photoelectric equation. [AP Mar. 19, 15, May 17; TS May 18, 16]
Answer:
Maximum kinetic energy of photo electron Kmax is the difference of energy of incident radiation (hυ) and work function (Φ)
K_max= hυ – Φ (when υ > υ₀)
OR
K_max = eV₀ = hυ – Φ Or V₀ = \(\frac{h}{e}\)υ – \(\frac{\phi_{0}}{e}\)

Question 9.
Write down de Broglie’s relation and explain the terms there in. [AP & TS Mar. 18, 16; TS May 17]
Answer:
de -Broglie assumed that matter will also exhibit wave nature when it is in motion.
de – Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{mυ}\)
P = mo, momentum of the body, h = Planck’s constant, λ = wave length of moving particle.

Question 10.
State Heisenberg’s uncertainty principle. [TS Mar. 17; AP Mar. 14]
Answer:
Heisenberg’s uncertainty principle :
We cannot exactly find both momentum and position of an electron at the same time. This is called Heisenberg’s uncertainty principle.

Short Answer Questions

Question 1.
What is the effect of (i) intensity of light (ii) potential on photoelectric current? [TS Mar. ’19, June ’15]
Answer:
Effect of intensity :
The number of photo electrons liberated is directly proportional to intensity of incident radiation. So photo current increases linearly with increase of intensity of incident light.

Effect of potential :
When positive potential given to collector, photocurrent is gradually increased up to certain limit called saturation current. In this stage all the photo electrons liberated from photo surface reached the collector.

When negative potential on collector is gradually increased electrons are repelled by collector and photo current decreases.

At a particular negative voltage photo current is zero.

Stopping potential :
The minimum negative potential required by collector to stop photo current (or) becomes zero is called cut-off voltage (V₀) stopping potential.

Question 2.
How is the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 volts? [AP Mar. ’15]
Answer:
de-Broglie wavelength λ = \(\frac{h}{\sqrt{2 \mathrm{~m}} \mathrm{eV}}\)
Potential difference V = 100V; h = 6.63 × 10^-34
Mass of electron m = 9.1 × 10^-31 kg;
charge of electron e = 1.6 × 10^-19c

Question 3.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volt? [TS May. ’16]
Answer:
Applied potential V = 100 V.
de Broglie wavelength

∴ de Broglie wavelength 1 = 0.1227 nm.

Question 4.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of photo electric effect :
To study photo electric effect a photo sensitive surface and a metallic plate called collector are arranged on an evacuated glass tube. An arrangement is made to give required positive or negative potential to collector. By changing the filters placed in the path of incident light we will allow light rays of required frequency to fall on given photo surface.

Stopping potential :
The minimum negative potential required by collector to stop photo current or photo current to become zero is called “cut off voltage V₀“.

Effect of frequency on stopping potential are

1. Stopping potential varies linearly with frequency of incident light.
2. Every photo surface has a minimum cut off frequency υ₀ for which stopping potential V₀ = 0

Question 5.
Summarise the photon picture of electromagnetic radiation.
Answer:
Radiation consists of discrete units of energy called quanta.
Energy of quanta (E) = hυ = \(\frac{hc}{\lambda}\)

In case of light energy quanta is called photon.

Properties of photons:

1. Energy of photon E = ho Momentum ho P= \(\frac{hυ}{e}\)
2. In interaction of radiation with matter light quanta will behave like particles
3. Photons are electrically neutral. So they are not deflected by electric and magnetic fields.
4. In photons-particle collision total energy and total mometum of are conserved is in collision photon will totally loose its enery and momentum.

Question 6.
What is the deBroglie wavelength of a ball of mass 0.12Kg moving with a speed of 20ms^-1? What can we infer from this result? [AP June ’15]
Answer:
Mass of ball, m = 0.12kg ;
Speed of ball, v = 20 m/s
Plancks’ constant, h = 6.63 × 10^-34J
But de Broglie wave length λ = \(\frac{h}{p}=\frac{h}{mv}\)

∴ de Broglie wave length of given moving ball X = 2.762 × 10^-34m

Question 7.
The work function of cesium is 2.14 eV. Find the threshold frequency for cesium. (Take h = 6.6 × 10^-34Js) [IMP]
Answer:
Work function Φ₀ = 2.14 eV; h = 6.6 × 10^-34 JS

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation Explain the effect of intensity and potential on photoelectric current? How did this equation account for die effect of frequency of incident light on stopping potential?
Answer:
Einstein’s photo electric equation :
According to Einstein radiation consists of discrete units of energy called quanta of energy radiation.

Energy of quanta called photon in light E = hυ

Maximum kinetic energy of photo electron (K_max) is the difference of energy of incident radiation (hυ) and work function (Φ)
∴ K_max= hυ – Φ(when υ > υ₀)
Photo electric equation can be written as

Effect of intensity :
As per Einstein’s photo electric equation energy of photon decides weather a photon will come out (or) not from metal surface. If frequency of incident light υ > υ₀, then electron will come out from surface.

Number of electrons liberated depends on the number of photons striking the surface i.e., on intensity of light. So as per Einstein’s equation photo current liberated must be linearly proportional to intensity of light which is practically proved.

Effect of voltage on photo current :
When positive potential on collector is gradually increased then photo current i.e., also gradually increased up to certain limit called saturation current. In this stage all the photo electrons liberated from photo surface reached the collector.

When υ > υ₀ photo electron Is released. The positive potential on collector will accelerate the electron. So it reaches the collector quickly. So photo current increases. But collector potential deals nothing with liberation of electron from photo surface.

Effect of frequency and stopping potential :
⇒ Kinetic energy of photo electron is directly proportional to frequency of incident light.
From Einstein’s photo electric equation
∴ K_max= hυ – Φ

Stopping potential :
The minimum negative potential required by the collector to stop photo current is called stopping potential.

At this potential even the fastest electron (or) electron with maximum kinetic energy is prevented to reach the collector.

When frequency incident light increases then K_max of electron increases. Hence stopping potential V₀ will also increase.

∴ The graph between stopping potential V₀ and frequency o must be a straight line.

Then slope of the line is \(\frac{h}{e}\). This is experimentally proved by Millikan.

In this way Einstein’s photo electric equation successfully explained photo electric effect.

Question 2.
Describe the Davisson and Germer experiment. What did this experiment conclusively prove?
Answer:
Davisson and Germer experiment :
In this experiment, electrons are produced by heating a tungsten filament coated with barium oxide with the help of a low voltage battery.

These electrons are focussed on to a nickel target in the form of a sharp beam.

This electron beam is accelerated by the strong positive potential on nickel target.

After colliding the target electron beam will get scattered.

The scattered electron beam is collected by electron beam detector called collector.

In this experiment intensity of electron beam I for different angles of scattering is measured. A graph is plotted between angle of scattering 0 and intensity I with different target potentials of 44V to 68V.

Intensity is found to be maximum at a target potential of 54V with a scattering angle of 50°

This value coincides with the de Broglie wave length of electron.

Importance :
This experiment proved the existence of matter waves practically. It gave a strong support to de Broglie’s hypothesis of matter wave concept.

Intext Question and Amswers

Question 1.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Photoelectric cut-off voltage, V₀ = 1.5 V ;
maximum kinetic energy K_e = eV₀
Where, e = Charge on an electron = 1.6 × 10^-19 C
∴ k_e = 1.6 × 10^-19 × 15 = 2.4 × 10^-19J
∴ The maximum kinetic energy of the photoelectrons emitted K = 2.4 × 10^-19 J.

Question 2.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 10³ W/m²
Hence, power of sunlight per square metre, P = 1.388 × 10³W
Speed of light, c = 3 × 10⁸ m/s ; Planck’s constant, h = 6.626 × 10^-34 Js
Average wavelength of photons present in sunlight, λ = 550 nm = 550 × 10^-19 m
Number of photons per square metre incident on earth per second = n
∴ Power p = n E

∴ 3.84 × 10²¹ photons are incident per square metre / sec on earth.

Question 3.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Answer:
The slope of the cut-off voltage (V) – frequency (v) graph is ; \(\frac{V}{v}\) = 4.12 × 10^-15 Vs
But hv = eV
Where, e = Charge on an electron = 1.6 × 10^-19 C; h = Planck’s constant
∴ h = e×\(\frac{V}{v}\) = 1.6 × 10^-19 × 4.12 × 10^-15
= 6.592 × 10^-34Js
∴ Planck’s constant h = 6.592 × 10^-34 Js.

Question 4.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Answer:
Power of the sodium lamp, P = 100 W ;
Wavelength of the emitted sodium light,
λ = 589 nm = 589 × 10^-9 m
Planck’s constant, h = 6.626 × 10^-34 Js ;
Speed of light, c = 3 × 10⁸ m/s
(a) Energy of photon E = \(\frac{hc}{\lambda}\)

(b) Number of photons delivered to the sphere = n
Power P = nE

Photons delivered per second = 2.96 × 10²⁰

Question 5.
The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:
Threshold frequency of the metal,
v₀ = 3.3 × 10¹⁴ Hz;
Charge on an electron, e = 1.6 × 10^-19 C ;
Frequency of light incident on the metal, v₀ = 8.2 × 10¹⁴ HZ;
Planck’s constant, h = 6.626 × 10^-34 Js
Cut-off voltage for the photoelectric emission from the metal = V₀; But eV₀ = h(υ – υ₀)

∴ The cut-off voltage for the photoelectric emission V₀ = 2.0292 V.

Question 6.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wave-length 330 nm?
Answer:
No.
Work function of the metal, Φ₀ = 4.2 eV;
Charge on an electron, e = 1.6 × 10^-19 C
Planck’s constant, h = 6.626 × 10^-34 Js ;
Wavelength of the incident radiation,
λ = 330 nm = 330 × 10^-9 m
Speed of light, c = 3 × 10⁸ m/s ;
The energy of the incident photon E = \(\frac{hc}{\lambda}\)

Question 7.
light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Frequency of photon, v = 488 nm
= 488 × 10^-9 m ;
Planck’s constant, h = 6.626 × 10^-34 Js
Maximum speed of the electrons, v = 6.0 × 10⁵ m/s ;
Mass of an electron, m = 9.1 × 10^-31 kg
Relation between v and K.E, \(\frac{1}{2}\) mv²
= h(ν – ν₀)⇒ ν₀ = ν –\(\frac{mv^2}{2h}\)

Threshold frequency ν₀ = 4.738 × 10⁵ Hz

Question 8.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer:
Wavelength of light produced by the argon laser, λ = 488 nm = 488 × 10^-9 m
Stopping potential, V₀ = 0.38 V ;
But leV = 1.6 × 10^-19 J
∴ V₀ = \(\frac{0.38}{1.6\times10^{-19}}\)eV
Planck’s constant, h = 6.6 × 10^-34 Js ;
Charge on an electron, e = 1.6 × 10^-19 C
Speed of light, c = 3 × 10 m/s
From Einstein’s photoelectric effect,

Question 9.
Calculate the
(a) Momentum/and
(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Potential difference, V = 56 V;
Planck’s constant, h = 6.6 × 10^-34 Js
Mass of an electron, m = 9.1 × 10^-31 kg ;
Charge on an electron, e = 1.6 × 10^-19 C

(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, for

The momentum of each accelerated electron
p = mv = 9.1 × 10^-31 × 4.44 × 10⁶
Momentum of each electron p
= 4.04 × 10^-24 kg m s^-1
(b)De Broglie wavelength of an electron,

Question 10.
What is the
(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Kinetic energy of the electron, E_k = 120 eV;
Planck’s constant, h = 6.6 × 10^-34 Js
Mass of an electron, m = 9.1 × 10^-31 kg ;
Charge on an electron, e = 1.6 × 10^-19 C
(a) Kinetic energy of electron E_k = \(\frac{1}{2}\)mv²
Where, υ = Speed of the electron

Momentum of the electron, p = mv
∴ P = 9.1 × 10^-31 × 6.496 × 10⁶
= 5.91 × 10^-24 kg m s^-1.
(b) Speed of the electron, v = 6.496 × 10⁶ m/s (from eq 1)

(c) De Broglie wavelength of an electron having a momentum p, is given as:

∴ de Broglie wavelength of the electron is 0.112 nm.

Question 11.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10^-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Mass of the bullet, m = 0.040 kg ;
Speed of the bullet, v = 1.0 km/s
= 1000 m/s
Planck’s constant, h = 6.6 × 10^-34Js
But De Broglie wavelength λ = \(\frac{h}{mv}\)

(b)Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength λ = \(\frac{h}{mv}\)

(c) Mass of the dust particle, m =1 × 10^-9 kg;
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength λ = \(\frac{h}{mv}\)

Question 12.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:
Wavelength of an electron (λ^e) and a photon (λ^p), λ^e = λ^p = λ = 1nm = 1 × 10^-9 m
Planck’s constant, h = 6.63 × 10^-34 Js
(a) The momentum of an elementary partide is given by de Broglie relation:

(b) The energy of a photon is given by the relation:

(c) The kinetic energy (A) of an electron having momentum p,is given by the relation:
m = Mass of the electron = 9.1 × 10^-31 kg;
p = 6.63 × 10^-25 kg ms^-1

Telangana TSBIE TS Inter 2nd Year Physics Study Material 10th Lesson Alternating Current Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 10th Lesson Alternating Current

Very Short Answer Type Questions

Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the secondary if the primary has 10 turns. [TS Mar. ’16; AP Mar. 19, ’18] [IMP]
Answer:
Primary voltage V_p = 200 V ;
Secondary voltage V_s = 2000 V
Number of turns in Primary N_p = 10 ;
Number of turns in Secondary = N_s = ?

Question 2.
What type of transformer is used in a 6V bed lamp? [AP Mar. ’17]
Answer:
6V Bed lamp ⇒ V_s = 6V. But our mains supply is 220V (called primary voltage V_p).
So V_s < V_p.
Hence transformer used in bed lamp is step down transformer.

Question 3.
What is the phenomenon involved in the working of a transformer? [AP May ’18, ’17, ’16; Mar. ’16, ’14; June ’15]
Answer:
A transformer uses the principle of “mutual Induction”.
It also obeys law of conservation of energy.
In a transformer V_pI_p = V_sI_s.

Question 4.
What is transformer ratio? [TS May ’18]
Answer:
The ratio of number of turns in secondary (N_s) to number of turns in primary (N_p) is called “transformer turns ratio”.
Transformer turns ratio = \(\frac{N_s}{N_p}\)

Question 5.
What is the phase difference between A.C. emf and current in the following? Pure resistor and pure inductor. [TS Mar. ’15]
Answer:

1. In pure resistor AC emf and current are in same phase.
   ∴ Phase difference f = 0.
2. In pure inductor current lags behind emf by a phase angle Φ = 90° or \(\frac{\pi}{2}\) radians.
3. In pure capacitor current leads emf by a phase angle f = 90° or \(\frac{\pi}{2}\) radians.

Question 6.
What is Step-up transformer? How it differs from Step-down transformer?
Answer:
A Step-up transformer will convert low voltage input into high voltage output.

A Step-down transformer will convert high voltage input into low voltage output.

Question 7.
Write the expression for the reactance of (I) an inductor and (ii) a capacitor.
Answer:

1. Reactance of an Inductor X_L = ωL
2. Reactance of Capacitor X_c = \(\frac{1}{\omega C}\)

Where ω = 2πυ is the angular frequency of applied voltage.

Question 8.
What is the phase difference between AC emf and current in the following : Pure resistor, pure inductor and pure capacitor. [TS Mar. ’15]
Answer:

1. In pure resistor AC emf and current are in same phase.
   ∴ Phase difference Φ = 0.
2. In pure inductor current lags behind emf by a phase angle Φ = 90° or \(\frac{\pi}{2}\) radians.
3. In pure capacitor current leads emf by a phase angle Φ = 90° or \(\frac{\pi}{2}\) radians.

Question 9.
Define power factor. On which factors does power factor depend?
Answer:
Power of AC circuit and power factor :
Power of AC circuit P = I²Z cos Φ. It indicates that power depends not only on current I and Impedance Z of circuit, but also cosine of phase angle between I and Z.

The term cos Φ is called power factor.

Question 10.
What is meant by wattless component of current? [TS Mar. ’17]
Answer:
Wattless current :
In a circuit with pure inductance or pure capacitance the phase angle between voltage and currents are Φ= \(\frac{\pi}{2}\) , so cos Φ = 0. Hence no power is dissipated through then even though current passes through them. This current is referred as wattless current.

Question 11.
When does a LCR series circuit have minimum impedance?
Answer:
Impedance of series LCR circuit

In series LCR circuit Impedance is minimum at Resonance condition.

Question 12.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity?
Answer:
Power in a circuit P = VI cos Φ
Power factor is unity cos Φ = 1 ⇒ Φ = 0°
∴ Phase difference between voltage and current Φ = 0.

Short Answer Questions

Question 1.
Obtain an expression for the current through an inductor when an AC emf is applied.
Answer:
Let an inductor (L) of negligible resistance is connected in an AC circuit. This is treated as purely inductive circuit.

Voltage applied V = V_m sin ωt
From Kirchhoff’s loop rule
V – L \(\frac{di}{dt}\) = 0 → (1) Because induced emf
ε = – L\(\frac{di}{dt}\) is in opposite direction.
∴ \(\frac{di}{dt}=\frac{V}{L}=\frac{v_m}{L}\) sin ωt → (2)
Total current through inductor

ωL is the reactance offered by inductor for the flow of current in circuit.
Reactance of Inductor X_L = ωL, similar to Resistance.
In a pure inductor current, I lags behind applied voltage V by an angle Φ = 90° or \(\frac{\pi}{2}\) radians.

Question 2.
Obtain an expression for the current in a capacitor when an AC emf is applied.
Answer:
Let an ac voltage V = Vm sin cot is applied across the plates of a capacitor. Then charge begin to accumulate on the capacitor plates then gradually its voltage increases.

By using ac voltage the capacitor will be charged and discharged alternately due to reversal of current in each half cycle.
Potential on capacitor V = \(\frac{q}{C}\) → (1)
But V = V_m sin ωt

In a capacitor current in circuit leads the applied voltage by an angle Φ = 90° or \(\frac{\pi}{2}\) radians.

Question 3.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Principle :
A transformer works on the principle of Mutual induction.

A transformer consists of two coils wound on a core made with soft iron.

The coil which is connected to mains supply is called primary coil. Let number of turns in primary are say N_p. The coil through which output is taken is called secondary coil. Let number of turns in secondary are say N_s.

Induced emf in secondary

Ratio of turns in secondary to primary \(\frac{N_s}{N_p}\) is called transformer turns ratio.

In a transformer output voltage V_s (\(\frac{N_s}{N_p}\))V_p → (4)
If V_s > V_p it is called step up transformer.
V_s < V_p it is called step down transformer.

The equation V_s = \(\frac{N_s}{N_p}\)V_p is applicable for Ideal transformers. But in real case a small amount of energy (less than 5%) is wasted due to 1) Flux leakage 2) Resistance of windings 3) Eddy currents and 4) Hysteresis,

Long Answer Questions

Question 1.
Obtain an expression for impedance and current in series LCR circuit. Deduce an expression for the resonating frequency of an LCR series resonating circuit.
Answer:
In series LCR circuit an inductance ‘L’, capacitance ‘C’ and a resistance ‘R’ are connected in series with an ac source.
Let voltage applied, V = V_m sin ωt → (1)
In series combination, total voltage in circuit is sum of voltage drops on each component

Here V_L and V_c are always in same phase but in opposite direction.

The term R² + (X_C – X_L)² is called impedance Z’. Its behaviour is like Resistance R in DC circuit.

The phasor diagram of V_R, V_c and VL are as shown.

Resonating frequency :
Resonance is a physical phenomenon at which a system tends to oscillate freely. This particular frequency is called Resonating frequency.
The reactance of Inductance X_L = ωL and
Reactance of capacitance X_c = \(\frac{1}{\omega C}\) opposite in nature.

When X_C = X_L then impedance Z = R has lowest value and current in LCR circuit is

(At resonating frequency potential across ‘L’ and potential across ‘C’ will cancell each other.

At resonance current in series LCR circuit is maximum i_m = \(\frac{V_m}{Z}=\frac{V_m}{R}\).)

Problems

Question 1.
A transformer converts 200 Vac into 2000 V ac. Calculate the number of turns in the secondaryif the primary has 10 turns.
Solution:
V₁ = 200 V, V₂ = 2000 V, n₁ = 10, n₂ = ?

∴ Number of turns on secondary n₂ = 100.

Question 2.
An ideal inductor (no internal resistance for the coil) of 20 mH is connected in series with an AC ammeter to an AC source whose emf is given by e = 20√2 sin(200t + π/3) V, where t is in seconds. Find the reading of the ammeter.
Solution:
Inductance L = 20 mH = 20 × 10^-3H
emf. e = 20√2 sin (200t + \(\frac{\pi}{3}\))V.
From standard equation e = e_m sin (ωt + Φ)
Maximum voltage em = 20√2 V
Average voltage = \(\frac{e_{\mathrm{m}}}{\sqrt{2}}=\frac{20 \sqrt{2}}{\sqrt{2}} \) = 20 V.
Impedance, Z_L = ωL. But ω = 200
∴ Z_L = 200 × 2 × 10^-3 = 4
Instantaneous current 1 = \(\frac{e}{Z}=\frac{20}{4}\) = 5A.

Question 3.
The instantaneous current and instantaneous ( voltage across a series circuit containing resistance and inductance are given by i = √2 sin (100t – π/4) A and V = 40 sin (100t) V. Calculate the resistance.
Solution:
Instantaneous current I = √2 sin(100t + \(\frac{\pi}{4}\))A
Instantaneous voltage V = 40 sin (100t) V

Question 4.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
Solution:
Voltage across Condenser V_C = 20 V
Voltage across Resistor V_R = 35 V
Voltage across Inductor V_L = 20V
In LCR circuit total potential difference =

Question 5.
An AC circuit contains a resistance R, an inductance L and a capacitance C connected in series across an alternator of constant voltage and variable frequency. At resonant frequency, it is found that the inductive reactance, the capacitive reactance and the resistance are equal and the current in the circuit is ifl. Find die current in the circuit at a frequency twice that of the resonant frequency.
Solution:
In LCR circuit at resonance Impedance Z = R
Current at resonance i₀ = \(\frac{V_0}{Z}=\frac{V_0}{R}\)
Reactance of inductor X_L = ωL and
Reactance of condenser X_C = – \(\frac{1}{\omega C}\)
When frequency is doubled (say ω₁ = 2ω)
X_L = 2 . ωL and X_C = \(\frac{1}{2\omega C}\)

Question 6.
A series resonant circuit contains L₁, R₁ and C₁. The resonant frequency is f. Another series resonant circuit contains L₂, R₂ and C₂. The resonant frequency is also f. If these two circuits are connected in series, calculate the resonant frequency.
Solution:
For first series LCR circuit
C = C₁, L = L₁ and R = R₁ and resonant
frequency f = \(\frac{1}{\sqrt{L_1 C_1}}\)

For second series LCR circuit
L = C₂, C = C₂ and R = R₂ resonant frequency f = \(\frac{1}{\sqrt{L_2 C_2}}\)

Given that f₁ = f₂ and two circuits are again in series with source.

In series combination same AC current will flow with same frequency through all the parts.

Circuits 1 and 2 are at resonance with the applied frequency f.

So when the two LCR circuits of same resonant frequency are connected in series still then there is no change in resonant frequency.

Question 7.
In a series, LCR circuit R = 200 Ω and the voltage and the frequency of the mains supply is 200 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 45°. On taking out the inductor from the circuit the current leads the voltage by 45°. Calculate the power dissipated in the LCR circuit.
Solution:
Given Resistance R = 200 Ω,
Voltage V = 200 V, Frequency u = 50 Hz.
When capacitor is removed current lags behind voltage by 45°
When Inductance is removed current leads voltage by 45°.
From above statements, the circuit is at resonance.
At resonance power dissipated P = \(\frac{V^2}{R}=\frac{200\times200}{200}\) = 200 W

Question 8.
The primary of a transformer with primary to secondary turns ratio of 1 : 2, is connected to an alternator of voltage 200 V. A current of 4 A is flowing though the primary coil: Assuming that the transformer has no losses, find the secondary voltage and current are respectively.
Solution:
Turns ratio of primary to secondary n_p : n_s = 1 : 2
Input voltage V₁ = 200 V, Input current I_i = 4A
Output voltage V₀ = V_i \(\frac{Ns}{np}\) = 200 × 2 = 400 V
In transformer V I = constant i.e., V_i.I_i = V₀I₀
∴ Output current I₀ = \(\frac{V_iI_i}{V_0}=\frac{200\times4}{400}\) = 2A

Exercises

Question 1.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
a) What is therms value of current in the circuit?
b) What is the net power consumed over a full cycle?
Answer:
Resistance R = 100 Ω ; Voltage, V = 220 V; Frequency, υ = 50 Hz.
a) The rms value of current in the circuit is given as:
I = \(\frac{V}{R}=\frac{220}{100}\) = 2.20A

b)The net power consumed over a full cycle is given as:
P = VI ⇒ p = 220 × 2.2 = 484 W

Question 2.
a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer:
a) Peak voltage of the ac supply, V₀ = 300 V
Rmsvoltage V = \(\frac{V_0}{\sqrt{2}}=\frac{300}{\sqrt{2}}\) = 212.1 V

b) The rms value of current I = 10 A Now, peak current
I₀ = √2I = 10√2 = 14.1 A

Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit
Answer:
Inductance of inductor, L = 44 mH = 44 × 10^-3 H ;
Supply voltage, V = 220 V
Frequency, υ = 50 Hz ;
Angular frequency, ω = 2πυ
Inductive reactance, X_L = ωL = 2πυL
= 2π × 50 × 44 × 10^-3Ω

Hence, the rms value of current in the circuit is 15.92 A.

Question 4.
A 60 µF capacitor is connected to a 110V, 60Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Capacitance of capacitor, C = 60 µF
= 60 × 10^-3 F;
Supply voltage, V = 110 V
Frequency, υ = 60 Hz;
Angular frequency, ω = 2πυ

Hence, the rms value of current is 2.49 A.

Question 5.
Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0 H, C = 32 µF, and R = 10Ω. What is the Q-value of this circuit?
Answer:
Inductance, L = 2.0 H;
Capacitance, C = 32 µF = 32 × 10^-6 F
Resistance, R = 10 W ; Resonant frequency

Hence, the Q-Value of this circuit is 25.

Question 6.
A transformer converts 200 Vac into 2000 V ac. Calculate the number of turns in the secondary coil if the primary coil has 10 turns.
Answer:
Primary voltage V_p = 200 V
Secondary voltage V_g = 2000 V
Number of turns in Primary N_p = 10
Number of turns in Secondary = N_s = ?

Question 7.
A charged 30 pF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
Capacitance, C = 30µF = 30 × 10^-6 F;
Inductance, L = 27 mH = 27 × 10^-3H

Hence, the angular frequency of free oscillations of the circuit is 1.11 x 103 rad/s.

Question 8.
Suppose the initial charge on the capacitor in Exercise 7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
Capacitance of the capacitor, C = 30 µF
= 30 × 10^-6F ;

Inductance of the inductor, L = 27 mH
= 27 × 10^-3 H ;

Charge on the capacitor, Q = 6 mC = 6 × 10^-3 C
Total energy stored in the capacitor E = \(\frac{1}{2}\frac{Q^2}{C}\)

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 9th Lesson Electromagnetic Induction

Very Short Answer Type Questions

Question 1.
What did the experiments of Faraday and Henry show?
Answer:

1. Faraday and Henry experiments showed that the relative motion between the magnet and a coil is responsible for generation of electric current in the coil.
2. The relative motion is not an absolute requirement to induce the current in a coil. If the current in a coil changes then also emf is induced in the nearby coil.

Question 2.
Define magnetic flux.
Answer:
Magnetic flux :
The number of magnetic field lines crossing unit area when placed perpendicular to the field is defined as “Magnetic flux”.
Magnetic flux, Φ = \(\overline{\mathrm{B}}.\overline{\mathrm{A}}\) = B A cos θ

Question 3.
State Faraday’s law of electromagnetic induction.
Answer:
Faraday’s law of Induction :
The rate of change of magnetic flux through a circular coil induces emf in it.

Question 4.
State Lenz’s law. [TS Mar. 19, June 15]
Answer:
Lenz’s Law :
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces current in that coil.

Question 5.
What happens to the mechanical energy (of motion) when a conductor is moved in a uniform magnetic field?
Answer:
When a condutor is moved in a magnetic field
Power of this motion, P = Fv F = Bil

The work done is in the form of mechanical energy.
This is dissipated into the form of joule heat.
∴ Joule heat = Power P_j = I²r = \(\frac{B^2l^2v^2}{r}\)

Question 6.
What are Eddy currents? [TS Mar. ’19; AP June ’15]
Answer:
Eddy currents :
When large pieces of conductors are subjected to changing magnetic flux then current is induced in them. These induced currents are called ”Eddy currents”.

Eddy currents will oppose the motion of the coil (or) they oppose the change in magnetic flux.

Question 7.
Define ‘inductance’.
Answer:
Inductance :
The process of producing emf in a coil due to changing current in that coil or in a coil nearby it is called “Inductance”.

Flux associated with a coil Φ_B is proportional to current i.e., Φ_B ∝ I

This constant of proportionality is called Inductance.

Question 8.
What do you understand by ‘self induetance’? [AP & TS June ’15]
Answer:
Self inductance :
If emf is induced in a single isolated coil due to change of flux in that coil by means of changing current through that coil then that phenomenon is called “Self Inductance L”.
In Self inductance, ε = -L\(\frac{dI}{dt}\)

Short Answer Questions

Question 1.
Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion. [TS May ’16]
Answer:
Let a conductor of length T is moving with a velocity “v” in a uniform and time independent magnetic field B.

Consider a rectangular metallic frame PQRS in which the side PQ is free to move without friction.

Let PQ move with a velocity ‘v’ in a perpendicular magnetic field B.

Magnetic flux in the loop Φ_B = Bl . x, where x = RQ a time changing quantity.

Here \(\frac{dx}{dt}\) = -v = Velocity of the rod.
Induced emf, ε = Blv is called Motional emf.

Question 2.
Describe the ways in which Eddy currents are used to advantage. [AP Mar. ’19, ’18, ’17, ’16, ’15, May ’18, ’17, ’16; TS Mar. ’18, ’15. May ’17]
Answer:
Advantages of Eddy currents:

1. Electromagnetic breaking : In some electrically powered trains strong electromagnets are placed above rails. When these electromagnets are activated eddy currents induced in rails will oppose motion of train. These breaks are smooth.
2. In galvanometers, a fixed core is made with non-magnetic material. When coil oscillates eddy currents induced in core will oppose the motion. As a result, the coil will come to rest quickly.
3. In induction furnaces high frequency oscillating currents are passed through a coil which surrounds the metal to be melted. These currents will produce eddy currents in the metal and it is heated sufficiently to melt it.
4. In electric power meters a metal disc is made to rotate due to eddy currents with some arrangement. Rotation of this disc is made to measure power consumed.

Question 3.
Obtain an expression for the mutual inductance of two long co-axial solenoids.
Answer:
Consider two long solenoids S₁ and S₂ each of length ‘l’, radius r₁ and r₂ and number of turns n₁ and n₂ respectively. When a current I₂ is sent through S₂ it will set up a magnetic flux Φ₁ through S₁.

Flux linkage with S₁ is N₁ Φ₁ = M₁₂I₂ where
M₁₂ is mutual inductance between the coils.
But Φ₁ = N₁A₁B where N₁ = n₁l ; A = πr²₁ and B = µ₀n₂I₂.
∴ N₁Φ₁ =(n₁l)(πr²₁ )(µ₀n₂I₂) = µ₀n₁n₂r²₁ …………. (1)

This approximation is highly valid when l > > r₂.

If current I₁ is passed through S₁ then
N₂Φ₂ = M₂₁I₁ where Φ₂ = N₂A₂B and
B = µ₀n₁I₁ and N₂ = n₂l.
∴ N₂Φ₂ =(n₂l)(πr²₁ )(µ₀n₁I₁) = µ₀n₁n₂πr²₁l …………. (2)

From eq. (1) & (2) Mutual inductance bet-ween co-axial solenoids M₁₂ = M₂₁. If the solenoid is on a core of permeability µ_r then
[M₁₂ = M₂₁ = µ₀µ_rn₁n₂πr²₁l]
Mutual inductance of a pair of coils or solenoids etc., depends on seperation between them and also on their orientaton.

Question 4.
Obtain an expression for the magnetic energy stored in a solenoid in terms of the magnetic field, area and length of the solenoid.
Answer:
When current is passed through a single isolated coil or solenoid changing magnetic flux can be developed by changing current through it. This changing flux will induce emf in that coil.
This phenomenon is called self induction (L).
Flux linkage NΦ_B ∝ I or NΦ_B = L . I
Induced emf, ε = \(\frac{d}{dt}\)(NΦ_B) = -L\(\frac{dI}{dt}\) ………. (1)

– ve sign indicates that induced emf will always oppose the flux changes in that coil (or) solenoid.

Let length of solenoid is ‘l’, area of cross section = A
then NΦ_B = (nl)(µ₀nI) (I) (∵ Φ_B =nµ₀I) ……….. (2)
and total number of turns N = n × l.
i.e., turns per unit length ‘n’ × length of solenoid ‘l’.
∴ L = \(\frac{\mathrm{N} \phi_{\mathrm{B}}}{\mathrm{I}}\) = µ₀n²Al …………. (3)

This self induced emf also called back emf will oppose any change in current in the coil. So to drive current in the circuit we must do some work.
Rate of work done = \(\frac{dW}{dt}\) = |ε|I = LI.\(\frac{dI}{dt}\) …………. (4)
∴ Energy required to send the currrent or Energy stored in inductor

Long Answer Questions

Question 1.
Outline the path-breaking experiments of Faraday and Henry and highlight the contributions of these experiments to our understanding of electromagnetism.
Answer:
Faraday and Henry conducted a series of experiments to understand electromagnetic inductioin.

First Experiment:
In this experiment, a galvanometer is connected to a coil. A magnet is moved towards the coil. They observed that current is flowing in the coil when magnet is in motion.

The direction of induced current is in opposite direction when the direction of motion of magnet is changed.

So they concluded that relative motion between magnet and coil is responsible for generation of electric current in the coil.

Second Experiment:
In this experiment a steady current is passed through one coil with the help of a battery and the second coil is connected to a galvanometer. When one of the coil is moved then current is induced in the coil. This current losts as far as there is relative motion between them.

Again they concluded that relative motion between the coils is responsible for induced electric current.

Third Experiment:
In this experiment, they connected one coil to a battery and a tapping key to make and break electric contact in that coil. The second coil is placed near the first coil. When electric contact is established current is induced in the second coil and momentary deflection is observed in galvanometer.

When electric contact is breaked again they got deflection in galvanometer in opposite direction.

So they concluded that it is not the relative motion between the coil and magnet or relative motion between the coils that induces the current. The changing magnetic flux is responsible for induced emf or current in the coil.

Finally Faraday proposed that induced emf ε = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\) ⇒ ε = N.\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
But from Lenz’s explanation he corrected this equation as ε = –\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\) ⇒ ε = -N\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)

Question 2.
Describe the working of a AC generator with the aid of a simple diagram and necessary expressions.
Answer:
AC generator consists of a coil of N turns placed in a magnetic field B produced by magnetic poles when the coil is rotated its effective area changes so flux linked with the coil changes. This changing flux will induce emf in the coil.

Electric generator converts mechanical energy into electrical energy.

Flux associated with the coil,
Φ_B = (\(\overline{\mathrm{B}}\) • \(\overline{\mathrm{A}}\)) = BA cos θ, where θ = ωt
Induced emf, ε = -N\(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\) = -NBA\(\frac{d}{dt}\)cos ωt
∴ Induced emf, ε = -NBAω sin ωt
The term NBAω is called maximum emf produced (ε_m).
∴ ε_m = NBAω
ε = ε_m sin ωt
Induced emf at any time E = ε_m sin ωt
When θ = 0, Induced emf is zero i.e., ε = 0.
The induced emf is maximum when θ = 90°
i.e., the plane of the coil is perpendicular
to magnetic field.
When θ = 90° ⇒ emf ε = ε_m
When θ = 180° ⇒ induced emf ε = 0.
When θ = 270° ⇒ induced emf ε = – ε_m.
again for θ = 360° ⇒ emf ε = 0.
∴ The induced emf varies sinusoidally in AC generator.

The coil is mounted on a rotor shaft. The axis of rotation of coil is perpendicular to magnetic field. The coil is connected to external circuit by means of slip rings and brushes.

Induced emf at any time is given by ε = ε_m sin ωt = ε_m sin 2πυt.

Depending on the method of supplying mechanical energy to rotate shaft these AC generators are classified as 1) Hydroelectric generators, 2) Thermal generators and 3) Nuclear generators.

Exercises

Question 1.
Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion. [AP May ’14]
Answer:
Let a conductor of length ‘l’ is moving with a velocity “v” in a uniform and time independent magnetic field B.

Consider a rectangular metallic frame PQRS in which the side PQ is free to move without friction.

Let the wire PQ is moved with a velocity ‘v’ in a perpendicular magnetic field B.

Magnetic flux in the loop Φ_B = Bl . x, where x = RQ a time changing quantity.

Here \(\frac{dx}{dt}\) = -v = Velocity of the rod.
Induced emf, ε = Blv is called Motional emf.

Question 2.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self inductance of the circuit. [AP Mar. 14; TS Mar. 16]
Answer:
Initial current, I₁ = 5.0 A ;
Final current, I₂ = 0.0 A ;
Change in current, dl = I₁ – I₂ = 5 A
Time taken for the change, t = 0.1 s;
Average emf, ε = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as

Hence, the self induction of the coil is 4H.

Question 3.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? [TS May ’18, Mar. ’17]
Answer:
Mutual inductance of a pair of coils, µ = 1.5 H;
Initial current, I₁ = 0 A
Final current I₂ = 20 A ;
Change in current, dl – I₂ – I₁ = 20 – 0 = 20 A
Time taken for the change, t = 0.5 s

Where dΦ is the change in the flux linkage with the coil.
Equating equations (1) and (2), we get
\(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = µ\(\frac{dI}{dt}\) ; dΦ = 1.5 × (20) = 30 Wb
Hence, the change in the flux linkage is 30 Wb.

Question 4.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Answer:
Speed of the jet plane, v = 1800 km/h = 500 m/s ;
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength,
B = 5.0 × 10^-4 T ; Angle of dip, δ = 30°

Vertical component of Earth’s magnetic field,
B_v = B sin δ = 5 × 10^-4 sin 30° = 2.5 × 10^-4T

Voltage difference between the ends of the wing can be calculated as
ε = (B_v) × l × v = 2.5 × 10^-4 × 25 × 500 = 3.125 V

Hence, the voltage difference developed between the ends of the wings is 3.125 V.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter

Very Short Answer Type Questions

Question 1.
A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field?
Answer:
For a magnetic dipole placed in a magnetic field, some net force is experienced It implies that force on the two poles of dipole is not equal.

This will happen only when magnetic dipole is in non-uniform magnetic field.

Question 2.
There is no question in text book. [TS Mar. 19, 17, May 14]
Question 3.
What happens to compass needles at the Earth’s poles? [TS Mar. 19, 17, May 14]
Answer:
When a compass is taken to earth poles say north pole then south pole of compass will adhere to north pole. It will align it self along magnetic meridian line.

Similarly when it is taken to south pole then north pole of compass is attracted by south pole and it will align itself along magnetic meridian line.

Question 4.
What do you understand by the ‘magnetisation’ of a sample?
Answer:
Magnetisation (T) :
It is the ratio of magnetic moment per unit volume.

I = (\(\frac{M}{V}\)) where M = the magnetic moments and V = volume of the given material.

Magnetic intensity is a vector, dimensions L^-1 A.
Unit : Ampere/metre : Am^-1

Question 5.
What is the magnetic moment associated with a solenoid?
Answer:
Magnetic moment associated with a solenoid (M) = nIA. Where
n = Number of turns in solenoid;
I = Current through it;
A = Area vector

Question 6.
What are the units of magnetic moment, magnetic induction and magnetic field? [AP Mar. 16. May 17, 16; TS Mar. 16]
Answer:
1. Magnetic moment m is a vector. Unit A-m², dimensions L^-2 A.
2. Magnetic induction (B) and magnetic field (B) are used with same meaning. Magnetic induction B is a vector.
Unit: Tesla (T), Dimension : MT^-2A^-1.

Question 7.
Magnetic lines form continuous closed loops. Why? [AP Mar. ’19, ’16, May ’18; TS May ’18, Mar. ’17]
Answer:
In magnetism magnetic monopole (single pole) is not existing. The simple possible way is to take a magnetic dipole. So the path a free magnetic needle or compass starts from north pole and terminates at south pole forms a loop.

Hence magnetic field lines are always closed loops.

Question 8.
Define magnetic declination. [TS Mar. 18, May 18, 17, 16; AP Mar. 18, 14, May 17, 16]
Answer:
Magnetic declination (D) :
The magnetic meridian at a place makes some angle (D) with true geographic north and south direction.

The angle between true geographic north to the north shown by magnetic compass is called “magnetic declination or simply declinations (D).”

Question 9.
Define magnetic inclination or angle of dip. [AP Mar. ’17, ’15; TS Mar. ’15]
Answer:
Magnetic inclination or angle of dip (I) :
It is the angle of total magnetic field BE at a given place with the surface of earth.
(OR)
The angle between horizontal to earth’s surface and net magnetic field of earth BE at that point.

Question 10.
Classify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [AP Mar. 19. 18. 17, 16, 15; TS Mar. 16. 15]
Answer:
Manganese : Paramagnetic substance
Cobalt : Ferromagnetic substance
Nickel : Ferromagnetic substance
Bismuth : Diamagnetic substance
Oxygen : Paramagnetic substance
Copper : Diamagnetic substance

Question 11.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What Is the magnetic field of the earth at this location?
Answer:
Given HE = 0.26 G; Dip angle = 60
But Dip angle = \(\frac{H_E}{B_E}\) = cos θ ⇒ B_E = H_E cos θ
∴ Magnetic field of earth = 0.26 × cos 60° = 2 × 0.26 = 0.52 G

Question 12.
Define Magnetisation of a sample. What is its SI unit?
Answer:
Magnetisation (I) : It is the ratio of net magnetic moment per unit volume.
I = \(\frac{m_{net}}{V}\) where m_net = the vectorial sum of magnetic moments of atoms in bulk material and V is volume of the given material.
Magnetic intensity is a vector, dimensions L^-1 A.
Unit: Ampere/metre : A m^-1.

Question 13.
Define Magnetic susceptibility. Mention its unit. [AP Mar. ’15]
Answer:
Magnetic susceptibility (χ) :
It is a measure for the response of magnetic materials to an external field.

It is a dimensionless quantity.

Short Answer Questions

Question 1.
What are Ferromagnetic materials? Give examples. What happens to a ferromagnetic material at Curie temperature?
Answer:
Ferromagnetism:

1. These substances are strongly attracted by magnets.
2. The susceptibility (χ) is +ve and very large.
3. Individual atoms of these substances will spontaneously align in a common direction over a small volume called domain.
4. Size of domain is nearly 1 mm³ or a domain may contain nearly 10¹¹ atoms.
5. In these substances, magnetic field lines are very crowded.
6. Every ferromagnetic substance will transform into paramagnetic substance at a temperature called Curie Temperature (T_c).
   Ex: Manganese, Iron, Cobalt, Nickel etc.

Effect of temperature on Ferromagnetic substances :
When ferromagnetic substances are heated upto Curie temperature, they will be converted into paramagnetic substances.

Question 2.
Derive an expression for the axial field of a solenoid of radius “r”, containing “n” turns per unit length and carrying current “I”.
Answer:
The behaviour of a magnetic dipole and a current carrying solenoid are similar.

Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.

Magnetic moment of solenoid (M) = nlA.

Consider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance ‘r’ from centre of the axis.

Magnetic field at point P

This is similar to magnetic field at any point on the axial line of magnetic dipole.

Question 3.
The force between two magnet poles separated by a distance ‘d’ in air is ‘F. At what distance between them does the force become doubled?
Answer:
Force between two magnetic poles F = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{m_1m_2}}{\mathrm{d^2}}\)


When separation between the poles is reduced by √2 times their force between them is doubled.

Question 4.
Compare the properties of para, dia and ferromagnetic substances. [TS & AP June ’15]
Answer:

Paramagnetic substances	Diamagnetic substances	Ferromagnetic substances
1. Feebly attracted by magnets.	1. Repelled by magnets.	1. Strongly attract by magnets.
2. Susceptibility is +ve and nearly equals to one. χ = i	2. Susceptibility is -ve and less than one. χ < 1	2. Susceptibility is + ve and large. χ > > 1
3. In a magnetic filed they move from weak field to strong field.	3. They move from strong field to weak field.	3. They move from weak field to strong field.
4. They have individual atomic magnetic moments but total magnetic moment is zero. Ex: Aluminium, sodium etc.	4. Individual atomic magnetic moment is zero. Ex: Bismuth, copper, lead.	4. They have individual atomic magnetic moments. These atoms will form domains. Magnetic moment of all atoms in adomain is in same direction. Ex: Iron, cobalt, nickel.

Question 5.
Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.
Answer:
Earth’s magnetism :
The magnetic field of earth is believed to arise due to electrical currents produced by convective motion of metallic fluids in outer core of earth. This effect is also known as the “dynamo effect”.

• The magnetic north pole of earth is at a latitude of 79.74° N and at a longitude of 71.8° W. It is some where in North Canada.
• The magnetic south pole of earth is at 79.74° S and 108.22° E in the Antarctica.

Magnetic declination (D) :
The angle between true geographic north to the mag-netic north shown by magnetic compass is called “magnetic declination or simply declination (D).”

Angle of dip or inclination (I):
The angle of dip is the angle of total magnetic field BE at a given place with the surface of earth.

At a given place horizontal component of earth’s magnetic field H_E = B_E cos I.

Vertical component of earth’s magnetic field Z_E = B_E sin I.

Tangent of dip tan I = \(\frac{Z_E}{H_E}\)

Question 6.
Define retentivity and coercivity. Draw the hysteresis curve for soft iron and steel. What do you infer from these curves?
Answer:
Hysteresis loop :
Magnetic hysteresis loop a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.

i) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point a’. It indicates that all atomic magnets of the sample are parallel to applied field.

ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.

Retentivity or Remanence :
The magnetic intensity (H) of a material at applied magnetic field B – 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.

iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point’d’.

Coercivity :
The -ve value of magnetic field (B^applied (i.e., in opposite direction of mag netisation) at which the magnetic intensity (H) inside the sample is zero is called “coer-civity”.

In hysteresis diagram the-value of B on -ve X-axis gives coercivity.

iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’.

Area of hysteresis loop is large for ferromagnetic substances with high permeability value.

Question 7.
If B is the magnetic field produced at the centre of adrcular coil of one turn of length L carrying current I then what is the manetic field at the centre of the same coil which is made into 10 turns?
Answer:
One turn coil means it is a circular loop.

For 1^st Case:
Magnetic field at the centre of a loop B₁ = \(\frac{\mu_0}{2} \frac{\mathrm{I}}{\mathrm{r_1}}\)
Given that length of the wire = L.

For 2^nd Case :
Given wire is made into a coil of 10 turns ⇒ n = 10
∴ 2πr₂10 = L ⇒ r₂ = \(\frac{L}{2 \pi}.\frac{1}{10}\)
For a coil of n turns magnetic field at its centre B₂ = \(\frac{\mu_0}{2} \frac{\mathrm{nI}}{\mathrm{r_2}}\)


When same length of wire is made into a coil of n turns then B₂ at centre = n² times previous value B₁.
⇒ B₂ = n² B₁ for given current T.

Question 8.
If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic held at the axis of the solenoid change?
Answer:
A solenoid will produce almost uniform magnetic field (B) along its axis.

Magnetic field along the axis of a solenoid B = µ₀nI.

Where n is number of turns per unit length.
In our case number of turns of a solenoid is doubled keeping others as constant i.e., length of solenoid L is not changed and permeability p0, and current T not changed. So new number of turns n₂ = 2n₁.
∴ New magnetic field at the same given point B₂ = µ₀n₂I.
But n₂ = 2n₁; ∴ B₂ = µ₀2n₁I = 2B₁

When number of turns of a solenoid is doubled then magnetic field at the given point on the axis of solenoid will also double.

Long Answer Questions

Question 1.
Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop.
Answer:
Consider a circular loop of radius R’ carrying a steady current i. Consider any point P’ on the axis of the coil (say X – axis).

From Biot – Savart’s law magnetic field at

The magnetic field at ‘P’ makes some angle ‘θ’ with X – axis. So resolve \(\mathrm{d} \bar{B}\) into components \(\mathrm{d} \bar{B}_x\) and \(\mathrm{d} \bar{B}\)⊥. Sum of \(\mathrm{d} \bar{B}\)⊥ is zero. Because \(\mathrm{d} \bar{B}\)⊥ component by an element d/ is cancelled by another diametrically opposite component. From fig \(\mathrm{d} \bar{B}_x\) = dB. cos θ. Where

Total magnetic field due to all elements on the circular loop

Question 2.
Prove that a bar magnet and a solenoid produce similar fields. (IMP)
Answer:
Magnetic field lines suggest that the behaviour of a current-carrying solenoid and a bar magnet are similar.

When a bar magnet is cut into two parts it will behave like two weak bar magnets. Similarly when a solenoid is cut into two parts and current is circulated through them they will also act as two solenoids of weak magnetic properties. Analogy between solenoid and bar magnet.

Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.

Magnetic moment of solenoid M = nlA.
Consider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance r from centre of the axis.

∴ Total magnetic field B is obtained by integrating dB with in the limits -1 to 1.

This value is similar to magnetic field at any point on the axial line of magnetic dipole. Thus, a bar magnet and a solenoid produce ! similar magnetic fields.

Question 3.
A small magnetic needle is set into oscillations in a magnetic field B. Obtain an expression for the time period of oscillation.
Answer:
Let a small compass of magnetic moment m and moment of inertia ‘I’ is placed in a uniform magnetic field ’B’.
Let the compass is set into oscillation in a horizontal plane.
Torque on the needle is τ = MB sin θ.
Where 0 angle between M and B.
At equilibrium the magnitude of deflecting torque and restoring torque are equal.
∴ Restoring torque τ = I\(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – MB sin θ.
– ve sign indicates that restoring torque is in opposite direction of deflecting torque.
θ is small; sin θ = θ.

Magnetic needle in a uniform field

From principles of angular.
Simple harmonic motion \(\frac{MB}{I}\) = ω²
⇒ ω = \(\sqrt{\frac{MB}{I}}\)

∴ Time period of oscillation of a magnetic needle placed in a magnetic field T = 2π\(\sqrt{\frac{I}{MB}}\)
and magnetic field at that point B = 4π²\(\frac{I}{MT^2}\).

Question 4.
A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T₁ and T₂ at two places, where the angles of dip are θ₁ and θ₂ respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.
Answer:
Let a bar magnet is held horizontally at a given place where earth’s magnetic field is B_B. When it is set into vibration in a horizontal plane it will oscillate with a time period T = 2π\(\sqrt{\frac{I}{MH_E}}\) ……….. (1)

Where H_E is horizontal component of earth’s magnetic field.

The relation between resultant magnetic field B_E, horizontal magnetic field H_E and angle of dip or inclination ‘I’ of earth’s magnetic field is H_E = B_E cos I. — (2)
Given at place ‘1’ angle of dip = θ₁
Given at place ‘2’ angle of dip = θ₂.

Question 5.
Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility. [AP Mar. ’15]
Answer:
Susceptibility χ :
The ratio of magnetisation of a sample (I) to the magnetic intensity (H) is called “susceptibility”.
Susceptibility χ = \(\frac{I}{H}\)
It is a dimensionless quantity.
It is a measure of how a magnetic material responds to external magnetic field.

For ferromagnetic materials susceptibility is +ve and χ >> 1 and χ is nearly in the order of 1000.
Ex: Iron, Cobalt, Nickel.
For paramagnetic substances χ is positive and nearly equals to one (χ ≅ 1).
Ex : Calcium, Aluminium, Platinum.

For diamagnetic substances χ is small and negative, magnetisation M and magnetic intensity H are in opposite direction.
Ex : Bismuth, Copper, Mercury, Gold.

Question 6.
Obtain Gauss’ Law for magnetism and explain it.
Answer:
Gauss Law in magnetism: Gauss law states that the net magnetic flux through any closed loop is zero.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{ds}}=0\)
Explanation :
Consider Gaussian surface I & II as shown in figure. In both cases the number of magnetic field lines entering the surface is equal to number of magnetic field lines leaving the surface.

The net magnetic flux is zero for both surfaces. This law is true for any surface of any shape. Consider an irregular Gaussian surface as shown in figure. Consider a smdl vector area element ∆S of closed surface ‘S’ placed in a Φ magnetic field B. Flux through ∆s is say ∆Φ.

When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.

Now flux (i.e., Number of magnetic field lines through unit area) through the element ∆S is given by ∆Φ_B = B . ∆S.

Let us divided the total Gaussian surface ‘S’ into number of small areas ∆S₁, ∆S₂, ∆S₃, ………… ∆S_n.

The total flux through the Gaussian surface

Where all’ term includes surface area of all surface elements ∆S₁, ∆S₂, ………… ∆S_n.

Magnetic monopole is not existing so there is no source or sinks of B’ in Gaussian surface. The simplest possible source is magnetic dipole i.e., bar magnet. Magnetic field lines of a bar magnet are closed curves of loops. So all the lines entering the Gaussian surface must leave from it.

Hence net magnetic flux through a closed surface is zero.

Question 7.
What do you understand by “hysteresis”? How does this property influence the choice of materials used in different appliances where electromagnets are used?
Answer:
Hysteresis loop :
Magnetic hysteresis loop is a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.
i) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.

ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.

Retentivity or Remanence :
The magnetic intensity (H) of a material at applied magnetic field B = 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.

iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point d’.

Coercivity :
The -ve value of magnetic field (B) applied (i.e., in opposite direction of magnetisation) at which the magnetic inten-sity (H) inside the sample is zero is called “coercivity”.

In hysteresis diagram the value of B on -ve X-axis gives coercivity.

iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’. Area of hysteresis loop is large for ferromagnetic substances with high permeability value.

Application of hysteresis in electromagnets. Hysteresis curve allows us to select suitable materials for permanent magnets and for electromagnets.

For materials to use as permanent magnets they must have high retentivity and high coercivity.

For electromagnets ferromagnetic substances of high permeability and low retentivity are used because when current is switched off it must loose magnetic properties quickly.

In case of transformers and telephone diaphragms they are subjected to prolonged AC cycles. For these applications the hysteresis curve of ferromagnetic materials must be narrow.

In this way hysteresis loop helps us to select magnetic materials for various applications.

Problems

Question 1.
What is the torque acting on a plane coil of “n” turns carrying a current “i” and having an area A, when placed in a constant magnetic field B?
Solution:
Number of turns = n ; Current = i;
Area of coil = A; Magnetic field = B.
Torque on a current carrying coil in a magnetic field τ = ni (\(\overline{\mathrm{A}}\times\overline{\mathrm{B}}\)) = niAB sin θ.

Question 2.
Acoilof 20 turns has an area of 800 mm² and carries a current of 0.5 A. If it is placed in a magnetic field of intensity 0.3 T with its plane parallel to the field, what is the torque that it experiences?
Solution:
Number of turns n = 200; Current i = 0.5 A;
Area A = 800 mm² = 800 × 10^-6 m² (∵ 1 mm² = 10^-6 m²) ;
Magnetic field B = 0.3 T.
Area of coil parallel to the field ⇒ Angle between area vector \(\overline{\mathrm{A}}\) and magnetic filed \(\overline{\mathrm{B}}\) =90°. Since area vector \(\overline{\mathrm{A}}\) is perpendicular to area of the coil.
∴ τ = niBA = 200 × 0.5 × 0.3 × 800 × 10^-6
= 3 × 800 × 10^-6 = 2.4 × 10^-3 N-m.

Question 3.
In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression for the magnetic moment (µ) of the electron in a Hydrogen atom in terms of its angular momentum L.
Solution:
Charge of electron = e ;
Angular momentum = L;
Mass of electron = m_e;
Magnetic moment of electron = µ = ?
When electron revolves in orbit current i = \(\frac{e}{T}\) ; Where time period T = 2π \(\frac{r}{υ}\)
∴ Current i = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\)
Magnetic moment of electron in orbit µ = iA

But m_e υr = L
∴ Magnetic moment of electron in orbit
µ = \(\frac{e}{2m_e}\)L

Question 4.
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid?
Solution:
Length of solenoid l = 22.5 cm = 22.5 × 10^-2m
Number of turns N = 900; Current I = 0.8 A.
a) Magnetising field near the centre H = ?
The behaviour of a solenoid is equal to that of a bar magnet.
Magnetic field due to a solenoid B = µ₀nI.
Magnetising field H = \(\frac{B}{\mu_0}\) = nI.
Where n = Nil.
∴ H = \(\frac{900}{22.5 \times 10^-2}\) × 0.8 = 3200 Am^-1.
A solenoid will give a uniform magnetic field along its axis.
Magnetising field far away from ends = 3200 Am^-1.

Question 5.
A bar magnet of length 0.1 m and with a magnetic moment of 5 Am² is placed in a uniform magnetic field of intensity 0.4 T, with its axis making an angle of 60° with the field. What is the torque on the magnet? [Mar. ’14]
Solution:
Length of bar magnet l = 0.1 m.;
Magnetic moment m = 5 Am².
Magnetic field B = 0.4 T ;
Angle with field θ = 60°.
Torque on the magnet τ = mB sin θ.
= 5 × 0.4 × sin 60°
= 5 × 0.4 × 0.8660 = 1.732 N-m.

Question 6.
If the Earth’s magnetic field at the equator is about 4 × 10^-5 T, what is its approximate magnetic dipole moment? (radius ofi Earth = 6.4 × 10⁶ m)
Solution:
Radius of earth r = 6.4 × 10⁶ m ; Magnetic field near equator B = 4 × 10^-5 T
Magnetic dipolemoment of earth = M.

Question 7.
The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10^-5 T and the angle of dip is 60°. What is the magnetic field of the earth at this location?
Solution:
Horizontal component of earth’s magnetic field HE = 2.6 × 10^-5 T

Angle of dip ‘I’ = 60°.
Earth’s magnetic field B_E = ?
Angle between B_E and H_E is called dip angle ‘I’.
H_E = B_E = cos θ ⇒ B_E = H_E/cos θ = 2.6 × 10^-5/ cos 60°
∴ B_E = 2.6 × 10^-5/ (0.5) = 5.2 × 10^-5T.

Question 8.
A solenoid, of insulated wire, is wound on a core with relative permeability 400. If the number of turns per metre is 1000 and the solenoid carries a current of 2A, calculate H, B and the magnetisation M.
Solution:
Relative permeability µ_r = 400 ;
Current I = 2A;
Number of turns / metre = n = 1000.
Magnetising force H = nl = 1000 × 2 = 2000 Am^-1.
Magnetic field along axis of solenoid B = ?
When solenoid is on a magnetic material B
= µ_r nl = 400 × 1000 × 2 = 8 × 10⁵ Am^-1

Intext Questions and Answer

Question 1.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^-2 J. What is the magnitude of magnetic moment of the magnet?
Solution:
Magnetic field strength, B = 0.25 T; Torque on the bar magnet, τ = 4.5 × 10^-2 J
Angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as : τ = MB sin θ

Hence, the magnetic moment of the magnet is 0.36 J T^-1.

Question 2.
A short bar magnet of magnetic moment m = 0.32 J T^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution:
Moment of the bar magnet, M = 0.32 J T^-1 ;
External magnetic field, B = 0.15 T
a) The bar magnet is aligned along the magnetic field. This system is considered as being instable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°.
Potential energy of the system
= -MB cos θ
= – 0.32 × 0.15 cos 0° = – 4.8 × 10^-2 J

b) The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium.
Potential energy = – MB cos θ ;
where θ = 180°
= -0.32 × 0.15 cos 180° = 4.8 × 10^-2 J.

Question 3.
A closely wound solenoid of800 turns and area of cross-section 2.5 × 10^-4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Solution:
Number of turns in the solenoid, n = 800 ;
Area of cross-section, A = 2.5 × 10^-4 m²
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length. The magnetic moment associated with the given current-carrying solenoid is calculated as :
M = nIA = 800 × 3 × 2.5 × 10^-4 = 0.6 J T^-1.

Question 4.
If the solenoid in exercise 8.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Solution:
Magnetic field strength, B = 0.25 T ;
Magnetic moment, M = 0.6 T^-1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as : τ = MB sin θ
τ = 0.6 × 0.25 sin 30° = 7.5 × 10^-2 J.

Question 5.
A bar magnet of magnetic moment 1.5 JT^-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment : (i) normal to die field direction, (ii) opposite to the field direction?
b) What is the torque on the magnet in cases (i) and (ii)?
Solution:
a) Magnetic moment, M = 1.5 J T^-1 ;
Magnetic field strength, B = 0.22 T

i) Initial angle between the axis and the magnetic field, θ₁ = 0°.
Final angle between the axis and the magnetic field, θ₂ = 90°.
The work required to make the magnetic moment normal to the direction of magnetic field is given as :
W = -MB (cos θ₂ – cos θ₁)
∴ B = -1.5 × 0.22 (cos 90° – cos 0°)
= -0.33 (0 – 1) = 0.33 J.

ii) Initial angle between the axis and the magnetic field, θ₁ = 0°.
Final angle between the axis and the magnetic field, θ₂ = 180°.
The work required to make the magnetic moment opposite to the direction of magnetic field is given as :
W = – MB (cos θ₂ – cos θ₁)
∴ W = – 1.5 × 0.22 (cos 180° – cos 0°)
= -0.33 (-1 – 1) = 0.66 J.

b) For case CD : θ = θ₂ = 90° then
Torque, τ = MB sin θ τ = 1.5 × 0.22 sin 90°
= 0.33 J

For case (ii): θ = θ₂ = 180° then
Torque, τ = MB sin θ τ = MB sin 180° = 0 J

Question 6.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
a) What is the magnetic moment associated with the solenoid?
b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 T is set up at an angle of 30° with the axis of the solenoid?
Solution:
Number of turns on the solenoid, n = 2000,
Area of cross-section of the solenoid, A = 1.6 × 10^-4 m²;
Current in the solenoid, I = 4 A

a) The magnetic moment along the axis of the solenoid is calculated as :
M = nAI = 2000 × 1.6 × 10^-4 × 4 = 1.28 Am²

b) Magnetic field, B = 7.5 × 10^-2 T
Angle between the magnetic field and the axis of the solenoid, 0 = 30°
Torque, τ = MB sin θ
∴ τ = 1.28 × 7.5 × 10^-2 sin 30°
= 4.8 × 10^-2 Nm.

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 × 10^-2 Nm.

Question 7.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 8.0 × 10^-2T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^-1. What is the moment of inertia of the coil about its axis of rotation?
Solution:
Number of turns in the circular coil, N = 16;
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A = πr² = π × (0.1)² m² ;
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 × 10^-2 T ;
Frequency of oscillations of the coil, v = 2.0 s^-1
∴ Magnetic moment, M = NIA = Nlπr²
= 16 × 0.75 × π × (0.1)²
∴ M = 0.377 J T^-1

Hence, the moment of inertia of the coil about its axis of rotation is 1.19 × 10^-4 kg m².

Question 8.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Solution:
Horizontal component of earth’s magnetic field, B_H = 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip = δ = 22°
Relation between B and B_H is B_H = B cos δ

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

Question 9.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Solution:
Angle of declination, θ = 12°;
Angle of dip, δ = 60°
Horizontal component of earth’s magnetic field, B_H = 0.16 G
Earth’s magnetic field at the given location = B
Relation between B and BH is B_H = B cos θ

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

Question 10.
A short bar magnet has a magnetic moment of 0.48 J T^-1. Give the direction and mag-nitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Solution:
Magnetic moment of the bar magnet,
M = 0.48 J T^-1.
a) Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is

The magnetic field is along the S – N direction.

b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is

The magnetic field is along the N – S direction.

Question 11.
A short bar magnet of magnetic moment 5.25 × 10^-2 J T^-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on
a) its normal bisector and
b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Solution:
Magnetic moment of the bar magnet,
M = 5.25 × 10^-2 J T^-1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10^-4 T
a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation :
B = \(\frac{\mu_0M}{4\pi R^3}\) ; When the resultant field is inclined at 45° with earth’s field, B = H

(OR) R = 0.05 m = 5 cm

b) The magnetic field at a distance R’ from the centre of the magnet on its axis is given as:
\(\frac{\mu_02M}{4\pi R^3}\)
B = The resultant field is inclined at 45° with earth’s field. ∴ B’ = H.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism

Very Short Answer Type Questions

Question 1.
What is the importance of Oersted’s experiment? [AP May ’18; TS Mar. ’17, May ’14]
Answer:
Oersted concluded that moving charges (or) currents produces a magnetic field in the surrounding space.

Question 2.
State Ampere’s law and Biot – Savart law.
Answer:
Ampere’s circuital law :-
The total magnetic flux coming out of a current carrying conductors enclosed in a perpendicular plane is p0 times greater than the alzebraic sum of currents (I_encl) enclosed by all the conductors in that plane.
\(\oint \mathrm{B} . \mathrm{d} l=\mu_0 \mathrm{I_{encl}}\)

Biot – Savart’s Law :
The magnetic field (dB) due to a current carrying element (dl) at any point r from the conductor is given by

Question 3.
Write the expression for the magnetic induction at any point on the axis of a circular current – carrying coil. Hence, obtain an expression for the magnetic induction at the centre of the circular coil.
Answer:
(i) Magnetic induction of any point x on the axis of a coil of radius (R) is

(ii) At centre of the coil (x = 0);

Question 4.
A circular coil of radius ‘r’ having N turns carries a current “i”. What is its magnetic moment?
Answer:
Magnetic induction at the centre of the coil
B = \(\frac{\mu_0}{2} \frac{\mathrm{i}}{\mathrm{R}}\)
For n turns the total magnetic field = (n times more)
∴ B = \(\frac{\mu_0}{2} \frac{\mathrm{ni}}{\mathrm{R}}\)

Question 5.
What is the force on a conductor of length L carrying a current “i” placed in a magnetic field of induction B? When does it become maximum?
Answer:
Let a conductor of length L’ and current i through it is placed in a magnetic field B then force on it F = i (\(\overline{\mathrm{L}}\times\overline{\mathrm{B}}\)) = i LB sin θ where θ = the angle between the length of conductor (L) and magnetic field direction \(\overline{\mathrm{B}}\). Force on current carrying conductor is maximum when tlie conductor makes an angle θ = 90° with magnetic field (F_man) = iLB.

Question 6.
What is the force on a charged particle of charge “q” moving with a velocity “v” in a uniform mangnetic field of induction B? When does it become maximum?
Answer:
Let a charge q is moving with a velocity v in a magnetic field B then force on the charged particle F = q(\(\overline{\mathrm{v}}\times\overline{\mathrm{B}}\)) =q\(\overline{\mathrm{v}}\)B sin θ. Where ‘θ’ is the angle between the direction of velocity (\(\overline{\mathrm{v}}\)) and direction of magnetic field (\(\overline{\mathrm{B}}\)).

Force on charged particle is maximum when it moves perpendicular to the magnetic field.
i. e., when θ = 90° ⇒ F_max = qvB

Question 7.
Distinguish between ammeter and voltmeter. [AP Mar. 18, 17, 15; May 17, 16; TS June 15]
Answer:

Ammeter	Voltmeter
1. It is used to measure current in a circuit.	1. It is used to measure potential difference between two points.
2. Its resistance must be low.	2. Its resistance must be high.
3. It is connected in series in the circuit.	3. It is connected in parallel between given points.

Question 8.
What is the principle of a moving coil galvanometer? [TS May ’16]
Answer:
Principle of moving coil galvanometer :
When a current carrying coil placed in a radial magnetic field is free to rotate then torque acting on it is τ = NIAB.

In M. C. G deflection torque (τ = NIBA) is produced by the current in the coil. Restoring torque is produced by torsional constant (K) of the spring. At equilibrium deflection θ = (\(\frac{NAB}{k}\))I

Question 9.
What is the smallest value of current that can be measured with a moving coil galvanometer?
Answer:
The smallest value of current that can be measured with moving coil galvanometer is in the order of 10^-6 to 10^-12 amperes.

Generally galvanometers will give full scale deflections for few micro amperes (µA)

Question 10.
How do you convert a moving coil galva-nometer into an ammeter? [AP Mar. 19, May 18, June 15; TS May 18, Mar. 18]
Answer:
A M.C.G is converted into an ammeter by connecting a low resistance (shunt) in parallel to it.
Shunt S = \(\frac{R_G}{n-1}\) where R_G = Resistance of galvanometer.
n = Ratio of currents \(\frac{i}{i_g}\)

Question 11.
How do you convert a moving coil galvanometer into a voltmeter? [AP May. 18. Mar. 16, 15; TS Mar. 16. 15]
Answer:
A M.C.G can be converted into a voltmeter by connecting a high resistance in series with it.
Series resistance R_s = (V/i_g – R_G)
V = Potential to be measured; i = maximum current through M.C.G
R_g = Resistance of M.C.G

Question 12.
What is the relation between the permittivity of free space ε₀, the permeability of free space µ₀, and the speed of light in vacuum?
Answer:
Relation between permittivity of free space ε₀ and permeability of free space µ₀is

Question 13.
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns about the vertical axis?
Answer:
No.

Question 14.
A current carrying circular loop is placed in a uniform external magnetic field. If the loop is free to turn, what is its orientation when it is achieves stable equilibrium?
Answer:
When a current carrying loop is placed in a uniform magnetic field the most stable state of equilibrium is magnetic moment of the loop (\(\overline{\mathrm{m}}\) = iA) and magnetic field \(\overline{\mathrm{B}}\) are parallel to each other, (i.e., θ = 0 between \(\overline{\mathrm{m}}\) and \(\overline{\mathrm{B}}\))

Question 15.
A wire loop of irregular shape carrying current is placed in an external magnetic field. If the wire is flexible, what shape will the loop change to? Why?
Answer:
If a flexible wire loop carrying current is placed in an external magnetic field it will attain circular shape.

Due to flow of charges (i.e., current) in the loop a force F = Bqv is acting at every point of flexible loop. So it attains circular shape.

Short Answer Questions

Question 1.
State and explain Biot – Savart’s law. [AP Mar. 18. 17, 16, May 14; TS Mar. 17, 16]
Answer:
According to Biot – Savart’s Law,

1. The magnitude of magnetic field dB is proportional to the current [dB α I]
2. Magnetic field dB is proportional to length of current carrying element [dB α dl]
3. Since of the angle between the current carrying element and the line joining the midpoint of the element and the given point and
4. Magnetic field dB is inversely proportional to the square of the distance ‘r’ of given point from the current carrying element dB α \(\frac{1}{r^2}\).

2) θ = the angle between the current carrying element d/ and the line joining the conductor and the given point.

Question 2.
State and explain Ampere’s law. [TS Mar. ’18]
Answer:
Ampere’s law :
The magnetic field lines emerging out of a long straight current carrying conductor are in the form of anti clock wise concentric circles with the wire at their centre.

The magnetic field emerging out through a small element dl is given by \(\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l}=\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l} \cos \theta\)

Here B and d/ are parallel to each other, so cos θ = 1.
Total magnetic field due to the entire loop \(\oint \mathrm{B} . \mathrm{d} l=\mu_0 \mathrm{i}\)

Ampere’s circuital Law :
Total magnetic field coming out of a current carrying conductor in a perpendicular plane is times greater than the current flowing through it.

1. If there are number of conductors ‘n’ in an enclosure then i is the alzebraic sum of currents in that enclosure and total magnetic field \(\oint \mathrm{B} . \mathrm{d} l=\mu_0 \mathrm{I_{encl}}\)
2. Ampere’s circuital law is valid to a closed loop of any shape.

Question 3.
Find the magnetic induction due to a long current carrying conductor. [AP June ’15]
Answer:
Consider a long straight conductor perpendicular to plane of paper, carrying a current I. Then the magnetic field lines emerging out of the conductor are in the plane of the paper. These magnetic field lines will form concentric circles with the conductor at the centre.

Take a small element d/ on the circle which is at a distance ‘r’ from the wire \(\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l}\)
= B dl cos θ where (\(\overline{\mathrm{B}}\)) and d\(\overline{\mathrm{l}}\) are in same direction. (θ = 0°)

Total magnetic field due to a straight current carrying conductor B = \(\oint \mathrm{B} . \mathrm{d} l=\mu_0 \mathrm{I}\)
From Ampere’s law total magnetic field = µ₀ I

∴ Magnetic field of any point due to a straight current carrying conductor B = \(\frac{\mu_0}{2 \pi} \frac{I}{r}\)

Question 4.
Derive an expression for the magnetic induction at the centre of a current carry-ing circular coil using Biot – Savart’s law. [TS May ’16]
Answer:
Consider a circular loop of radius ‘R’ carrying a current i. P is a point on the axis of the coil (say X – axis).
From Biot – Savart’s law magnetic field at any point can be written as

Element d/ is on Y axis and point P is in XY – plane.
∴ \(|\mathrm{d} \bar{l} \times \overline{\mathbf{r}}|\) = rdl

The magnetic field at ‘P’ makes some angle ‘θ’ with X – axis. So resolve d \(\overline{\mathrm{B}}\) into two perpendicular components say d\(\overline{\mathrm{B}}_{x}\) and d \(\overline{\mathrm{B}}\)⊥. Sum of d\(\overline{\mathrm{B}}\)⊥ is zero. Because d\(\overline{\mathrm{B}}\)⊥ component by an element d/ is cancelled by another diametrically opposite component.

Question 5.
Derive an expression for the magnetic induction (B) at a point on the axis of a current carrying circular coil using Biot – Savart’s law. [TS May ’16]
Answer:
Consider a circular loop of radius ‘R’ carrying a current i and P is a point on the axis of the coil (say X – axis).
From Biot – Savart’s law magnetic field at

The magnetic field at ‘P’ makes some angle ‘0’ with X – axis. So resolve d\(\overline{\mathrm{B}}\) into two perpendicular components say d\(\overline{\mathrm{B}}_{x}\) and d\(\overline{\mathrm{B}}\)⊥. Sum of d\(\overline{\mathrm{B}}\)⊥ is zero. Because d\(\overline{\mathrm{B}}\)⊥ component by an element dl is cancelled by another diametrically opposite component. From fig d\(\overline{\mathrm{B}}_{x}\) = dB. cos θ. Where

Question 6.
Obtain an expression for the magnetic dipole moment of a current loop.
Answer:
Let a rectangular loop of length ‘l’, breadth ‘b’ is placed in a uniform magnetic field B. Let a current I is passed through the loop. Now force on each side F₁ = F₂ = I b. B

Current passes through the loop in opposite direction, so F₁ and F₂ are opposite. So F₁ and F₂ forms a couple.

Let the coil is rotated by an angle θ then the perpendicular distance between F and B is a/2 cos θ.
∴ Total torque
τ = IbB.\(\frac{a}{2}\)sin θ + IbB.\(\frac{a}{2}\)sinθ = I(ab)B sinθ = IAB sinθ
Torque τ = moment of force couple = Force x ⊥^lr distance
Here we are defining a new term called a magnetic dipole moment of the loop \(\overline{\mathrm{m}}\) = IA.
∴ Torque τ = \(\overline{\mathrm{mB}}\)sinθ = \(\overline{\mathrm{m}}\times\overline{\mathrm{B}}\)
This is similar to torque τ = \(\overline{\mathrm{r}}\times\overline{\mathrm{F}}\)
∴ Magnetic dipole moment \(\overline{\mathrm{m}}\) = IA for a loop of one turn.
\(\overline{\mathrm{m}}\) = nIA for a coil of n turns.

Question 7.
Derive an expression for the magnetic dipole moment of a revolving electron. [AP Mar. ’16]
Answer:
Let an electron (e) be revolving around the nucleus in a circular path of radius r’. Current I = number of electrons flowing per second.

Dipolement ot electron
Let time period of revolution of electron = T
Then current I = \(\frac{e}{T}\) But T = \(\frac{2\pi r}{υ}\)
∴ I = eν/2πr
Magnetic moment is associated with a circulating current in a loop or closed path. Magnetic moment (M) = IA.
Here magnetic moment is represented by µ_r.

The direction of this magnetic moment is into the plane of the paper.

Question 8.
Explain how crossed E and B fields serve as a velocity selector.
Answer:
Let a charge ‘q’ is moving with a velocity in the presence of both electric field E and magnetic field B. Let these two fields are perpendicular to each other and also perpendicular to the velocity of the particle.
Force due to electric field F_E = qE
Force due to magnetic field F_B = qvB
Let the two fields are adjusted such that force applied by the two fields are equal and opposite. Then total force on the charged particle is zero.

Now qE = qvB (or) v = \(\frac{E}{B}\)

∴ Velocity of charged particle (v) is the ratio of strength of electric field E and magnetic field B.

So crossed electric field E and magnetic field B will serve as a velocity selector because only particles with a velocity v = \(\frac{E}{B}\) will pass undeflected through crossed electric and magnetic fields.

Question 9.
What are the basic components of a cyclotron? Mention its uses? [AP May ’16]
Answer:
Cyclotron is used to accelerate the charged particles.

The main parts of cyclotron are

1. ‘D’ shaped metal boxes called dees
2. Variable electric field
3. Variable magnetic field and
4. A Radio frequency oscillator.

1) Dees : Two ‘D’ shaped metal boxes are placed side by side with a small gap between them. An exit port is provided to one of the Dee’.

2) Electric field E is useful to accelerate the charged particles in the gap between the dees, electric field is shield by the metallic dees. Inside D’ there is no effect of electric field.

3) Magnetic field B is used to rotate the charged particle. In magnetic field B time

m = mass of charged particle ;
q = charge on particle
Time period T is independent of velocity of the particle.

4) Radio frequency oscillator is used to obtain resonance condition. When frequency of applied voltage (V_a) is equal to frequency of cyclotron υ_c then cyclotron is said to be in resonance condition. By adjusting phase difference between dees the charged particles can be accelerated upto required energy. Kinetic energy of charged particle depends on its velocity ‘v’.

Long Answer Questions

Question 1.
Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive an expression for the force per unit length between two parallel current carrying conductors.
Answer:
Consider a rod of uniform cross section A and length ‘l’. Let the density of the mobile charge carriers per unit volume is ‘n’.

For a steady current i the total number of mobile charge carriers in it nAl.

Let average drift velocity of each charge = v_d

When this conductor is placed in a magnetic field B. Force on it F = total charge

Force between two long parallel conductors :
If two long conductors ‘a’ and ‘b’ are carrying the currents I_a and I_b. Separated by the distance ‘d’. Conductor ‘A’ produces a magnetic field B_a at all points along conductor ‘B’ due to the current flowing through it.
Magnetic field due to ‘A’ = B_a = \(\frac{\mu_0}{2\pi } \frac{\mathrm{I_a}}{\mathrm{d}}\) …………… (1)

Conductor ‘B’ carrying a current Ib is in the magnetic field produced by ‘A’. Force acting on conductor ‘B’ due to ‘A’ is F_ba = I_b L B_a = \(\frac{\mu_0}{2\pi } \frac{\mathrm{I_aI_b}}{\mathrm{d}}\).L

Where L is length of conductor ‘B’.

Now conductor ‘A’ is in the magnetic field produced by ‘B’.
Force due to ‘B’ on ‘A’ = F_ab.
From Newton’s 3^rd law F_ab = – F_ba
∴ Force between two parallel conductors F_ab = F_ba = \(\frac{\mu_0}{2\pi } \frac{\mathrm{I_aI_b}}{\mathrm{d}}\).L
∴ Force between parallel conductors per unit lenght = \(\frac{F_{ab}}{L }=\frac{\mu_0}{2\pi } \frac{\mathrm{I_aI_b}}{\mathrm{d}}\) ……….. (2)

Question 2.
Obtain an expression for the torque on a current carrying loop placed in a uniform magnetic field. Describe the construction and working of a moving coil galvanometer.
Answer:
Let a rectangular loop of length b’, breadth a’ is placed in a uniform magnetic field B. Let a current I is passed through the loop. Now force on each side F₁ = F₂ = I b. B
Current passes through the loop in opposite direction so F₁ and F₂ are opposite. So F₁ and F₂ forms a force couple.

Where A = ab = Area of coil
Let the coil is rotated by an angle 0 then the perpendicular distance between F and B is a/2 sinG.
∴ Total torque τ = IbB\(\frac{a}{2}\) sinθ + IbB.\(\frac{a}{2}\) sinθ = I(ab)B sinθ = IAB sinθ
∴ Torque on a coil placed in a magnetic field τ = IAB sinθ

Construction and working of a moving coil galvanometer :
A moving coil galvanometer consists of a rectangular coil of n’ turns. It is placed in a uniform radial magnetic field. So \(\overline{\mathrm{B}}\) is perpendicular to area vector \(\overline{\mathrm{A}}\).

Hence torque is maximum.
Torque on coil τ = NiAB
This torque tends to rotate the coil. So it is called deflecting torque. A spring attached to the coil will provide the restoring torque. At equilibrium deflection (say Φ) is given by kΦ = NIAB (or) Φ = \(\frac{(NAB)}{k}\)I
Where k is torsional constant of spring. NAB
The term \(\frac{(NAB)}{k}\)is called constant of galvanometer.

Current sensitivity of Galvanometer is defined as deflection for unit current
\(\frac{\phi}{I}=\frac{(NAB)}{k}\) = constant of galvanometer.

In this galvanometer the coil is moving so it is called moving coil galvanometer. Its sensitivity is high for few microamperes (µA) of current it gives full deflection.

Question 3.
How can a galvanometer be converted to an ammeter? Why is the parallel resistance smaller that the galvanometer resistance? [Mar. ’14]
Answer:
A galvanometer can be converted into an ammeter by connecting a low resistance called shunt resistance in parallel to the galvanometer.

Every galvanometer has two important properties.

1. Resistance of galvanometer R_G,
2. Maximum current tolerable by it say I_G.

Ammeter :
The block diagram of ammeter is as shown. Let I is the current to be measured. When ammeter is connected in a circuit current I_G will flow through galvanometer and current I_s will flow through shunt.

Let ‘r’ is the total resistance of ammeter, G is resistance of galvanometer and R_s is shunt connected in parallel combination of

ammeter is equals to resistance of shunt.

Necessity of small shunt resistance :
Ammeter is used to measure current in a circuit. In a circuit current is constant for series combination only. So ammeter must be connected in series in a circuit. To measure the value of current exactly our ammeter must have zero resistance because in series combination (R = R₁ + R₂ + ………… etc) if resistance of shunt Rs is small effective resistance of ammeter is nearly equals to resistance of shunt and error in the value of current measured is also less.

Question 4.
How can a galvanometer be converted to a voltmeter? Why is the series resistance greater than the galvanometer resistance?
Answer:
A galvanometer can be converted into voltmeter by connecting a high resistance in series with galvanometer.

Conversion of galvanometer into voltmeter :
Every galvanometer has two important properties

1. Resistance of galvanometer R_G
2. Maximum current tolerable by it say I_G.

The block diagram of a voltmeter is as shown in figure.

Let voltage to be measured is ‘V’.
Total resistance of voltmeter R = \(\frac{V}{I_G}\)

But from block diagram R = R_G + R_s because they are in series. If R_s > > R_G then resistance of voltmeter R ≅ R_s

Necessity of high series resistance :
A voltmeter is used to measure potential difference between two given points. So it must always be connected parallelly in a circuit. In parallel combination, high current will flow through low resistance. To measure potential difference exactly the voltmeter should not draw any current from circuit. Theoritically resistance of ideal voltmeter is infinity. Practically it is kept as high as possible when series resistance of volt-meter is high it draws very little current from the circuit and measurement of voltage with it is more accurate.

Question 5.
Derive an expression for the force acting between two very long parallel current-carrying conductors and hence define the ampere.
Answer:
Let two long conductors say ‘a’ and ‘b’ are carrying the currents I_a and I_b. Separation between them is say ‘d’. Conductor ‘a’ produces a magnetic field B_a at all points a long conductor b’ due to the current flowing through it.

Magnetic field due to ‘a” = B_a = \(\frac{\mu_0}{2 \pi} \frac{I_a}{d}\) ……(1)
Conductor ‘b’ carrying a current I_b is in the magnetic field produced by ‘a’. Force acting on conductor ‘b‘ due to ‘a’ is F_ba = I_bLB_a = \(\frac{\mu_0}{2 \pi} \frac{I_aI_b}{a}\).L

Where L is length of conductor ‘b’.
Now conductor ‘a’ is in the magnetic field produced by ‘b’. Force due to ‘b’ on ‘a’ is F_ab From Newton’s 3^rd law F_ab = – F_ba
∴ Force between two parallel conductors

Force between parallel conductors per unit

If the currents in the two wires are parallel then force between them is attractive force because at the point of intersection of the two fields their directions are opposite so polarity is opposite.

Hence if currents are in same direction the conductors will attract. If currents are in opposite direction then the conductors will repel.

Definition of ampere :
From the force between two parallel conductors ampere is defined as follows.

Ampere is that value of steady current which when maintained through each conductor separated by a distance of 1 metre a part in vacuum produces a force of 2 × 10^-7 newton per metre between them.

Problems

Question 1.
A current of 10A passes through two very long wires held parallel to each other and separated by a distance of lm. What is the force per unit length between them? [TS Mar. 19. 15; AF Mar. 15]
Solution:
Current i₁ = i₂ = 10A
Separation d = lm
Force per unit length L = lm
Force between two parallel conductors

Question 2.
A moving coil galvanometer can measure a current of 10^-6A. What is the resistance of the shunt required if it is to measure 1A?
Solution:
Old range i.e., maximum current to be measured i₁ = 10^-6 A
New range i.e., maximum current to be measured i₂ = 1A
Ratio of ranges or currents to be measured,

Where G is resistance of galvanometer.

Question 3.
A circular wire loop of radius 30cm carries a current of 3.5 A. Find the magnetic field at a point on its axis 40 cm away from the centre.
Solution:
Radius of loop r = 30cm = 0.3m = 3 × 10^-1m.
Current i = 3.5A ; Distance of point on the axis r = 40cm = 0.4m = 4 × 10^-1m.
Magnetic induction field on the axis of a

Intext Question and Answer

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? [TS May ’18]
Answer:
Number of turns on the circular coil,
n = 100 ; Radius of each turn,
r = 8.0 cm = 0.08 m ; Current I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,

∴ Magnitude of the magnetic field
B = 3.14 × 10^-4T.

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? [TS June ’15]
Answer:
Current in the wire, 1 = 35 A ; Distance of a point from the wire, r = 20 cm = 0.2 m ; Magnitude of the magnetic field at this point is given as:

∴ Magnitude of the magnetic field at 20 cm from the wire B = 3.5 × 10^-5T.

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of Bat a point 2.5m east of the wire.
Answer:
Current in the wire, I = 50A ; ∴ Distance of the point from the wire, r = 2.5 m.
Given point is 2.5 m away from the east of the wire.


∴ Direction of magnetic field B is vertically upwards.
Note : The point is located normal to the . wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Question 4.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer:
Current I = 90 A; Distance of point r = 1 .5 m

Note : The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Answer:
Current in the wire, I = 8 A ; uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, θ = 30°;
Magnetic force per unit length f = BI sinθ
∴ f = 0.15 × 8 × 1 × sin30° = 0.6 N m^-1

Question 6.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Answer:
Length of the wire, l = 3 cm = 0.03 m;
Current flowing in the wire, I = 10 A
Magnetic field, B = 0.27 T ; Angle between the current and magnetic field, θ = 90°
From F = BIl sinθ; F = 0.27 × 10 × 0.03 sin90°
= 8.1 × 10^-2 N
∴ Magnetic force on the wire is 8.1 × 10^-2 N.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Answer:
Current flowing in wire A, I_A = 8.0 A ;
Current flowing in wire B, l_B = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m;
Length of wire A, l = 10 cm = 0.1 m

The magnitude of force is 2 × 10^-5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Answer:
Length of the solenbid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 × 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m ; Current in the solenoid, I = 8.0 A
Magnetic field inside the solenoid near its

∴ Magnetic field inside the solenoid near its centre is 2.512 × 10^-2 T.

Question 9.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Answer:
Length of a side of the square coil, 1 = 10 cm = 0.1 m; Current in the coil, l = 12 A

Number of turns n = 20 ; Angle between plane of the coil and magnetic field, θ = 30°
Strength of magnetic field, B = 0.80 T ; But τ = n BI A sin θ
Where, A = Area of the square coil = l × l = 0.1 × 0.1 = 0.01 m²
∴ τ = 20 × 0.8 × 12 × 0.01 × sin 30° = 0.96 N m

Question 10.
Two moving coil meters, Mj and M2 have the following particulars:
R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T
R₂ = 14 n, N₂ = 42, A₂ = 1.8 × 10^-3 m² , B₂ = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Answer:
a) Current sensitivity of M.C.G is given as:
I = NBA/k

Hence, the ratio of current sensitivity of M₂ to M₁ is 1.4.

b) Voltage sensitivity for M.C.G is given as:
Vs₂ = \(\frac{N_2B_2A_2}{k_2R_2}\)

Hence, the ratio of voltage sensitivity of M₂ to M₁ is 1.

Question 11.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.6 × 10^-19 C,m_e = 9.1 × 10^-31 kg)
Answer:
Magnetic field strength, B = 6.5 G = 6.5 × 10^-4 ;
Speed of the electron, v = 4.8 × 10⁶ m/s
Charge on the electron, e = 1.6 × 10^-19 C ;
Mass of the electron, mg = 9.1 × 10^-31 kg
Angle between the shot electron and magnetic field, θ = 90°;
Force on electron, F = evB sin θ
This force will make the electron in circular path.
∴ Centripetal force exerted on the electron, F_c = \(\frac{mυ^2}{r}\) at equilibrium F_c = F

Hence, the radius of the circular orbit of the electron is 4.2 cm.

Question 12.
In exercise 11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Answer:
Magnetic field strength, B = 6.5 × 10^-4 T ;
Charge of the electron, e = 1.6 × 10^-19 C
Mass of the electron, m_e = 9.1 × 10^-31 kg ;
Velocity of the electron, v = 4.8 × 10⁶ m/s
Radius of the orbit, r = 4.2 cm = 0.042 m ;
Frequency of revolution of the electron = υ
Angular frequency of the electron = co = 2πν
But v = rω

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Question 13.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is sus-pended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of die counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Answer:
(a) Number of turns, n = 30 ; Radius of the coil, r = 8.0 cm = 0.08 m
Area, A = πr² = π(0.08)² = 0.0201 m² ;
Current, I = 6.0 A
Magnetic field B = 1 T
Angle between the field lines and normal with the coil, θ = 60°
∴ Torque τ = n IBA sinθ …
∴ τ = 30 × 6 × 1 × 0.0201 × sin 60° = 3.133 N m

(b) Magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. So if the circular coil is replaced by a planar coil of some irregular shape of same area still then torque does not change.

Question 14.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Answer:
Inner radius of the toroid, r₁ = 25 cm = 0.25 m ;
Outer radius of the toroid, r² = 26 cm = 0.26 m
Number of turns , N = 3500 ; Current in the coil, I = 11 A
(a) Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.
(b) Magnetic field inside the core of a toroid

(c) Magnetic field in the empty space surrounded by the toroid is zero.

Question 15.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Answer:
Current in both wires, I = 300 A ;
Distance between the wires, r = 1.5 cm = 0.015 m
Length of the two wires, / = 70 cm = 0.7 m

Since the direction of the current in the wires is opposite, a repulsive force exists between them.

Question 16.
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Answer:
Resistance of the galvanometer coil, G = 12 Ω
Current for which there is full scale deflection, I_g = 3 mA = 3 × 10^-3 A
Range of the voltmeter is 0, which needs to be converted to 18 V. ∴ V = 18 V
Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:

Hence, a resistor of resistance 5988Ω is to be connected in series with the galvanometer.

Question 17.
A galvanometer coil has a resistance of 15Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Answer:
Resistance of the galvanometer coil, G = 15 Ω Current for which the galvanometer shows full scale deflection,
I_g = 4 mA = 4 × 10^-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.; ∴ Current, I = 6 A

A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter.

But shunt resistance is given by,

Hance, 10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 6th Lesson Current Electricity Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 6th Lesson Current Electricity

Very Short Answer Type Questions

Question 1.
Define mean free path of electron in a conductor.
Answer:
Mean free path :
It is defined as the average distance that an electron can travel between two successive collisions.

Question 2.
State Ohm’s law and write its mathematical form.
Answer:
Ohm’s law :
At constant temperature current (I) flowing through a conductor is proportional to the potential difference between the ends of that conductor.
V ∝ I ⇒ V = RI where R = constant called resistance. Unit: Ohm (Ω).

Question 3.
Define resistivity (or) specific resistance.
Answer:
Resistivity :
Resistivity of a substance ρ = \(\frac{RA}{l}\)

It is defined as the resistance of a unit cube between its opposite parallel surfaces.

It depends on the nature of substance but not on its dimensions.
Unit: Ohm – metre (Ωm).

Question 4.
Define temperature coefficient of resistance.
Answer:
Temperature coefficient of resistivity :
The resistivity of a substance changes with temperature. ρ_τ = ρ₀ [1 + ∝ (T – T₀)]. Where a is temperature coefficient of resistivity.

Question 5.
Under what conditions is the current through the mixed grouping of cells maximum?
(Note: Mixed grouping of cells not given in New Syllabus.)
Answer:
Let m cells are in series and there are n such rows this arrangement is called mixed grouping.
In mixed grouping current (i) = \(\frac{mnE}{mr + nR}\)
Current will be maximum when mr = nR
∴ i_max = \(\frac{nE}{2r}\)

Question 6.
If a wire is stretched to double its original length without loss of mass, how will the resistivity of the wire be influenced?
Answer:
Resistivity (ρ) is independent of dimensions of the material. It depends only on nature of material.
So even though the wire is stretched its resistivity does not change.

Question 7.
Why is manganin used for making standard resistors?
Answer:
Temperature coefficient of resistance of manganin is very less. So its resistance is almost constant over a wide range of temperature.

Due to this reason manganin is used to prepare standard resistances.

Question 8.
The sequence of bands marked on a carbon resistor are : Red, Red, Red, Silver. What is its resistance and tolerance?
Answer:
Given sequence of colour band Red Red Red Silver
Colour code for Red = 2
1st band red = 2 ; ‘
2nd band Red = 2
3rd band = No. of zeros = 2 (∵ Red) :
∴ Resistance R = 2200 Ω
tolerance band (4th band) Silver = 10%
∴ For silver resistor R = 2200 Ω with 10% tolerance.

Question 9.
Write the colour code of a carbon resistor of resistance 23 kilo ohms.
Answer:
Reistance 23 kilo ohms = 23000
∴ 1st band = 2 ⇒ Red ;
2nd band = 3 ⇒ orange
3rd band = No. of zeroes = 3 ⇒ orange
∴ Colour code for 23 kΩ = Red Orange, Orange.

Question 10.
If the voltage V applied across a conductor is increased to 2V, how will the drift velocity of the electrons change?
Answer:
Drift velocity will also double.
Drift velocity vd = \(\frac{-p}{m}\)Eτ
Where E = intensity of electric field. It depends on applied potential. When applied potential is doubled (V to 2V). Intensity of electric field E is also doubled.

(∵ E = \(\frac{V}{d}\) , where d is not charged). So drift d velocity is doubled.

Question 11.
Two wires of equal length, of copper and manganin, have the same resistance. Which wire is thicker?
Answer:
Manganin wire is thicker.
Resistance R = ρ\(\frac{l}{A}\) given l is same.
For copper specific resistance ρ is less than manganin.
∵ R is constant material with high ‘ρ’ must have larger area A.
Hence manganin wire is thicker.

Question 12.
What is the magnetic moment associated with a solenoid of ‘N’ turns having radius of cross-section ‘r’ carrying a current I?
Answer:
Magnetic moment of solenoid (N) = πr² In × 2l where ‘2l’ = length of solenoid, r = radius, I = current
n = number of turns.

Question 13.
Why are household appliances connected in parallel?
Answer:
In parallel combination potential drop is constant. Different amounts of current is allowed through each component separately. i.e., we may switch off unwanted connections without disturbing other.

Hence parallel connection of appliances is prefered in household connections.

Question 14.
The electron drift speed in metals is small (~ m^-1) and the charge of the electron is also very small (~ 10^-19) Q, but we can still obtain a large amount of current in a metal. Why?
Answer:
Eventhough charge of electron ‘e’, drift velocity v_d are less number of electrons n is very high, current i = ne v_d. (i.e., charges flowing per second)
∵ n is extremely high we are getting large current.

Short Answer Questions

Question 1.
A battery of emf 10 V and internal resistance 3Ω is connected to a resistor R.
(i) If the current in the circuit is 0.5 A. Calculate the value of R.
(ii) What is the terminal voltage of the battery when the circuit is closed? [TS Mar. ’15]
Answer:
emf E = 10V, Internal resistance r = 3Ω

ii) Terminal voltage V = E – ir
∴ V = 10 – 0.5 × 3 = 10 – 1.5 = 8.5V

Question 2.
Draw a circuit diagram showing how a, potentiometer may be used to find internal resistance of a cell and establish a formula for it.
Answer:
Circuit diagram to find internal resistance; of a cell is

Theory :
i) In circuit key ‘k₂’ is open and potentiometer jockey is adjusted to zero deflection.
Balancing length l₁ is measured.
Now ε = Φl₁ → (1)

ii) Key ‘k₂‘ is closed. Balancing length l₁ is measured.
Now V = Φl₁ → (2)
But ε = I (R + r) and V = IR so from eq (1) & (2)

Question 3.
Derive an expression for the effective resistance when three resistors are connected i) series ii) parallel.
Answer:
i) Series combination :
Let three resistors R₁, R2 and R₃ be connected in series as shown in figure.

Same current i flows through all resistors.
Potential drop across R₁ ⇒ (V₁) = iR₁
Similarly V₂ = iR₂ and V₃ = iR₃ are P.D across R₂ + R₃
Total potential across AB (V) = V₁ + V₂ + V₃
∴ V = iR₁ + iR₂ + iR₃ = i (R₁ + R₂ + R₃), But V = i R
∴ V = i R_eq = i(R₁ + R₂ + R₃)
In series combination R_eq = R₁ + R₂ + R₃
∴ Equivalent resistance is the sum of individual resistors.

ii) Parallel combination of resistors :
Let three resistors R₁, R₂ and R₃ be connected in parallel as shown in figure.

In this combination potential drop across AB is constant. But different values of current flows though each resistor. Total current i = i₁ + i₂+ i₃

Current through R₁ ⇒ (i₁) = \(\frac{V}{R_1}\) Similarly,
⇒ i₂ = \(\frac{V}{R_2}\) and i₃ = \(\frac{V}{R_3}\).

Question 4.
‘m’ cells each of emf E and internal resistance ‘r’ are connected in parallel. What is the total emf and internal resistance? Under what conditions is the current drawn from the mixed grouping of cells a maximum?
Answer:
i) Let m’ identical cells each of emf E’ and internal resistance V be connected in parallel as shown in the figure.
ii) Applying Kirchoffs voltage law to the circuit, we have for the first cell,

iii) Adding the above m’ equations, we get, – m(iR) – m(\(\frac{i}{m}\))r + mE = 0
(mR + r)i = mE
∴ I = \(\frac{mE}{mR+r}\)

iv) The above expression can be written as

Mixed grouping of cells :
Note : Mixed grouping of cells not given in New Syllabus.
Answer:
Let m cells are in series and there are n such rows this arrangement is called mixed grouping.
In mixed grouping current i = \(\frac{mnE}{mr+nR}\)
Current will be maximum when mr = nR
∴ i_max = \(\frac{nE}{2r}\)

Question 5.
Define electric resistance and write it’s SI unit. How does the resistance of a conductor vary if
a) Conductor is stretched to 4 times of it’s length.
b) Temperature of conductor is increased?
Answer:
Resistance :
The obstruction created by a conductor for the mobility of charges through it is known as resistance.
i) The resistance of a conductor (R) is proportional to length R ∝ l → (1)
ii) and inversely proportional to area of cross-section of the conductor.

where “ρ” = resistivity of the conductor.

a) When conductor is stretched by four times its new length l₂ = 4l₁ or \(\frac{l_1}{l_2}\) = 4
When a wire is stretched by keeping its mass
constant then Rg = R₁[latex]\frac{l_2}{l_1}[/latex]²
∴ R₂ = R₁ . (4)² = 16R₁
So when a wire is stretched by 4 time its resistance is increased by 16 times.

b) When temperature of conductor is increased its resistance will increase.
Resistance at t°C is R_t = R₀ (1 + αt)
where α’ = temperature coefficient of resistance of that conductor.

Question 6.
When the resistance connected in series with a cell is havled, the current is equal to or slightly less or slightly greater than double. Why?
Answer:
Let a cell of emf ‘E’ and internal resistance ‘r’ is connected with external resistance R as shown in figure.

Total resistance in circuit R_T = R + r → (1)
Current in circuit i₁ = \(\frac{E}{R+r}\) → (2)

a) Now external resistance R is reduced to \(\frac{R}{2}\).
Total resistance in circuit R_T1 = \(\frac{R}{2}\) + r
\(\frac{R+2r}{2}\) → (2)
From eq (1) and (3)
\(\frac{R+2r}{2}\) is not equals to \(\frac{R+r}{2}\).
i.e., when R is reduced to \(\frac{R}{2}\) total resistance R_T1 is not equals to \(\frac{R_T}{2}\).
So current in the circuit i₂ ≠ 2i₁. But i₂ is slightly less than 2i₁ it depends on the value of internal resistance ‘r’.

b) In case of ideal battery where r = 0 then R_T2 = \(\frac{R}{2}\) in this case R_T2 = \(\frac{R_{T_1}}{2}\)and
current i₂ = 2i₁.

So when reistance R in the circuit is reduced to half of its value current in circuit is doubled or slightly less than double.

Question 7.
Two cells of emfs 4.5V and 6.0V and internal resistance 6Ω and 3Ω respectively have their negative terminals joined by a wire of 18Ω and positive terminals by a wire of 12Ω resistance. A third resistance wire of 24Ω connects middle points of these wires. Using Kirchhoff s laws, find the potential difference at the ends of this third wire.
Answer:
The circuit diagram as per given data is as shown

From fig for loop ABCDFGJA
6i₁ + 24 (i₁ + i₂) + 9i₁ – 6i₁ = 4.5
33i₁ + 24i₂ = 4.5 → (1)

For loop IBCDFGHI
6i₂ + 24(i₁ + i₂) + 9i₂ – 3i₂ = 4.5
24i₁ + 36i₂ = 6.0 → (2)

Question 8.
Three resistors each of resistance 10 ohm are connected, in turn, to obtain (i) minimum resistance (ii) maximum resistance. Compute (a) The effective resistance in each case (b) The ratio of minimum to maximum resistance so obtained.
Answer:
Resistance of each resistor R = 10Ω ;
Number of Resistors n = 3
a) In parallel combination resistance is minimum.

b) In series combination resultant resistance is maximum.

Question 9.
State Kirchhoff s law for an electrical network. Using these laws deduce the condition for balance in a Wheatstone bridge. [AP Mar. ’19. ’18. ’14. May ’16. ’14; TS May ’18. Mar. ’18, ’16, June ’15]
Answer:
Kirchhoffs Laws:
i) Junction rule :
At any junction, the sum of currents towards the junction is equal ‘ to sum of currents away from the junction.
(OR)
Alzebraic sum of currents around a junction is zero.

ii) Loop rule :
Alzebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.

Applying Kirchoff’s Law to Wheatstones bridge:
Wheatstones bridge consists of four resistances P, Q, R, S connected as shown in the figure and are referred as arms of the bridge.

A battery of emf ‘E’ is connected between two junctions A and B and a galvanometer of resistance ‘G’ is connected between ‘C’ and ‘D’.

Applying Kirchoff’s first law,
at the junction C; i₁ = i_g + i₃ — (1)
at the junction D; i₂ + i_g = i₄ — (2)
Applying Kirchoff’s second law for the loop ACDA
– i₁ P – i_g G + i₂ R = 0 — (3)
For the loop CBDC,
– i₃ Q + i₄ S + i_g G = 0 — (4)

If no current passes through the galvanometer then the bridge is said to be balanced.
So, i_g = 0, then (1), (2), (3) and (4) becomes
i₁ = i₃, i₂ = i₄, i₁P = i₂R and i₃ Q = i₄ S.
By adjusting the above equations we get \(\frac{P}{Q}=\frac{R}{S}\).
This is the balancing conditions of a Wheatstones bridge.

Question 10.
State the working principle of potentiometer explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer. [AP Mar. 17,16, May ‘ 17; June ’15; TS Mar. 19, May 16]
Answer:
Potentiometer working principle :
Potentiometer consists of 10 meters length of wire with uniform internal resistance. Let total length of wire is L and its total resistance is R_p.

where Φ = Potential drop per unit length i.e., potential gradient.

Comparison of e.m.f. of two cells :
The circuit diagram used to compare emf of two cells is as shown in figure.
Circuit connection were given as per diagram.

The potentiometer contains two circuits, primary and secondary.

The primary circuit consists of a cell of emf ‘E’ that a plug key (K) and a rheostat (Rh). These are connected in series with potentiometer wire AB.

The secondary circuit consists of two cells of emf e₁ and e₂, two plug keys K₁ and K₂, a galvanometer (G) and a Jockey (J) as shown in the figure.

The positive terminals of the two cells in primary and secondary are connected to terminal A’.

Procedure:
i) The plug keys K and K₁ are closed and by adjusting Jockey on the wire, the balancing length ‘l₁‘ is determined.
e₁ ∝ l₁ ——- (1)

ii) Now plug key K₁ is opened and K₂ is closed. Again the Jockey is adjusted and balancing length ‘l₂‘ is determined.
e₂ ∝ l₂ ——- (2)
from (1) and (2 ) \(\frac{e_1}{e_2}=\frac{l_1}{l_2}\) Ratio of emf of two cells.

Question 11.
State the working principle of potentiometer explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell. I [AP May 18, Mar. 15; TS Mar. 17. 15, May 17]
Answer:
Potentiometer working principle :
Potentiometer consists of 10 meters length of wire with uniform internal resistance. Let total length of wire is L and its total resistance is R_p.

Current through potentiometer wire i

where Φ = Potential drop per unit length i.e., potential gradient.

Determination of internal resistance (r) :
Circuit diagram used to find internal resistance of battery is as shown in figure.

Battery E is connected in primary circuit through plus key k₁.

Another battery E₁ is connected in secondary circuit. A low resistance R’ is connected in parallel to Battery E₁ along with key k₂.

Plug k₂ is open. Jockey J is moved on potentiometer until zero deflection is obtained. Balancing length l₁ is measured.

Plug key k₂ is closed so resistance R will come into operatioin. Again Jockey J is moved on potentiometer wire to find balancing point. When zero deflection is obtained balancing length l₂ is measured.

Question 12.
Show the variation of current versus voltage graph for GaAs and mark the (i) Nonlinear region (ii) Negative resistance region.
Answer:
The voltage-current characteristic graph of galium arsinide (GaAs) is as shown in figure.

GaAs is a semiconducting substance. Its V-I characteristics are non-ohmic. So V-I graph is not linear.
The region OA is linear. In which Ohm’s Law is obeyed.
The region AB is a curve (i). Here voltage V is not linear with currrent (i). So It is a non-linear region.

(ii) In the region BC current I decreases even though voltage ‘V’ increases.
Resistance R = \(\frac{V}{I}\)
∵ I decreases resistance in the region BC is negative.
The regions (1) and (2) are shown in figure.

Question 13.
A student has two wires of iron and copper of equal length and diameter. He first joins two wires in series and passes an electric current through the combination which increases gradually. After that, he joins two wires in parallel and repeats the process of passing current. Which wire will glow first in each case?
Answer:
Iron and copper wires of equal length and diameter are taken. For Iron, resistivity is ρ_i high. For copper resistivity, ρ_c is less.
∴ Resistance of Iron wire R_i is high and Resistance of copper wire R_c is less.

a) When connected in series same current flows through Iron and Copper wires. Heat produced Q = I²R. So heat produced in copper wire is less and that of Iron wire is high.

In series combination when current increases gradually than Iron wire will glow first.

b) When the two wires are connected in parallel and current is increased gradually.

In parallel combination high current flows through low resistance and low current through high resistance. They follow the relation I_c/I_i = R_i/R_c.

So current through copper wire is high. Heat produced Q = i²R ⇒ Q_c > Q_t
Heat produced in copper wire is more than that in iron wire.
So in parallel combination copper wire will glow first.

Question 14.
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance.
Answer:
In parallel combination when ‘n’ identical wires are connected parallelly equivalent resistance is given by R_eq = R/n.

Given three identical wires ⇒ n = 3.
∴ R_eq = R/3 where R is resistance of each wire.
∴ Resistance of each wire used in parallel combination is R.

Hence resistance of each wire used in parallel combination is R.

Long Answer Questions

Question 1.
Under what condition is the heat produced in an electric circuit (a) directly pro-portional (b) inversely proportional to the resistance of the circuit? Compute the ratio of the total quantity of heat produced in the two cases.
Answer:
Let a current I is flowing through a conductor between its ends say A and B. Let potential at A is V(A) and potential at B is V(B).

Potential difference across AB is say V = V(A) – V(B)

Let charge flowing in time ∆t is ∆Q = I∆T
Change in potential energy ∆V = Final P.E – Initial P.E
= ∆Q[V(B) – V(A)] = -∆QV = -TV∆t < 0 → (1)
Let charges are moving without collision with atoms.
Then kinetic energy of charges will also increase.
∆K = -∆U_Pot or ∆K = IV ∆t > 0 → (2)
Work done in this process ∆W = IV ∆t → (3)
But power

From Ohm’s Law V = IR
∴ Power P = I.I.R = I²R
In this case power p ∝ R.

In this case energy of charge carriers is useful to heat the conductor. Amount of energy dissipated in conductor per second is power.

Where priority is given to current through conductor, at ordinary voltages we will say that p ∝ R.

i) In case of transmission lines the primary purpose is to transmit electrical energy from one place to another place. While doing so a part of energy is wasted in conductor in the form of heat. Let total power to be transmitted = P. Resistance of conductor = R_c,
Line voltage = V

Power wasted in transmission or transmission losses P_c = \(\frac{P^2R_c}{V^2}\). Due to this reason to reduce transmission losses we are transmitting electric power at very high voltages.

When voltage is given priority power P = \(\frac{V^2}{R}\)

In this case power P is said to be inversely proportional to resistance ‘R’.
Ratio of heats produced in the two cases is I²R: V²/R
But V = IR
∴ Ratio is I²R = \(\frac{I^2R^2}{R}\) = 1 : 1

Question 2.
Two metallic wires A and B are connected in parallel. Wire A has length L and radius r wire B has a length 2Land radius 2r. Compute the ratio of the total resistance of the parallel combination and resistance of wire A.
Answer:
Let resistivity of metallic wire A is ρ_A and that of B is ρ_B.

For wire A :
Length = L, radius = r, resistivity = ρ_R
Resistance of wire A is R_A = \(\frac{\rho_{\mathrm{A}} \mathrm{L}}{\mathrm{A}}=\frac{\rho_{\mathrm{A}} \cdot \mathrm{L}}{\pi r^2}\) → (1)

For wire B :
Length = 2L; radius = 2r ; resistivity = ρ_B.
Resistance of wire B is

a) Ratio of resistance of the wires = R_A : R_B

b) When resistors RA and Rg are connected in parallel effective resistance

Question 3.
In a house three bulbs of 100 W each are lighted for 4 hours daily and six tube lights of 20W each are lighted for 5 hours daily and a refrigerator of 400 W is worked for 10 hours daily for a month of 30 days. Calculate the electricity bill if the cost of one unit is Rs. 4.00.
Answer:
Electric consumption in KWH per one month

Electricity bill = No. of units × cost of unit = 174 × 4 = Rs. 696/-.
∴ Current bill, for that month is Rs. 696/-

Question 4.
Three resistors of 4 ohms, 6 ohms and 12 ohms are connected in parallel. The combination of above resistors is connected in series to a resistance of 2 ohms and then to a battery of 6 volts. Draw a circuit diagram and calculate
a) Cureent in main circuit
b) Current flowing through each of the resistors in parallel
c) P.D and the power used by the 2 ohm resistor.
Answer:
Given R₁ = 4Ω, R₂ = 6Ω and R₃ = 12Ω
i) When R₁, R₂ and R₃ are connected in parallel effective resistance

ii) R_eq is connected to a 2Ω resistor in series and then to a battery of 6V. The circuit diagram is as shown.

Total Resistance in circuit = R_eq + 2 = 2 + 2 = 4Ω
a) Current in main circuit I = \(\frac{V}{R}=\frac{6}{4}\) = 1.5 amp
b) Current through

c) RD across 2Ω Resistor
i = 1.5A, R = 2Ω ∴ P.D = iR= 1.5 × 2 = 3V
Power used P = i²R = 1.5 × 1.5 × 2 = 4.5 watt.

Question 5.
Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V are connected in parallel to a 220 volt supply. What current is drawn from the supply line?
Answer:
For 1st Lamp power P = 100 W at potential V = 220 V
Supply voltage V = 220 V ; Power P = VI

∴ I₁ = \(\frac{100}{200}\)
For 2nd lamp power P = 60 W at 220 V.
Supply voltage = 220 V
Current i₂ = \(\frac{P^2}{V}=\frac{60}{200}\)
Total current drawn by parallel combination
I = I₁ + I₂⇒ I = 0.4545 + 2727 = 0.7272 A

Question 6.
A light bulb is rated at 100W for a 220V supply. Find the resistance of the bulb. [IMP]
Answer:
Power P = 100 W; Potential V = 100 V

Question 7.
Estimate the average drift speed of conduction electrons in a copper wire of crosssectional area 3.0 × 10^-7 m² carrying a current of 5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³ and its atomic mass is 63.5 u.
Answer:
Area A = 3.0 × 10^-7m² ;
Current i = 5A
Density ρ = 9.0 × 10³ kgm^-3 = 9.0 × 10⁶gm^-3
Atomic mass m = 63.5 U
Avagadro number NA = 6.022 × 10²³;
Charge on electron e = 1.6 × 10^-19 C
Drift velocity v_d = i/neA. Where n = total number of electrons / unit volume

Question 8.
Compare the drift speed obtained above with
i) Thermal speed of copper atoms at ordinary temperatures.
ii) Speed of propagation of electric field along the conductor which causes the drift motion.
Answer:
Drift velocity of electrons in copper = 1.22 m.m/s
i) Thermal speed of copper atoms V_T = \(\sqrt{k_{B}T/m}\).
Where k_B is Boltzman’s constant.
k_B = 1.381 × 10^-25
Let T = 300 K average temperature or ordinary temperature
m = mass of copper atom, By subtituting the above values V_T = 200 m/ sec
V_d < < V_rms

Thermal speed of copper atom is 10⁵ times more than drift velocity of electrons.

ii) An electric field travels with velocity of light along the conductor is velocity of electromagnetic wave 3 × 10⁸m/s.

Speed of electric field travelling along a conductor is 10¹¹ times more than drift velocity of electrons.

Problems

Question 1.
A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there is no change in its density on stretching, calculate the resistance of the stretched wire.
Solution:
Resistance R = 10Ω ; Final length l₂ = 3l₁
When a wire is stretched its total volume is constant.

Question 2.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter? [AP Mar. 19,14. May 16; TS Mar. 16]
Solution:
Resistance = 4R
The wire is bent in the form of a circle and Resistance is to be calculated along its diameter.

Between the point AB the wire is a parallel
combination of 2R and 2R.
∴ Resultant resistance \(\frac{1}{R_R}=\frac{1}{2R}+\frac{1}{2R}=\frac{1}{R}\)
∴ Effective resistance between A & B = R.

Question 3.
Find the resistivity of a conductor which carries a current of density of 2.5 × 10⁶ A m^-2 when an electric field of 15 Vm^-1 is applied across it.
Solution:
Current density j = 2.5 × 10⁶ A
Electric field E = 15 V m^-1
Resistivity ρ = ? Resistivity p = E/j

Question 4.
What is the color code for a resistor of resistance 3500Ω with 5% tolerance?
Solution:
Resistance R = 3500 Ω. Tolerance = 5%
R = 3500 First digit 4 ⇒ Orange
2nd digit 5 ⇒ Green
Last two digits represent number of zeros = 2 ⇒ Silver
Tolerance = 5% ⇒ gold colour So colour code of that resistor is orange, green, silver and gold bands.

Question 5.
You are given 8Ω resistor. What length of wire of resistivity 120 ftm should be joined in parallel with it to get a value of 6Ω?
Solution:
Resistance Rj = 8Ω, Resistivity ρ = 12Ω m
Resistance across parallel combination R_p = 6Ω
Let parallel resistance to be connected
x = 120 / metres

Question 6.
Three resistors 3Ω, 6Ω and 9Ω are connected to a battery. In which of them will the power dissipation be maximum if: (a) they all are connected in parallel (b) they all are connected in series? Give reasons.
Solution:
Values of resistors R₁ = 3 Ω ; R₂ = 6 Ω ; R₃ = 9 Ω
a) When in series R_eff = R₁</sub. + R₂ + R₃ = 3 + 6 + 9 = 18
Power consumed P = \(\frac{V^2}{R}=\frac{V^2}{18}\) → (1)

b) When connected in parallel total power consumed is the sum of powers in each resistor.

From eq 1 & 2 power dissipation is maximum when they are connected in parallel.

Question 7.
A silver wire has a resistance of 2.1Ω at 27.5°C and a resistance of 2.7Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:
Temperature t₁ = 27.5°C ;
Resistance R₁ = 2.1 Ω
Temperature t₂ = 100°C ;
Resistance R₂ = 2.7 Ω
Temperature coefficient of resistivity

Question 8.
If the length of a wire conductor is doubled by stretching it while keeping the potential difference constant, by what factor will the drift speed of the electrons change?
Solution:
Length of wire is doubled ⇒ l₂ = 2l₁
Potential V = constant ; Same material is used ⇒ electron density per unit volume is same.
Drift velocity v_d = \(\frac{i}{neA}\) here i, n and e are constants.

∴ Drift velocity is doubled.

Question 9.
Two 120V light bulbs, one of 25W and another of 200W are connected in series. One bulb burnt out almost instantaneously. Which one was burnt and why?
Solution:
For 1st bulb, Power P₁ = 25 W; For 2nd bulb P₂ = 200 W
When they are connected to mains supply of 240 V in series

In series combination voltage drop is high on high resistance and less on low resistance.

Since voltage on 25W bulb is more than rated voltage it blows off instantly.

Question 10.
A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance.
Solution:
% Increase in length = 5%

∴ Percentage change in resistance = 2 × 5 = 10%

Question 11.
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance.
Solution:
In parallel combination when ‘n’ identical wires are connected parallelly, equivalent resistance is given by R_eq = R/n.

Given three identical wires ⇒ n = 3.
∴ R^eq = R/3 where R is resistance of each wire.
∴ Resistance of each wire used in parallel combination is R.

Hence resistance of each wire used in parallel combination is R.

Question 12.
Two wires A and B of same length and same material, have their cross sectional areas in the ratio 1 : 4. What would be the ratio of heat produced in these wires when the voltage across each is constant?
Solution:
Given lengths of wire are same ⇒ l₁ = l₂
Same material is used ⇒ ρ₁ = ρ₂
Ratio of area of cross sections A₁ : A₂ = 1 : 4 ⇒ A₂ = 4A₁
Potentail V is same.

∴ Ratio of Heat produced = 1 : 4

Question 13.
Two bulbs whose resistances are in the ratio of 1 : 2 are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation in these?
Solution:
Ratio of resistances ⇒ R₁ : R₂ = 1 : 2
⇒ R₂ = 2R₁
When in parallel P.D is same on each bulb.

Question 14.
A potentiometer wire is 5 m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180 cm of the potentiometer wire. [AP Mar. 17, 16. June 15; TS May 16]
Solution:
Total length of potentiometer wire L = 5 m
= 500 cm
Balancing length l₁ = 180 cm
P.D across the terminals = 6V
In potentiometer e.m.f at balancing point
V = \(\frac{El}{L}\frac{6\times180}{500}\)
∴ E.m.f of cell in secondary E₁ = 2.16 V

Question 15.
In a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution:
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁ = 35 cm
The cell is replaced by another cell of emf E₂.
New balance point of the potentiometer, l₂ = 63 cm.
The balance condition is given by the

Therefore, emf of the second cell is 2.25V.

Question 16.
If the balancing point in a meter-bridge from the left is 60 cm, compare the resistances in left and right gaps of meter-bridge.
Solution:
Balancing point distance l₁ = 60 cm
∴ l₂ = 100 – 60 = 40 cm
Ratio of resistance \(\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{60}{40}=\frac{3}{2}\)
(or) R₁ : R₂ = 3 : 2

Question 17.
A battery of emf 2.5V and internal resistance r is connected in series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r. (Internal resistance)
Solution:
E.m.f. of battery E₁ = 2.5 V ;
Internal resistance = r
Series Resistance R = 45 Ω;
Ammeter Resistance = 1Ω
Reading in ammeter = 50 mA = 50 × 10^-3 Amp

From Kirchhoff’s mesh rule
E = iR + iR₁ – ir
2.5 = i 45 + i × 1 – ir
⇒ ir = 45 i + i – ir ⇒ E – 46i = – ir
∴ 2.5 – 46 × 50 × 10^-3 = -50 × 10^-3 r ⇒ 2.5 – 2.3 = -ir
∴ r = \(\frac{-0.2}{50\times10_{-3}}\) -ve sign indicates current in opposite direction.
∴ Internal resistance of battery
r = \(\frac{200}{50}\) = 4Ω

Question 18.
Amount of charge passing through the cross section of a wire is q(t) = at² + bt + c. Write the dimensional formula for a, b and c. If the values of a, b and c in SI unit are 6, 4, 2 respectively, find the value of current at t = 6 seconds.
Solution:
Given q (t) = at² + bt + c
But current i = \(\frac{d}{dt}\) q(t) = \(\frac{d}{dt}\) (at² + bt + c) = 2at + b → (1)
Put a = 6, b = 4, c = 2 and t = 6 in eq – (1)
∴ Current i = 2x6x6 + 4 = 72 + 4 = 76 Amp

Intext Question and Answer

Question 1.
The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?
A. Emf of the battery, E = 12 V ;
Internal resistance of the battery, r = 0.4Ω
Maximum current drawn from the battery = I ; According to Ohm’s law,
E = Ir
I = \(\frac{E}{r}=\frac{12}{0.4}\) = 30 A
The maximum current drawn from the given battery is 30 A.

Question 2.
A potentiometer wire is 5 m long and a potential difference of 6V is maintained between its ends. Find the emf of a cell which balances against a length of 180 cm of the potentiometer wire. [TS May ’16]
Answer:
Total length of potentiometer wire L = 5 m = 500 cm
Balancing length l₁ = 180 cm
P.D across the terminals = 6V
In potentiometer e.m.f at balancing point
El 6×180
V = \(\frac{El}{L}=\frac{6\times180}{500}\)
∴ E.m.f of cell in secondary E₁ = 2.16 V

Question 3.
A battery of emf 10 V and internal resistance 3Ω is connected to a resistor (R). If the current in the circuit is 0.5 A, what is the resistance (K) of the resistor? What is the terminal voltage of the battery when the circuit is closed? [(IMP) TS Mar, ’15]
Answer:
Emf of the battery, E = 10 V ;
Internal resistance of the battery, r = 3 Ω
Current in the circuit, 1 = 0.5 A ;Resistance of the resistor = R
The relation for current using Ohm s law is,
I = \(\frac{E}{R+r}\) ⇒ R + r = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) = 20 Ω;
∴ R = 20 – 3 = 17 Ω
Terminal voltage of the resistor = V
According to Ohm’s law,
V = IR = 0.5 × 17 = 8.5 V
Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform crosssection 6.0 × 10^-7m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer:
Length of the wire, l = 15 m
Area of cross-section of the wire, a = 6.0 × 10^-7m²
Resistance of the material of the wire.
R = 5.0 Ω
Resistivity of the material of the wire = ρ
Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10^-7 Ω m.

Question 5.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Answer:
Temperature, T₁ = 27.5°C ;
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C ;
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0.0039°C^-1.

Question 6.
Determine the current in each branch of the network shown in fig:

Answer:
Current flowing through various branches of the circuit is represented in the given figure.
I₁ = Current flowing through the outer circuit
I₂ = Current flowing through branch AB
I₃ = Current flowing through branch AD
I₂ – I₄ = Current flowing through branch BC
I₃ – I₄ = Current flowing through branch CD
I₄ = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,
10I₂ + 5I₄ – 5I₃ = 0 (OR) ⇒ 2I₂ + I₄ – I₃ = 0
I₃ = 2I₂ + I₄ …… (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I₂ – I₄) – 10(I₃ + I₄) – 5I₄ = 0 (OR)
5I₂ + 5I₄ – 10I₃ – 10I₄ – 5I₄ = 0
5I₂ – 10I₃ – 20I₄ = 0 ; I₂ = 2I₃ + 4I₄ ….. (2)
For the closed circuit ABCFEA, potential is zero i.e.,
– 10 + 10(I₁) + 10(I₂) + 5(I₂ – I₄) = 0
10 = 15I₂ + 10I₂ – 5I₄ ⇒ 3I₂ + 2I₁ – I₄ = 2 ….. (3)
From equations (1) and (2), we obtain
I₃ = 2(2I₃ + 4I₄) + I₄; I₃ = 4I₃ + 8I₄ + I₄ – 3I₃ = 9I₄ ⇒ -3I₄ = + I₃ ….. (4)
Putting equation (4) in equation (1), we obtain
I₃ = 2I₂ + I₄ (OR) ⇒ -4I₄ = 2I₂ ; I₂ = – 2I₄ …… (5)
It is evident from the given figure that,
I₁ = I₃ + I₂ ……. (6)
Putting equation (6) in equation (1), we obtain
3I₂ +2(I₃ + I₂) – I₄ = 2
5I₂ + 2I₃ – I₄ = 2 ….. (7)
Putting equations (4) and (5) in equation (7), we obtain 5(-2I₄) + 2(-3I₄) – I₄ = 2 – 10I₄ – 6I₄ – I₄ = 2 (OR) 17I₄ = – 2
⇒ I₄ = \(\frac{-2}{17}\) A
Equation (4) reduces to I₃ = – 3(I₄) = -3\(\frac{-2}{17}=\frac{6}{17}\)A

Question 7.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell? [(IMP) AP Mar. ’15]
Answer:
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁ = 35 cm
The cells is replaced by another cell of emf E₂.
New balance point of the potentiometer, l₂ = 63 cm
The balance condition is given by the

Therefore, emf of the second cell is 2.25V.

Question 8.
The number density of free electrons in a copper conductor estimated in Example 6.1 is 8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Answer:
Number density of free electrons in a copper conductor, n = 8.5 × 10²⁸ m^-3,
Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 × 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where, e = Electric charge =1.6 × 10^-19 C
v_d = Drift velocit

Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 10⁴ s.