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NCERT Exemplar Problems Class 12 Physics Chapter 15 Communication Systems

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?
(a) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave
Solution: (b)
Key concept: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation.
• Ground wave propagation
• Sky wave propagation
• Space wave propagation
Mode of communication frequency range:
• Ground wave propagation— 500 kHz to 1710 kHz
• Sky wave propagation — 2 MHz to 40 MHz
• Space wave propagation— 54 MHz to 42 GHz
So, A is transmitted via ground wave, B via sky wave and C via space wave.

Question 2. A loom long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with λ.
(a) ~400m (b) -25 m (c) -150 m (d) -2400 m
Solution: (a) Length of the building (l) is
l = 500 m
and length of antenna = 100 m
and we know, wavelength of the wave which can be transmitted by
L =λ/4. So, λ~ 4l= 4 x 100 = 400 m
Wavelength (λ) is nearly equal to 400 m.

Question 3. A1 kW signal is transmitted using a communication channel which provides attenuation at the rate of -2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in d_B =10 log₁₀(p₀/p_i)]
(a) 900 W (b) 100 W (c) 990 W (d) 1010 W
Solution:

Question 4. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be
(a) 1.003 MHz and 0.997 MHz (b) 3001 kHz and 2997 kHz
(c) 1003 kHz and 1000 kHz (d) 1 MHz and 0.997 MHz
Solution: (a)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged.
Side band frequencies: The AM wave contains three frequencies f_c, (f_c + fm) and (f_c -f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c +f_m): Upper side band (USB) frequency
(f_c -f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
According to the problem, frequency of carrier signal is f_c = 1 MHz and frequency of speech signal = 3 kHz
= 3 x 10^-3 MHz
= 0.003 MHz
We know that, Frequencies of side bands = (f_c ± f_m) = (1 + 0.003) and (1 – 0.003)
So, side band frequencies are 1.003 MHz and 0.997 MHz.

Question 5. A message signal of frequency ω_m is superposed on a carrier wave of frequency ω_c to get an Amplitude Modulated Wave (AM). The frequency of the AM wave will be

Solution: (b)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged or we can say that the frequency of modulated wave is equal to the frequency of carrier wave. Now, according to the problem, frequency of carrier wave is f_c.
Thus the amplitude modulated wave also has frequency f_c.

Question 6. I-V Characteristics of 4 devices are shown in figure.

Solution: Key concept: A square law modulator is the device which can produce modulated waves by the application of the message signal and the carrier wave.
Square law modulator is used for modulation purpose. Characteristics shown by (i) and (iii) correspond to linear devices.
And by (ii) corresponds to square law device which shows non-linear relations. Some part of (iv) also follow square law.
Hence, (ii) and (iv) can be used for modulation.

Question 7. A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0 < m <1)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission.
Solution: (b) In this problem, the frequency of modulated signal received becomes more, due to improper selection of bandwidth.
This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal.
But, the frequency of male voice is less than that of a female.

Question 8. A basic communication system consists of
A. transmitter.
B. information source.
C. user of information.
D. channel.
E. receiver.
Choose the correct sequence in which these are arranged in a basic communication system.
(a) ABCDE (b) BADEC (c) BDACE (d) BEADC
Solution: (b) A basic communication system consists of an information source, a transmitter, a link (channel) and a receiver or a communication system is the set-up used in the transmission and reception of information from one place to another.
The whole system consist of several elements in a sequence. It can be represented as the diagram given below:

Question 9. Identify the mathematical expression for amplitude modulated wave,

Solution:

One or More Than One Correct Answer Type

Question 10. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal cannot be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km
Solution: (a, b, d)
Key concept: Size of the antenna or aerial. For transmitting a signal, we need an antenna or an aerial. This antenna should have a size comparable to the wavelength of the signal (at least 1/4 in dimension) so that the antenna properly senses the time variation of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Obviously, such a long antenna is not possible to construct and operate. Hence direct transmission of such baseband signals is not practical. We can obtain transmission with reasonable antenna s if transmission frequency is high (for example,
if n is 1 MHz, then λ is 300 m). Therefore, there is a need of translating the information contained in our original low frequency baseband signal into high or radio frequencies before transmission.
Effective power radiated by an antenna: A theoretical study of radiation from a linear antenna (length l) shows that the power radiated is proportional to (1/λ)² . This implies that for the same antenna length, the power radiated increases with decreasing λ, i.e., increasing frequency. Hence, the effective power radiated by a long wavelength baseband signal would be small. For a good transmission,we need high powers and hence this also points out to the need of using high frequency transmission.

Question 11. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?
(a) The side band frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6 kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The side band frequencies are 1503 kHz and 1497 kHz
Solution: (b, d)

Also,bandwidth =2 f_m=2 x 3=6 KHz

Question 12. A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be (6.4 x 10⁶ m)
(a) 100 km (b) 24 km (c) 55 km (d) 50 km
Solution: (b, c, d)

Therefore, the range of 55.4 km covers the distance 24 km, 55 km and 50 km.

Question 13. The frequency response curve (figure) for the filter circuit used for production of AM wave should be

Solution:(a, b, c)
Key concept:
(i) Side band frequencies-. The AM wave contains three frequencies f_c ,(f_c +f_m) and (f_c-f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c + f_m)- Upper side band (USB) frequency
(f_c – f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency,
(ii) Bandwidth: The two side bands lie on either side of the carrier frequency at equal frequency interval ω_m.
So, bandwidth = {(ω_c + ω_m) – (ω_c – ω_m)} = 2ω_m

To produce an amplitude modulated wave, bandwidth is given by the difference between upper side band frequency and lower side band frequency. Bandwidth = ω_USB – ω_LSB = (ω_c + ω_m) – (ω_c – ω_m)

Question 14. In amplitude modulation, the modulation index m is kept less than or equal to 1 because
(a) m> 1, will result in interference between carrier frequency and message frequency, resulting into distortion.
(b) m > 1, will result in overlapping of both side bands resulting into loss of information
(c) m > 1, will result in change in phase between carrier signal and message signal.
(d) m > 1, indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion.
Solution: (b, d)
Key concept: Modulation index: The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave Is called the modulation factor or degree of modulation or modulation index (m_a).


Very Short Answer Type Questions

Question 15. Which of the following would produce analog signals and which would produce digital signals?
(a) A vibrating tuning fork
(b) Musical sound due to a vibrating sitar string
(c) Light pulse
(d) Output of NAND gate
Solution: Analog and digital signals are the gateway of information or we can say that they are used to transmit information through electric signals. In both these signals, the information such as any audio or video is transformed into electric signals.
The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes. So, output of a NAND gate and a light pulse produces a digital signal.
Thus, (a) and (b) would produce analog signal and (c) and (d) would produce digital signals.

Question 16. Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?
Solution: A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.
But, here the frequency of TV signals are 60 MHz which is beyond the required range (frequency range: there is a maximum frequency of EM waves called critical frequency, above which wave cannot reflect back).
So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.
Important point: Sky wave propagation: These are the waves which are reflected back to the earth by ionosphere.
Ionosphere is a layer of atmosphere having charged particles, ions and electrons and extended above 80 km – 300 km from the earth’s surface.

Question 17. Two waves A and B of frequencies 2 MHz and 3 MHz, respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?
Solution: We know that refractive index p of a layer is

The refractive index of wave B is more than refractive index of wave A because frequency of wave B is more than wave A (as refractive index increases with frequency increases).
Sin i / sin r = µ (lesser the value of r larger the value of µ )
For higher frequency wave (i.e., higher refractive index) the angle of refraction is less, i.e., bending is less. So, wave B travels longer distance in the ionosphere before suffering total internal reflection.
Importance point: Refractive index of a medium is that characteristic which decides speed of light in it.
Dependence of Refractive index:
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media: Refractive index decreases with the increase in temperature.
Total internal reflection: When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90°. This angle of incidence is called critical angle (C).
When angle of incidence exceeds the critical angle then light ray comes back into the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

Question 18. The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index?
Solution:

Question 19. Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.
Solution:

Question 20. Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel?
Solution: An AM signal likely to be more noisy than FM signal through a channel because in case of AM, the instantaneous voltage of carrier waver waves is varied by the modulating wave voltage So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal transmitting in channel. So, noise does not affect FM signal or simply we can say that noise signals are difficult to filter out in AM reception whereas FM receivers easily filter out noise.
Important point: In frequency modulation m_f (frequency modulation index) is inversely proportional to modulating frequency f_m. While in PM it does not vary with modulating frequency. Moreover, FM is more noise immune.

Short Answer Type Questions

Question 21. Figure shows a communication system. What is the output power when input signal is of 1.01 mW? [Gain in d_B = 10 log₁₀ (P₀ / P₁)]

Solution:

Question 22. A TV transmission tower antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to case (i).
Solution:

Question 23. If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earth’s radius.?
Solution:
Key concept: Distance or range of transmission tower, d_T =√2Rh_T
where, R is the radius of the earth (approximately 6400 km). h_T is the height of transmission tower, .
d_T is also called the radio horizon of the transmitting antenna.
Let us consider the figure given below to solve this problem.
Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h₁, i.e. h_T = h_R and radius of earth is R. If d_M is the line-of-sight distance between the transmission and receiving antennas, then maximum distance

Question 24. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max= 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F₁ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere. Estimate the maximum electron densities of the F₁ and F₂ layers on that day.
Solution:

Question 25. On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c + ω_m). Suggest ways to minimise cost of radiation without compromising on information.
Solution:
Key concept: Side band frequencies. The AM wave contains three frequencies ω_c, (ω_c + ω_m) and (ω_c – ω_m), ω_c is called carrier frequency, (ω_c + ω_m) and ( ω_c – ω_m) are called side band frequencies.
(ω_c + ω_m) = Upper side band (USB) frequency
(ω_c – ω_m) =Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
Only side band frequencies contain information in amplitude modulated signal, [only (ω_c+ ω_m) and (ω_c + ω_m)].
Here, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c+ω_m)
For reduction of cost of radiation without compromising on information ω_c can be left and transmitting the frequencies (ω_c + ω_m), (ω_c – ω_m) or both (ω_c + ω_m) and (ω_c – ω_m).

Long Answer Type Questions

Question 26. The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I₀e^-ax, where I₀ is the intensity at x = 0 and α is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of (In4/α).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 log₁₀(I/I₀).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
Solution:

Question 27. A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above Earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
Solution: Let the receiver is at point A and source is at B.

Question 28. An amplitude modulated wave is as shown in figure. Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and
(iii) peak value of information voltage

Solution:

Question 29. (i) Draw the plot of amplitude versus ω for an amplitude modulated wave whose carrier wave (ω > ω_c) is carrying two modulating signals, ω1 and ω₂ (ω₂> ω₁).
(ii) Is the plot symmetrical about ω_c? Comment especially about plot in region (ω < ω_c ).
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
Solution:

(ii) In the plotted graph shown, we note that frequency spectrum is not symmetrical about ω_c. Crowding of spectrum is present for ω < ω_c.
(iii) If more modulating signals are present then there will be more crowding in the modulation signal in the region ω <ω_c. That will result more chances of mixing of signal.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves ω_c. This shows that large carrier frequency
enables to carry more information (i.e., more ω_m) and the same will in turn increase bandwidth.

Question 30. An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
(i) R = 1 kΩ, C= 0.01 µF?
(ii) R= 10 kΩ, C=0.01 µF?
(iii) R = 10 kΩ, C = 0.1 µF?
Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. The conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free current carries increases
(b) relaxation time increases
(c) both number density of carries and relaxation time increase
(d) number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density .
Solution: (d)

Question 2.

Solution: (b)
Key concept: P-N Junction Diode:
– When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting arrangement is called P-N junction or P-N junction diode.

(1) Depletion region: On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region.
Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions, appears near the junction in N-crystal. This layer is called depletion layer.

(2) Potential barrier: The potential difference created across the P-N junction due to the diffusion of electron and holes is called potential barrier.

Height of potential barrier is decreases when p-n junction is forward biased, it opposes the potential barrier junction, when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

Question 3. In figure given on next page, assuming the diodes to be ideal
(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa
(c) D₁ and D₂ are both forward biased and hence current flows from A to B
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice-versa

Solution:

Question 4. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor?

Solution: (d)
Key concept: Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called half wave rectifier.

Question 5. Hole is
(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created particle
Solution: (b) Concept of holes in the semiconductor:

1. .When an electron is removed from a covalent bond, it leaves a vacancy behind. An electron from a neighbouring atom can move into this vacancy, leaving the neighbour with a vacancy. In this way the vacancy formed is called a hole (or cotter), and can travel through the material and serve as an additional current carriers.
2. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron.
3. Holes acts as a virtual charge, although there is no physical charge on it.
4. Effective mass of hole is more than an electron.
5. Mobility of hole is less than an electron.

Question 6. The output of the given circuit in figure is given below.

(a) would be zero at all times
(b) would be like a half wave rectifier with positive cycles in output
(c) would be like a half wave rectifier with negative cycles in output
(d) would be like that of a full wave rectifier
Solution: (c)

When the diode is forward biased during positive half cycle of input AC voltage, the resistance of p-n junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero output voltage across p-n junction.
And when the diode is reverse biased during negative half cycle of AC voltage, the p-n junction is reverse biased. The resistance of p-n junction becomes high which will be more than resistance in series. That is why, there will be voltage across p-n junction with negative cycle in output, hence option (c) is correct.

Question 7. In the circuit shown in figure given below, if the diode forward voltage between A and B is

(a) 1.3 V (b) 2.3 V (c) 0 (d) 0.5 V
Solution:

Question 8. Truth table for the given circuit is

Solution:

One or More Than One Correct Answer Type

Question 9. When an electric field is applied across a semiconductor
(a) electrons move from lower energy level to higher energy level in the conduction band
(b) electrons move from higher energy level to lower energy level in the conduction band
(c) holes in the valence band move from higher energy level to lower energy level
(d) holes in the valence band move from lower energy level to higher energy level
Solution: (a, c)
In valence band electrons are not capable of gaining energy from external electric field. While in conduction band the electrons can gain energy from external electric field.
When electric field is applied across a semiconductor, the electrons in the conduction band (which is partially filled with electrons) get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Question 10. Consider an n-p-n transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
(a) Electrons crossover from emitter to collector
(b) Holes move from base to collector
(c) Electrons move from emitter to base
(d) Electrons from emitter move out of base without going to the collector.
Solution: (a, c)

In normal operation base-emitter is forward biased, i.e., the positive pole of emitter base battery is connected to base and its negative pole is connected to the emitter. And collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base. Thus, electron moves from emitter to base and crossover from emitter to collector.

Question 11.

Solution: (b, c, d) According to above graph transfer characteristics of a base biased common emitter transistor, we note that .
(a) when V_i= 0.4 V, output voltage remain same,there is no collection current. So, transistor circuit is not in active state.
(b) when V_i = 1 V (This is in between 0.6 V to 2 V), the transistor circuit is in active state and when input is increasing output is decreasing because when CE is used as an amplifier input and output voltages are 180° out of phase. Then it is used as an amplifier.
(c) when V_i = 0.5 V, there is no collector current. The transistor is in cut off state. The transistor circuit can be used as a switch to be turned off.
(d) when V_i = 2.5 V, the collector current becomes maximum and transistor is in a saturation state and can used as switch turned on state.

Question 12. In a n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
(a) The emitter current will be 8 mA
(b) The emitter current will be 10.53 mA
(c) The base current will be 0.53 mA
(d) The base current will be 2 mA
Solution: (b, c) According to the problem, the collector current is 95% of electrons reaching the collector after emission. And collector current, I_C = 10 mA

Question 13. In the depletion region of a diode
(a) there are no mobile charges
(b) equal number of holes and elections exist, making the region neutral
(c) recombination of holes and electrons has taken place
(d) immobile charged ions exist
Solution: (a, b, d) On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-rcgion diffuse through the junction into P-region and the hole from P-region diffuse into N-region.

Due to diffusion, neutrality of both N-and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This layer is called depletion layer.
The thickness of depletion layer is 1 micron = 10^-6 m.
Width of depletion layer ∞ 1/Dopping
Depletion is directly proportional to temperature.
Important point: The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a dielectric.

Question 14. What happens during regulation action of a Zener diode?
(a) The current and voltage across the Zener remains fixed
(b) The current through the series Resistance (R_s) changes
(c) The Zener resistance is constant
(d) The resistance offered by the Zener changes

Solution: (b, d) Symbolically zener diode represents like this:
In the forward bias, the zener diode acts as an ordinary diode. It can be used as a voltage regulator.

A zener diode when reverse biases offers constant voltage drop across in terminals as unregulated voltage is applied across circuit to regulate. Then during regulation action of a Zener diode, the current through the series resistance R_s changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

Question 15. To reduce the ripples in rectifier circuit with capacitor filter
(a) R_L should be increased
(b) input frequency should be decreased .
(c) input frequency should be increased
(d) capacitors with high capacitance should be used
Solution: (a, c, d)
Ripple factor may be defined as the ratio of r.m.s. value of the ripple voltage to the absolute value of the DC component of the output voltage, usually expressed as a percentage. However ripple voltage is also commonly expressed as the peak-to-peak value. Ripple factor (r) of a full wave rectifier using capacitor filter is given by

Ripple factor is inversely proportional to R_L, C and v.
Thus to reduce r, R_L should be increased, input frequency v should be increased and capacitance C should be increased.

Question 16. The breakdown in a reverse biased p-n junction is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strongelectricfieldinadepletionregionifthedopingconcentrationissmall
(d) strong electric field in the depletion region if the doping concentration is large
Solution: (a, d)
Reverse biasing: Positive terminal of the battery is connected to the N-crystal and negative terminal of the battery is connected to P-crystal.

(i) In reverse biasing width of depletion layer increases
(ii) In reverse biasing resistance offered R_Reverse = 10⁵ Ω
(iii) Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority carriers.
(A very small reverse current may exist in the circuit due to the drifting of minority carriers across the junction)
(iv) Break down voltage: Reverse voltage at which break down of semiconductor occurs. For Ge it is 25 V and for Si it is 35 V.

So, we conclude that in reverse biasing, ionization takes place because the minority charge carriers will be accelerated due to reverse biasing and striking with atoms which in turn cause secondary electrons and thus more number of charge carriers.
When doping concentration is large, there will be a large number of ions in the depletion region, which will give rise to a strong electric field.

Very Short Answer Type Questions

Question 17. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Solution: When pure semiconductor material is mixed with small amounts of certain specific impurities with valency different from that of the parent material, the number of mobile electrons/holes drastically changes. The process of addition of impurity is called doping. The size of the dopant atom should be compatible such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Silicon or Germanium atoms, which are provided by group XIII or group XV elements.

Question 18. Sn, C and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?
Solution: The conduction level of any element depends on the energy gap between its conduction band and valence band.
In conductors, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for semiconductor the energy gap is moderate.
The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV related to their atomic size. Therefore Sn is a conductor, C is an insulator, and Ge and Si are semiconductors

Question 19. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Solution: We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance. Potential of potential barrier for Ge is V_B = 0.3 V and for silicon is V_B = 0.7 V.
On the average the potential barrier in P-N junction is ~0.5 V.

Question 20. Draw the output waveform across the resistor in the given figure.

Solution: The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased. So the output is obtained only when positive input is given,so the output waveform is

Question 21. The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10,20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value).
(i) if DC supply voltage is 10 V?
(ii) if DC supply voltage is 5 V?
Solution: Total voltage amplification is defined as the ratio of output signal voltage and input signal voltage.

Question 22. In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?
Solution:


The power gain is very high in CE transistor amplifier. In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violate the law of conservation.

Short Answer Type Questions

Question 23. (i) Name the type of a diode whose characteristics are shown in figure (a) and (b).
(ii) What does the point P in fig. (a) represent?
(iii) What does the points P and Q in fig. (b) represent?

Solution:
(i) Fig. (a) represents the characteristics of Zener diode and curve (b) is of solar cell.
(ii) In fig. (a), point P represents Zener breakdown voltage.
(iii) In fig. (b), the point Q represents zero voltage and negative current. Which means the light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point Q the battery is short circuited. Hence it represents the short circuit current.
And the point Pin fig. (b) represents some open circuit, voltage on solar cell with zero current through solar cell.
It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.

Question 24. Three photo diodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å ?
Solution:Key concept: In Photo diodes electron and hole pairs are created by junction photoelectric effect. That is the covalent bonds are broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals.

The incident radiation which is detected by the photodiode D₂ because energy of incident radiation is greater than the band-gap.

Question 25. If the resistance R₁ is increased (see figure), how will the readings of the ammeter and voltmeter change?

Solution: Let us redrawn the circuit diagram to find the change in reading of ammeter and voltmeter.


So, R₁ is increased, I_B is decreased.
Now, the current in ammeter is collector current I_C.
I_C =βI_B as I_B is decreased, I_C is also decreased and the reading of voltmeter and ammeter also decreased.

Question 26. Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Draw a circuit that resembles this situation using diodes for this situation.
Solution: As car enters in either of the garages or both, the common gate opened automatically.
This means that if any one input is high, output will high otherwise low.
The device is shown like this:

Question 27. How would you set up a circuit to obtain NOT gate using a transistor?
Solution:
(1)It has only one input and only one output.
(2)Boolean expression is Y = Ᾱ and is read as “y equals not A” .
Logical symbol of NOT gate.

(3)Realization of NOT gate: The transistor is so biased that the collector voltage V_CC = V (Voltage corresponding to 1 state)
The resistors R and R_B are so chosen that if the input is low, i.e. 0, the transistor is in the cut off and hence the voltage appearing at the output will be the same as applied V = 5 V. Hence Y= V(or state I)
If the input is high, the transistor current is in saturation and the net voltage at the output Y is 0 (in state 0).

Question 28. Explain why elemental semiconductor cannot be used to make visible LEDs.
Solution:

Question 29. Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.

Solution:

Question 30.

Solution:

Long Answer Type Questions

Question 31. If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents I₁, I₂, I₃ and I₄?

Solution:

Question 32. In the circuit shown in figure, when the input voltage of the base resistance is 10 V, V_BE is zero and V_CE is also zero. Find the values of I_B, I_C and β .

Solution:

Question 33. Draw the output signals C₁ and C₂ in the given combination of gates.

Solution:

Question 34. Consider the circuit arrangement shown in figure for studying input and output characteristics of n-p-n transistor in.CE configuration.
Select the values of R_B and R_C for a transistor whose V_BE = 0.7 V so that the transistor is operating at point Q as shown in the characteristics (see figure).

Solution:

Question 35. Assuming the ideal diode, draw the output waveform for the circuit given in figure, explain the waveform.

Solution:
Key concept: An ideal diode is a diode in which it has a very large resistance in reverse biased and very, low resistance in forward biased. So, it acts like a perfect conductor when voltage is applied forward biased and like a perfect insulator when voltage is applied reverse biased.
In reverse biased when the input voltage is equal to or less than 5 V diode,then it will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.
If input voltage is greater than +5 V, diode is in conducting state, then it will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.
When input voltage is negative, there will be opposition to 5 V battery in p-n junction input voltage becomes more than -5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals.
The output waveform will be like this (as shown below).

Question 36. Suppose a n-type wafer is‘created by doping Si crystal having 5 x 10²⁸ atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering n_i = 1.5 x 10¹⁶ m^-3,
(i) Calculate the densities of the charge carriers in the n and p regions,
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
Solution: n-type wafer is created when As is implanted in Si crystal. The number of majority carriers electrons due to doping of As is

Question 37. An XOR gate has the following truth table.

It is represented by following logic relation Y = Ᾱ . B + A . B’ . Build this gate using AND, OR and NOT gates.
Solution: XOR gate can be realized by the combination of two NOT gates, two AND gates and one OR gate. According to the problem, the logic relation for the . given truth table is

Question 38. Consider a box with three terminals on top of it as shown in figure.

Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit as shown in figure.
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit. The graphs are
(i) when A is positive and B is negative .



Solution: The V-I characteristics of these graph is discussed in points:
(a)In V-I graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n-side of p-n junction I and B is connected top-side of p-n junction I with a resistance in series.
(b)In V-I graph of condition (ii), a forward characteristics is shown in fig. (d), where 0.7 V is the knee voltage of p-n junction I. 1/slope = (1/1000) Ω.
It means A is connected to n-side of p-n junction I and B is connected to p-side of p-n junction I and resistance R is in series of p-n junction I between A and B.
(c)In V-I graph of condition (iii), a forward characteristics is shown in figure (e) , where 0.7 V is the knee voltage. In this case p-side of p-n junction II is connected to C and n-side of p-n junction II to B.
(d)In V-I graphs of conditions (iv), (v), (vi) also concludes the above connection of p-n junctions I and II along with a resistance R.
Thus, the arrangement of p-n I, p-n II and resistance R between A, B and C will be as shown in the figure.

Question 39. For the transistor circuit shown in figure, evaluate V_E, R_B, R_E, given I_C= 1 mA, V_CE= 3, V_BE = 0.5 V and V_CC= 12 V, β= 100.

Solution: Let us redraw the circuit diagram given here to solve this problem.

Question 40. In the circuit shown in figure, find the value of R_{c .}

.

Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 13 Nuclei

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year,
(a) all the containers will have 5000 atoms of the material
(b) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000
(c) the containers will in general have different numbers of the atoms of the material but their average will be close to 5000
(d) none of the containers can have more than 5000 atoms
Solution: (c)
Key concept: Half life ( T_1/2):
Radioactivity is a process due to which a radioactive material spontaneously decays. Time interval in which the mass of a radioactive substance or the number of its atom reduces to half of its initial value is called the half life of the substance.

In half-life (t= 1 yr) of the material on the average half the number of atoms will decay. Therefore, the containers will in general have different number of atoms of the material, but their average will be approx 5000.

Question 2. The gravitational force between a H-atom and another particle of mass m will

Solution: (b)
Key concept: The gravitational force between a H-atom and another
particle of mass m will be given by Newton’s law, F = G M.m/r²
Here M is the effective mass of Hydrogen atom.
Let us learn how to find the effective mass of a Hydrogen atom.
Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just m_proton + m_electron.
Now let the proton and electron fall towards each other under their mutual electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they are moving with a high speed. Note that we haven’t added or removed any energy, so the mass/energy of the system is still m_proton + m_electron.
The trouble is that this would not form a hydrogen atom because the proton and electron will just speed past each other and fly away again. To form a hydrogen atom we have to take the kinetic energy of the electron and proton out of the system so we can bring them to a stop. Let’s call the kinetic energy E_k. This energy has a mass given by Einstein’s famous equation E = mc², so die mass of our atom is the mass we started with less the energy we’ve taken out:

Question 3. When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom
(a) do not change for any type of radioactivity
(b) change for α and β-radioactivity but not for γ-radioactivity
(c) change for α -radioactivity but not for others
(d) change for β-radioactivity but not for others
Solution: (b)
Key concept:

A /3-particle carries one unit of negative charge (-e), an α-particle carries 2 units of positive charge (+ 2e ) and γ (particle) carries no charge. Hence electronic energy levels of the atom charges for α and β decay, but not for γ-decay.

Question 4. M_x and M_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β^– decay is Q₁ and that for a β⁺decay is Q₂. If me denotes the mass of an electron, then which of the following statements is correct?

Solution: (a)
Key concept: Q value or energy of nuclear reaction: The energy absorbed or released during a nuclear reaction is known as Q-value of nuclear reaction.
Q-value = (Mass of reactants – mass of products)c² Joules
= (Mass of reactants – mass of products) amu
If Q < 0, the nuclear reaction is known as endothermic. (The energy is absorbed in the reaction)
If Q > 0, the nuclear reaction is known as exothermic. (The energy is released in the reaction)

Question 5. Tritium is an isotope of hydrogen whose nucleus triton contains 2 neutrons and 1 proton. Free neutrons decay into p + e + n . If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because
(a) Triton energy is less than that of a He³ nucleus
(b) The electron created in the beta decay process cannot remain in the nucleus
(c) Both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He³ nucleus.
(d) Free neutrons decay due to external perturbations which is absent in triton nucleus .
Solution: (a)

Question 6. Heavy stable nuclei have more neutrons than protons. This is because of the fact that .
(a) neutrons are heavier than protons
(b) electrostatic foree between protons are repulsive
(c) neutrons decay into protons through beta decay
(d) nuclear forces between neutrons are weaker than that between protons
Solution: (b)

Question 7. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose, because
(a) they will break up
(b) elastic collision of neutrons with heavy nuclei will not slow them down
(c) the net weight of the reactor would be unbearably high
(d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature
Solution: (b)
Key concept: A moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission. By slowing the neutrons down the probability of a neutron interacting with Uranium-235 nuclei is greatly increased thereby maintaining the chain reaction. Moderators are made from materials with light nuclei which do not absorb the neutrons but rather slow them down by a series of collisions.
The moderator only slows neutrons down in order to increase the interaction with Uranium nuclei. They do not give any protection if the reaction goes out of control. 1 fa chain reaction is heading out of control the reactors needs to be able to reduce the concentration of neutrons. For this the reactor uses control rods. Control rods are matte from material with the ability to absorb neutrons. Cadmium and Boron are examples of suitable materials. By inserting.control rods between the fuel rods the chain reaction can be slowed dowp-or shut down. Withdrawing the control rods can restart or speed up the reaction.
In our given question, the moderator used have light nuclei (like proton). When protons undergo perfectly elastic collision with the neutron emitted their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons.
Heavy nuclei will not serve the purpose because elastic collisions of neutrons with heavy nuclei will not slow them down.

One or More Than One Correct Answer Type

Question 8. Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact
(a) nuclear forces have short range
(b) nuclei are positively charged
(c) the original nuclei must be completely ionized before fusion can take place
(d) the original nuclei must first break up before combining with each other
Solution: (a, b)
Key coneept:
Nuclear Fusion: In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of a single nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in the release of tremendous amount of energy To achieve fusion, you need to create special conditions to overcome this tendency.
Here are the conditions that make fusion possible:
High Temperature: The high temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons.
• Fusion requires temperatures about 100 million Kelvin (approximately six times hotter titan the sun’s core).
• At these temperatures, hydrogen is a plasma, not a gas. Plasma is a high-energy state of matter in which all the electrons are stripped from atoms and move freely about.
• The sun achieves these temperatures by its large mass and the force of gravity compressing this mass in the core. We must use energy from microwaves, lasers and ion particles to achieve these temperatures.
High pressure: Pressure squeezes the hydrogen atoms together. They must be within 1 x 10^-15 metres of each other to fuse.
• The sun uses its mass and the force of gravity to squeeze hydrogen atoms together in its core.
• We must squeeze hydrogen atoms together by using intense magnetic fields, powerful lasers or ion beams.
Fusion processes are impossible at ordinary temperatures and pressures. The reason is that nuclei are positively charged and nuclear forces are short range strongest forces. In order to force two hydrogen nuclei together, we need to have a very high pressure, or a very high temperature, or both. A high pressure helps because it causes all the hydrogen nuclei in the sun to squeeze into a smaller space. Then there is more chance of one hydrogen bumping into another. A high temperature helps because it makes the hydrogen nuclei move faster. They need this extra speed so that they can get close together and join. It is as if the nucleus has to break through a barrier, and so the faster it is moving, the greater chance it has.
So, at the “normal” temperature and pressure on earth, a hydrogen nucleus has basically no chance of ever joining with another hydrogen nucleus.
Important point: We know that in the middle of the sun, where the temperature is about 16 million degrees, and the pressure is 250 billion atmospheres, hydrogen nuclei will sometimes have enough energy to join together. (An atmosphere is the “normal”, pressure of the air here on earth. A pressure of 250 billion atmospheres is like having a large mountain piled on top of you!)

Question 9. Samples of two radioactive nuclides A and B are taken. λ_A and λ_B are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?
(a) Initial rate of decay of A is twice the initial rate of decay of B and λ_A = λ_B
(b) Initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B
(c) Initial rate of decay of B is twice the initial rate of decay of A and λ_A > λ_B
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λ_A < λ_B
Solution: (b, d)
Key concept:
Law of radioactive disintegration : According to Rutherford and Soddy law for radioactive decay is as follows:
“At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant.” i.e.
dN/dt ∞ N => dN/dt = -λN
it can be proved that N=N₀e^-λ1
In terms of mass M— M₀e^-λ1
where N = Number of atoms remains undecayed after time t,
N₀ = Number of atoms present initially (i.e., at t = 0),
M = Mass of radioactive nuclei at time t,
M₀ = Mass ofradioactive nuclei at time t = 0,
N₀-N= Number of disintegrated nucleus in time t,
dN/dt= rate of decay, λ = Decay constant or disintegration constant or radioactivity constant or Rutherford Soddy’s constant or the probability of decay per unit time of a nucleus.
The samples of the two radioactive nuclides A and B can simultaneously have the same decay rate at any time if initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B.
Also, when initial rate of decay of B is the same as rate of decay of A at t = 2h and λ_B < λ_A.

Question 10.

Solution:

Hence at point P, rate of decay for both A and B is the same.

Very Short Answer Type Questions

Question 11. He₂³ and He₁³ nuclei have the same mass number. Do they have the same binding energy?
Solution: The nuclei He₂³ and He₁³ have the same mass number. He₂³ has two protons and one neutron. He₂³ has one proton and two neutrons. As He3 has only one proton hence the repulsive force between protons is missing in ₁He³, so the binding energy of ₁He³ is greater than that of ₂He³.

Question 12. Draw a graph showing the variation of decay rate with number of active nuclei.
Solution:

Question 13. Which sample AoxB shown in figure has shorter mean-life?

Solution:
Key concept:
Mean (or average) life (ґ) : The time for which a radioactive material remains active is defined as mean (average) life of that material.
• It is defined as the sum of lives of all atoms divided by the total number of atoms.



Question 14. Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.
Solution:
Key concept: The energy of internal motion of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typically of the order of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by emission of one or more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. This proceks is called gamma (γ) decay.
Excited electron cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV ( mega electron volt), y-radiations have energy of the order of MeV.

Question 15. In pair annihilation, an electron and a positron destroy each other to produce gamma radiations. How is the momentum conserved?
Solution: In pair annihilation, an electron and a positron destroy each other to produce 2yphotons which move in.opposite directions to conserve linear momentum. The annihilation is shown below:

Short Answer Type Questions

Question 16. Why do stable nuclei never have more protons than neutrons?
Solution: The reason is that protons, being charged particles, repel each other. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.
Important point: As you get to heavier elements, with each new proton you add, there is a larger repulsive force. The nuclear force is attractive and stronger than the electrostatic force, but it has a finite range. So you need to add extra neutrons, which do not repel each other, to add extra attractive force. You eventually reach a point where the nucleus is just too big, and tends to decay via alpha decay or spontaneous fission.
To view this in quantum mechanical terms, the proton potential well is not as deep as the neutron well due to the electrostatic repulsion. [Due to the Pauli exclusion principle, you only get two particles per level (spin up and spin down)]. If one well is filled higher than the other, you tend to get a beta decay to even them out. As the nuclei get larger, the neutron well gels deeper as compared to the proton well and you get more neutrons than protons.

Question 17. Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence:
A -> B -> C
Here B is an intermediate nuclei which is also radioactive. Considering that there are N₀ atoms of A initially, plot the graph showing the variation of number of atoms of A and B versus time.
Solution: Consider radioactive nucleus A have N₀ atoms of A initially; or at t = 0, N_A = N₀ (maximum) whole N_B = 0. As time increases, N_A decreases exponentially and the number of atoms of B increases. After some time N_B becomes maximum. As B is an intermediate nuclei which is also radioactive, it also start decaying and finally drop to zero exponentially by radioactive decay law. We can represent the situation as shown in the graph.

Question 18. A piece of wood from the ruins of an ancient building was found to have a ¹⁴ C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴ C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ¹⁴ C is 5760 yr.
Solution:
Key concept:
Carbon dating: Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method tor determining the age of an object containing organic material by using the properties of radiocarbon -14(¹⁴ C), a radioactive isotope of carbon.
Radiocarbon, or carbon 14, is an isotope of the element carbon that is unstable and weakly radioactive. The stable isotopes are carbon 12 and carbon 13.
Carbon 14 is continually being formed in the upper atmosphere by the effect of cosmic ray neutrons on nitrogen 14 atoms. It is rapidly oxidized in air to form carbon dioxide and enters the global carbon cycle.
Plants and animals assimilate carbon 14 from carbon dioxide throughout their lifetimes. When they die, they stop exchanging carbon with the biosphere and their carbon 14 content then starts to decrease at a rate determined by the law of radioactive decay.
Radiocarbon dating is essentially a method designed to measure residual radioactivity. By knowing how much carbon 14 is left in a sample, the age of the organism when it died can be known. It must be noted though that radiocarbon dating results indicate when the organism was alive but not when a material from that organism was used.

Question 19. Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10^-15 m.
Solution:
Key concept: A nucleon is one of the particles that makes up the atomic nucleus. Each atomic nucleus consists of one or more nucleons, and each atom in turn consists of a cluster of nucleons surrounded by one of more electrons. There are two known kinds of nucleon: the neutron and the proton. The mass number of a given atomic isotope is identical to its number of nucleons. Thus the term nucleon number may be used in place of the more common terms mass number or atomic mass number.
For resolving two objects separated by distance d, the wavelength A of the proving signal must be less than d. Therefore, to detect separate parts inside a nucleon, the electron must have a wavelength less than 10^-15 m.

Important point: Until the 1960s, nucleons were thought to be elementary particles, each of which would not then have been made up of smaller parts. Now they are known to be composite particles, made of three quarks bound together by the so-called strong interaction. The interaction between two or more nucleons is called intemucleon interactions or nuclear force, which is also ultimately caused by the strong interaction. (Before the discovery of quarks, the term “strong interaction” referred to just intemucleon interactions.)

Question 20. A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z₁=N₂ and Z₂ = N₁ (a) What nuclide is a mirror isobar of ₁₁²³Na ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?
Solution:
Key concept: Mirror nuclei are nuclei where the number of protons of element one (Z₁) equals the number of neutrons of element two (N₂), the number of protons of element two (Z₂) equal the number of neutrons in element one (N₁) and the mass number is the same.
Pairs of mirror nuclei have the same spin and parity. If we constrain to odd number of nuclcons(A), then w find mirror nuclei that differ one another by exchanging a proton by a neutron. Interesting to observe is their binding energy which is mainly due to the strong interaction and also due to Coulomb interaction. Since the strong interaction is invariant to protons and neutrons one can expect these mirror nuclei to have very similar binding energies.

Long Answer Type Questions

Question 21. Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is

Assume that we start with 1000 ³⁸ S nuclei at time t = 0. The number of ³⁸ Cl is of count zero at t = 0 and will again be zero at t =∞, At what value of t, would the number of counts be a maximum?
Solution:

Question 22. Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen.
Solution: Given the binding energy of a deuteron, B = 2.2 MeV Let kinetic energy and momentum of neutron and proton be K_n, K_P and p_n, p_p respectively.
From conservation of energy,

Question 23. The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in a deuteron as given in the form a coulomb potential but with an effective charge e’.

Solution:

Question 24. Before the neutrino hypothesis, the beta decay process was thought to be the transition.
n —> p + e
If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.
Solution:

Question 25. The activity R of an unknown radioactive nuclide is measured at hourly intervals. The result found are tabulated as follows:

Solution:

Question 26. Nuclei with magic number of proton Z = 2, 8, 20, 28, 50, 52 and magic number of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable.
(i) Verify this by calculating the proton, separation energy S_P for ¹²⁰ Sn (Z= 50) and ¹²¹ Sb(Z= 51).
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by

Solution:

Important point: “Magic Numbers” in Nuclear Structure Careful observation of the nuclear properties of elements, showed certain patterns that seemed to change abruptly at specific key elements. Mayer noticed that magic numbers applied whether one counts the number of neutrons (A). the atomic number (Z), or the sum of the two, known as mass number (A). I-xamples are Helium Z = 2, Lead Z = 82, Helium A = 2, Oxygen N = 8, Lead A = 126, Neon A = 20, Silicon A = 28.
Magic numbers in the nuclear structure have been coming up during all this time, but no plausible explanation for their existence has ever been given. Interestingly, there are peaks and dips for binding energy, repeating every fourth nucleon. This periodicity is one clear indication of a geometrical structure within the nucleus. In particular, those nuclei that can be thought of as containing an exact number of alpha particles (2P + 2A), are more tightly bound than their neighbours. This effect is more pronounced for the lightest nuclei, but is still perceptible up to A – 28. For those nuclei with A > 20, the number of neutrons exceeds the number of protons, so some sort of distortion occurs wi thin the cluster.
It is found that nuclei with even numbers of protons and neutrons are more stable than those with odd numbers. This comes from the fact that the physical structure must have an even number of vertices. A type of regular polyhedron would satisfy this condition, since no regular polyhedron exists with an odd number of vertices. These specific “magic numbers” of neutrons or protons which seem to be particularly favoured in terms of nuclear stability are:
2, 8. 20. 28, 50, 82, 126
. Note that the structure must apply to both protons and neutrons individually, so that we can speak of “magic nuclei” where any one nucleon type, or their sum, is at a magic number. .
The existence of these magic numbers suggests closed shell configurations, like the shells in atomic structure. They represent one line of reasoning which led to the development of a shell model of the nucleus. Other forms – of evidence suggesting shell structure include the following.

1. Enhanced abundance of those elements for which Z or N is a magic number.
2. The stable elements at the end of the naturally occurring radioactive series all have a “magic number” of neutrons or protons.
3. The neutron absorption cross-sections for isotopes where N = magic number are much lower than surrounding isotopes.
4. The binding energy for the last neutron is a maximum for a magic neutron number and drops sharply for the next neutron added.
5. Electric quadrupole moments are near zero for magic number nuclei.
6. The excitation energy from the ground nuclear state to the first excited state is greater for closed shells.

Visualizing the densely packed nucleus in terms of orbits and shells seems much less plausible than the corresponding shell model for atomic electrons. You can easily believe that an atomic electron can complete many orbits without running into anything, but you expect protons and neutrons in a nucleus to be in a continuous process of collision with each other. But dense-gas type models of nuclei with multiple collisions between particles didn’t fit the data, and remarkable patterns like the “magic numbers” in the stability of nuclei suggested the seemingly improbable shell structure.

NCERT Exemplar Problems Class 12 Physics Chapter 12 Atoms

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Taking the Bohr radius as a₀= 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model, will be about
(a) 53 pm (b) 27 pm (c) 18 pm (d) 13 pm
Solution: (c)

Question 2. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is

Solution: (c) In a hydrogen atom, electron revolving around a fixed proton nucleus have some centripetal acceleration. Therefore its frame of reference is non- inertial. If the frame of reference, where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference.

Question 3. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with man electrons. This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law
Solution: (a) The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons because when we derive the formula for radius/energy levels etc, we make the assumption that centripetal force is provided only by electrostatic force of attraction by the nucleus. Hence, this will only work for single electron atoms. In multi-electron atoms, there will also be repulsion due to other electrons. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons.

Question 4. For the ground state, the electron in the H-atom has an angular .momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) because Bohr model gives incorrect values of angular momentum
(b) because only one of these Would Have a minimum energy
(c) angular momentum must be in the direction of spin of electron
(d) because electrons go around only in horizontal orbits
Solution: (a)

Bohr’s model gives only the magnitude of angular momentum. The angular momentum is a vector quantity. Hence we cannot express angular momentum completely by Bohr model. Hence it does not give correct values of angular momentum of revolving electron.

Question 5. 02 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short-ranged
(b) is as important as electrostatic force for binding the two atoms
(c) cancels the repulsive electrostatic force between the nuclei
(d) is not important because oxygen nucleus have equal number of neutrons and protons
Solution: (a)
Key concept: Forces that keep the nucleons bound in the nucleus are called nuclear forces.
• Nuclear forces are short range forces. These do not exist at large distances greater than 10~15 m.
• Nuclear forces are the strongest forces in nature.
• These are attractive force and causes stability of the nucleus.
• These forces are charge independent.
• Nuclear forces arc non-central force.
The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. .
In O₂ molecule which consists of two oxygen atoms molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged and act inside the nucleus only.

Question 6. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(a) 10.20 eV (b) 20.40eV (c) 13.6 eV
Solution: (a)

Question 7. A set of atoms in an excited state decays
(a) in general to any of the states with lower energy
(b) into a lower state only when excited by an external electric field
(c) all together simultaneously into a lower state
(d) to emit photons only when they collide .
Solution: (a)

One or More Than One Correct Answer Type

Question 8. An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,
(a) the electron would not move in circular orbits
(b) the energy would be (2)⁴ times that of a H-atom
(c) the electrons, orbit would go around the protons
(d) the molecule will soon decay in a proton and a H-atom
Solution: (a, c) In a hydrogen atom, electron revolves around a fixed proton nucleus in circular path. This can be explained by Bohr model. But in case of ionised H-molecule which consists of two protons in nucleus and where protons are separated by a small distance of the order of angstrom, cannot be explained by Bohr model. Hence in this case the ground state the electron would not move in circular orbits, the electrons orbit would go around the protons.

Question 9. Consider aiming a beam of free electrons towards free protons. When-they scatter, an electron and a proton cannot combine to produce a H-atom.
(a) Because of energy conservation
(b) Without simultaneously releasing energy in the form of radiation
(c) Because of momentum conservation
(d) Because of angular momentum conservation
Solution: (a, b) A moving electron and proton cannot combine to produce a H-atom because of energy conservation and without simultaneously releasing energy in the form of radiation.

Question 10. The Bohr model for the spectra of a H-atom
(a) will not be applicable to hydrogen in the molecular form
(b) will not be applicable as it is for a He-atom
(c) is valid only at room temperature
(d) predicts continuous as well as discrete spectral lines
Solution: (a, b) Bohr proposed a-model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom, e.g.: H, He⁺, Li⁺², Na⁺¹ etc). It is not applicable to hydrogen in the molecular form and also, it will not be applicable as it is for a He-atom.

Question 11. The Balmer series for the H-atom can be observed
(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) if we measure the frequencies of light emitted due to transitions between
excited states and the first excited state
(c) in any transition in a H-atom .
(d) as a sequence of frequencies with the higher frequencies getting closely
Solution: (b, d)




From above discussion we can say Balmer series for the H-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the’higher frequencies getting closely packed.

Question 12.

Solution: (b, d) Let E₂ and E₁ be the energy corresponding to n = 2 and n = 1 respectively. If radiation of energy ∆E = (E₂ – E₁) = hf incident on a sample where all the H-atoms are in the ground state, according to Bohr model some of the atoms will move to the first excited state. As this energy is not sufficient for transition from n = 1 to n =3, hence no atoms will make a transition to the n = 3 state.

Question 13. The simple Bohr model is not applicable to He⁴ atom because
(a) He⁴ is an inert gas
(b) He⁴ has neutrons in the nucleus
(c) He⁴ has one more electron
(d) electrons are not subject to central forces
Solution: (c, d)
Key concept: Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom). It is valid only for one electron atoms, e.g. : H, He⁺, Li⁺², Na⁺¹ etc.
The Bohr model is not applicable to He⁴, as it has one more electron and electrons are not subjected to central forces.

Very Short Answer Type Questions

Question 14. The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?
Solution:

Question 15. Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.
Solution: Bohr model is applicable for hydrogen atom-and some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom). If we remove one electron from He⁴ and He³, atoms contain one electron and becomes hydrogen like atoms. Now we can apply Bohr model to define the energy levels.

Question 16. When an electron falls from a higher energy to a lower energy level, the
difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? .
Solution: The electrons are charged particles. When an electron falls from a higher energy to a lower energy level, it accelerates. We know accelerating charged particle radiates energy in the form of electromagnetic radiation.

Question 17. Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+ 4/3)e and electron a charge (-3/4)e, where e = 1.6 x 10^-19 C. Give reasons for your answer.
Solution: According to Bohr, for an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force.

Hence the magnitude of electrostatic force F ∞ q₁ x q₂
If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remains same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

Question 18. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
Solution: Bohr postulated that an electron in an atom can move around the nucleus in a certain circular stable orbits without emitting radiations.
Bohr found that the magnitude of the electron’s angular momentum is

Short Answer Type Questions

Question 19. Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
Solution:
Key concept: Positronium (Ps) is a system consisting of an electron and its anti-particle a positron, bound together into an exotic atom, specifically anonium. The system is unstable: the two particles annihilate each other to predominantly produce two or three gamma-rays, depending on the relative spin states. The orbit and energy levels of the two particles arc similar to that of the hydrogen atom (which is a bound slate of a proton and an electron). I lowever, because of the reduced mass, the frequencies of the spectral lines are less than half of the corresponding hydrogen lines.

Question 20. Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
Solution:

Question 21. Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.
Solution:

Question 22. Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than n, are approximate harmonics (i.e., in the ratio 1:2:3 …) when n >> 1.
Solution:

Question 23. What is the minimum energy that must be given to a H-atom in ground state so that it can emit an H_ґ Line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such H_ґ photon?
Solution:

Long Answer Type Questions

Question 24. The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.
Solution: Let µ_H and µ_D are the reduced masses of electron for hydrogen and deuterium respectively.

Question 25. Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ²H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre bf mass.
Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ=m_eMI(m_e + M), wfiere M is the nuclear mass and m_e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ‘H and ²H
(Mass of ¹H nucleus is 1.6725 x 10^-27 kg, mass of ²H nucleus is 3.3374 x 10^-27 kg, Mass of electron = 9.109 x 10^-31 kg.)
Solution:

Question 26. If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å and (ii) R = 10 Å .
Solution: In a H-atom in ground state, electron revolves round the point-size proton in a circular orbit of radius r_B (Bohr’s radius).

Question 27. In the Auger process, an atom maizes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n – 2 to n = 1 transition.
Solution:
Key concept:
Auger Effect: The Auger effect is a process by which electrons with characteristic energies are ejected from atoms in response to a downward transition by another electron in the atom. In Auger spectroscopy, the vacancy is produced by bombardment with high energy electrons, but the Auger effect can occur if the vacancy is produced by other interactions. It is observed as one of the methods of electron rearrangement after electron capture into the nucleus.
If an inner shell electron is removed from an atom, an electron from a higher level will quickly make the transition downward to fill the vacancy. Sometimes this transition will be accompanied by an emitted photon whose quantum energy matches the energy gap between the upper and lower level. Since for heavy atoms this quantum energy will be in the x-ray region, it is commonly called x.-ray fluorescence. This emission process for lighter atoms and outer electrons gives rise to line spectra.
In other cases, the energy released by the downward transition is given to one of the outer electrons instead of to a photon, and this electron is then ejected from the atom with an energy equal to. the energy lost by the electron which made the downward transition minus the binding energy of the electron that is ejected from the atom. Though more involved in interpretation than optical spectra, the analysis of the energy spectrum of these emitted electrons does give information about die atomic energy levels. The Auger effect bears some resemblance to internal conversion of the nucleus, which also ejects an electron
Sometimes an upper election drops to fill the vacancy, emitting a photon.

As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model.

Question 28.

Solution:

Question 29. The Bohr model for the H-atom relies on the Coulomb ’s law of electrostatics. Coulomb’s law has not directly been verified for very
short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge +q₁, – q₂ modified to

Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (a) H (b) H^1/2 (c) H⁰ (d) H^-1/2
Solution: (d)
Key concept: According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
The wave associated with a moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de-Broglie wave is given by

Question 2. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10^-3 nm
(c) 1.2 x 10^-6 nm (d). 1.2 x 10 nm
Solution: (b)
Key concept: According to Einstein’s quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy.
Energy of photon is given by

Question 3. Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then,
(a) no electrons will be emitted as only photons can emit electrons
(b) electrons can be emitted but all with an energy, E₀
(c) electrons can be emitted with any energy, with a maximum of E₀ – ɸ (ɸ is the work function)
(d) electron can be omitted with energy ,with a maximum of E₀
Solution: (d) If a beam of electrons of having energy E₀ is incident on a metal surface kept in an evacuated chamber.

The electrons can be emitted with maximum energy E₀ (due to elastic collision) and With any energy less than E₀, when part of incident energy of electron is used in liberating the electrons from the surface of metal.

Question 4. Consider the figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that (a) will be larger than the earlier value
(b) will be the same as the earlier value
(c) will be less than the earlier value
(d) will depend on the target

Solution: (c)
Key concept: Davision and Germer Experiment:
1. It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by an . electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

2. The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it .about the point of incidence. The energy of the incideni beam of electrons can also be varied by changing the applied voltage to the electron gun.
According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. It is maximum for diffracting angle 50° at 54 volt potential difference.
3. If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula 2d sinθ = nλ, we can determine the wavelength of these waves.
The de-Broglie wavelength associated with electron is where V is the applied voltage.
Using the Bragg’s formula we can determine the wavelength of these waves. If there is a maxima of the, diffracted electrons at an angle θ, then
2d sin θ = A (ii)
From Eq. (i), we note that if V is inversely proportional to the wavelength λ. i.e., V will increase with the decrease, in λ.
From Eq. (ii), we note that wavelength λ is directly proportional to sinθ and hence θ.
So, with the decrease in λ , θ will also decrease.
Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that will be less than the earlier value.

4. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as

Question 5. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as (a) λ_p = λ_n > λ_e > λ_α (b) λ_α < λ_p = λ_n > λ_e
(c) λ_e< λ_p = λn> λ_α (d) λ_e = λ_p = λn = λ_α
Solution: (b)
Key concept:
• Matter Waves (de-Broglie Waves)
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.

Question 6. An electron is moving with an initial velocity v = v₀i and is in a magnetic field B = B₀j. Then, its de-Broglie wavelength
(a) remains constant
(b) increases with time
(c) decreases with time
(d) increases and decreases periodically
Solution: (a)
Key concept: If a particle is carrying a positive charge q and moving with a velocity v enters a magnetic field .5 then it experiences a force F which is given by the expression
F = q(v x B)=$ F = qvB sin θ. As this force is perpendicular to v and B , so the magnitude of v will not change, i.e. momentum (p = mv) will remain constant in magnitude. Hence,

Question 7. An electron (mass m) with an initial velocity v = v₀i(v₀ > 0) is in an electric field E =E₀î (E₀ = constant > 0). Its de-Broglie wavelength at time t is given by

Solution: (a)
Key concept: The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de- Broglie wave is given by

Question 8. An electron (mass m) with an initial velocity v =v₀î is in an electric field

Solution: (c) According to the problem de-Broglie wavelength of electron at time t=0.

One or More Than One Correct Answer Type

Question 9. Relativistic corrections become necessary when the expression for the kinetic
energy 1/2 mv² , becomes comparable with mc². where m is the mass of the
particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?
(a) A=10nm (b) A =10^-1 nm (c) A=10^-4 nm (d) A=10^-6 nm
Solution: (c, d)
Key concept: De-Brogile or matter wave is independent of die charge on the material particle. It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
The de-Broglie wavelength at which relativistic corrections become important that the phase velocity of the matter waves can be greater than the speed of the light (3 x 10⁸ m/s).
The wavelength of de-Broglie wave is given by
λ = h/p = h/mv
Here, h = 6.6 x 10^-34 Js
and for electron, m = 9 x 10^-31 kg
To approach these types of problem we use hit and trial method by picking up each option one by one.

Question 10. Two particles A₁ and A₂ of masses m₁ , m₂ ( m₁> m₂) have the same de-Broglie
wavelength. Then,
(a) their momenta are the same (b) their energies are the same
(c) energy of A₁ is less than the energy of A₂
(d) energy of A₁ is more than the energy of A₂
Solution: (a. c)

Question 11. De-Broglie wavelength associated with uncharged particles: For Neutron efe-Broglie wavelength is given as v_e=c/100.Then

Solution:(b,c)

Question 12. Photons absorbed in a matter are converted to heat. A source emitting v photon/sec of frequency v is used to convert 1 kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion
(a) decreases with increasing n. with v fixed
(b) decreases with n fixed, v increasing
(c) remains constant with n and v changing such that nv = constant
(d) increases when the product nv increases
Solution: (a, b. c).

Question 13. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-BrOglie wavelength of the particle varies cyclically between two values λ₁, λ₂ with λ₁> λ₂ Which of the following statements are true?
(a) The particle could be moving in a circular orbit with origin as centre
(b) The particle could be moving in an elliptic orbit with origin as its focus
(c) When the de-Broglie wavetength is λ₁ the particle is nearer the origin than when its value is λ₂
(d) When the de-Broglie wavelength is λ₂ the particle is nearer the origin than when its value is λ₁
Solution:

Very Short Answer Type Questions

Question 14. Aproton and an a-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λ_p and λ_α related to each other?
Solution:

Question 15. (i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy E_max of the emitted electron as E_max = hv – ɸ₀
where ɸ₀ is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
Solution:

Question 16. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength?
Solution: In the first case, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is low when it has a longer wavelength or in short we can say that energy given out is less than the energy supplied.
But in second case, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light Of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.
But this is not possible for a stable substances.

Question 17. Do all the electrons that absorb a photon come out as photo electrons?
Solution:
Key concept: Photo-Electric Effect:
The photo-electtic effect is the emission of electrons (called photo-electrons when light strikes a surface. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface.
The photoelectric effect is based on the principle of conservation of energy.
1. Two conducting electrodes, the anode (Q) and cathode (P) are enclosed in an evacuated glass tube as shown on next page.
2. The battery or other source of potential difference creates an electric field in the direction from anode to cathode.
3. Light of certain wavelength or frequency falling on the surface of cathode causes a current in the external circuit called photoelectric current.
4. As potential difference increases, photoelectric current also increases till saturation is reached.

Question 18. There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.
Solution:
Key concept: X-Rays:
1. X-rays were discovered by scientist Roentgen that is why they are also called Roentgen rays.
2. Roentgen discovered that when pressure inside a discharge tube is kept 10“3 mm of Hg and potential difference is kept 25 kV, then some unknown radiations (X-rays) are emitted by anode.
3. There are three essential requirements for the production of X-rays.
(i) A source of electron
(ii) An arrangement to accelerate the electrons
(iii) A target of suitable material of high atomic weight and high melting point on which these high speed electrons strike.

Question 19. Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Solution:The momentum of incident photon is transferred to the metal ,during photo electric emission.
At microscopic level ,atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.The excited electron is emitted.Therefore,the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electron.

Question 20.Consider a metal exposed to light the wavelength of 600 nm. The maximum energy of electron doubles when light of wavelength 400 nm is used.Find the work function in eV?
Solution:

Question 21. Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg uncertainty principle (∆x x∆p=h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p = ∆p, find the energy of the electron in electron volts.
Solution:

Question 22. Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?
Solution:

Question 23. Two particles A and B of de-Broglie wavelengths A, and combine to form a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one-dimensional)
Solution:

Question 24. A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1 nm. The first maximum intensity in the reflected beam occurs at θ= 30°. What is the kinetic energy E of the beam in eV?
Solution:

Long Answer Type Questions

Question 25. Consider a thin target (10^-2 cm square, 10^-3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100 W/m² (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na = 0.97 kg/m³]
Solution:

Question 26. Consider an electron in front of metallic surface of a distance d.Assume the force of attraction by the plate is given as . Calculate work in taking the to an infinite distance from the plate .Taking d=0.1 nm. find the work done in electron volts?
Solution:

Question 27. A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stop versus v is given in figure.

Solution:

Question 28. A particle A with a mass m_A is moving with a velocity v and hits a particle B (mass m_B) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic.
Solution:

Question 29. Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?
Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 10 Wave Optics

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in figure.

A Polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.
(a) For a particular orientation, there shall be darkness as observed through the polaroid.
(b) The intensity of light as seen through the polaroid shall be independent of the rotation.
(c) The intensity of light as seen through the polaroid shall go minimum but not zero for two orientations of the polaroid.
(d) The intensity of light as seen through the polaroid shall go minimum for four orientations of the polaroid.
Solution: (c)
Key concept:
Brewster’s law: Brewster discovered that when a beam of unpolarised light is reflected from a transparent medium (refractive index = µ), the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarisation θ_p).
From figure, it is clear that θ_p+ θ_r = 90°
Also n = tan θ_p (Brewster’s law)
(i) For i < θ_p or i > θ_p
Both reflected and refracted rays becomes partially polarised.
(ii) For glass θ_p = 51° for water θ_p = 53°

If a light beam is incident on a glass slab at Brewster’s angle, the transmitted beaiji is unpolarised and reflected beam is polarised.
In the given figure, the light beam is incident from air to the glass slab at Brewster’s angle (i_p). The incident ray is unpolarised and is represented by dot (•). The reflected light is plane polarised represented by arrows.
As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

Important points:
Polarised light: The phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of light wave is called polarization of light.
(i) The plane in which oscillation occurs in the polarised light is called plane of oscillation.
(ii) The plane perpendicular to the plane of oscillation is called plane of polarisation.
(iii) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.
Polaroids: It is a device used to produce the plane polarised light. It is based on the principle of selective absorption and is more effective than the tourmaline crystal, or it is a thin film of ultramicroscopic crystals of quinine iodosulphate with their optic axis parallel to each other.

Question 2. Consider sunlight incident on a slit of width 104 Å. The image seen through the slit shall
(a) be a fine sharp slit white in colour at the centre
(b) a bright slit white at the centre diffusing to zero intensities at the edges
(c) a bright slit white at the centre diffusing to regions of different colours
(d) only be diffused slit white in colour.
Solution: (a)
Key concept: –
Diffraction of Light is the phenomenon of bending of light around the comers of an obstacle/aperture of the size of the wavelength of light.

In figure (A), no diffraction phenomenon is observed as the size of slit is weary large compared to wavelength. But in figure(B), there will be diffraction of light as size of slit is compared to the wavelength of light incident.
Here in the question it is given, width of the slit
b = 104 Å = 104 x 10^-10 m = 10^-6 m = 1 pm
Wavelength of (visible) sunlight varies from 4000 Å to 8000 Å.
Hence the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear, i.e., mixing of colours form white patch at the centre.

Question 3. Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Solution: (a)
Key concept: If slab of a glass is placed in air, the wave reflected from the upper surface (from a denser medium) sutlers a sudden phase change of π, while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change.
It is useful to draw an analogy between reflected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary:
Figure (a) shows that ray reflecting from a medium of higher refractive index undergoes a 180° phase change.

Question 4. In a Young’s double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case,
(a) there shall be alternate interference patterns of red and blue
(b) there shall be an interference pattern for red distinct from that for blue
(c) there shall be no interference fringes
(d) there shall be an interference pattern for red mixing with one for blue
Solution: (c)
Key concept:
Condition for Observing Interference
The initial phase difference between the interfering waves must remain constant. Otherwise the interference will not be sustained.
The frequency and wavelengths of two waves should be equal. If not the phase difference will not remain constant and so the interference will not be sustained.
The light must be monochromatic. This eliminates overlapping of patterns as each wavelength corresponds to one interference pattern.
Here in this problem of Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filtration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.
The wave front emitted by a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent sources so obtained are imaginary.

Question 5. Figure shows a standard two slit arrangement with slits S₁, S₂, P₁, P₂ are the two minima points on either side of P (figure).

At P₂ on the screen, there is a hole and behind P₂.is a second 2-slit arrangement with slits S₃, S₄ and a second screen behind them.
(a) There would be no interference pattern on the second screen but it would be lighted
(b) The second screen would be totally dark
(c) There would be a single bright point on the second screen
(d) There would be a regular two slit pattern on the second screen
Solution: (d)
Key concept:
Wave front
Every point on the given wave front acts as a source of new disturbance called secondary wavelets which travel in all directions with the .velocity of light in the medium.
A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wave front at that instant. This is called secondary wave front.In the given question, there is a hole at point which is a maxima point. From Huygen’s principle, wave will propagate from the sources S₁ and S₂. Each point on the screen will act as secondary sources of wavelets.

One or More Than One Correct Answer Type

Question 6. Two sources S₁ and S₂ of intensity I₁ and I₂ are placed in front of a screen .
(a) The pattern of intensity distribution seen in the central portion is given by Fig. (b).
In this case, which of the following statements are true?

Solution: Key concept:

Question 7. Consider sunlight incident on a pinhole of width 103 Å. The image of the pinhole seen on a screen shall be
(a) a sharp white ring
(b) different from a geometrical image
(c) a diffused central spot, white in colour
(d) diffused coloured region around a sharp central white spot
Solution: (b, d)
Key concept: Diffraction of Light can be observed only if the size of obstacle/aperture is less than the wavelength of light.
Given, width of pinhole = 103 Å = 1000 Å
We know that wavelength of sunlight ranges from 4000 Å to 8000 Å. Clearly, wavelength X < width of the slit.
Hence, light is diffracted from the hole. Due to diffraction from the slit the image formed on the screen will be different from the geometrical image.

Question 8. Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(a) the size decreases (b) the intensity increases
(c) the size increases (d) the intensity decreases
Solution: (a,b)

The central bright disc is known as Airy’s disc.
As the size of the hole is increased, AO will decrease and size of Airy’s disc will decrease.
As the size of the hole is increased, the width of the central maximum of the diffraction pattern of hole decreases. Since the same amount of light is now distributed over a small area, as intensity ∞ 1/area, the area is decreasing so area intensity increases.

Question 9. For light diverging from a point source,
(a) the wavefront is spherical
(b) the intensity decreases in proportion to the distance squared
(c) the wavefront is parabolic
(d) the intensity at the wavefront does not depend on the distance
Solution: (a, b)

Question 10. Is Huygen’s principle valid for longitudinal sound waves?
Solution: Yes it can. Huygen’s principle basically states that every point wave front can be considered as a secondary source of tiny wavelets that spread out in the forward direction of the wave itself. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen’s principle.
So, Huygen’s principle’is valid for longitudinal sound.waves also.

Question 11. Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
Solution: Orientation of wave front is perpendicular to ray. The ray diagram of the situation is as shown in figure.

Parallel rays incident on lens L₁ forms the image I₂ at the focal point of the lens. This image acts as object for the lens L₂ Now, due to the converging lens L₂ , let final image formed is I which is point image. Hence the wavefront for this image will be of spherical symmetry.

Question 12. What is the shape of the wavefront on earth for sunlight?
Solution: The sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. We can treat it like a point object as seen from the surface of earth.
Because of large distance, the radius of wavefront can be considered as Large (infinity) and hence, wavefront is almost plane.

Question 13. Why is the diffraction of Sound waves more evident in daily experience than that of light wave?
Solution: The frequencies of sound waves lie between 20 Hz to 20 kHz, their wavelength ranges between 15 m to 15 mm. The diffraction occurs if the wavelength of waves is nearly equal to slit width.
The wavelength of light waves is 7000 x 10^-10 m to 4000 x 10^-10 m. For observing diffraction of light we need very narrow slit width. In daily life experience we observe the slit width very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

Question 14. The human eye has an approximate angular resolution of ɸ = 5.8 x 10^-4 rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.
Solution: It is given, angular resolution of human eye ɸ = 5.8 x 10^-4 rad and printer prints 300 dots per inch.

Question 15. A polaroid (I) is placed in front,of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.
Solution: A polaroid (I) is placed in front of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) when the pass axis of (II) is parallel to (I), light passes through polaroid-II is unaffected.

Now polariod (II) is rotated till no light passes. In this situation the pass axis of polariod (II) is perpendicular to polariod (I), then (I) and (II) are set in crossed positions. No light passes through polaroid-II.
Now third polaroid (III) is now placed in between (I) and (II). Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Short Answer Type Questions

Question 16. Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
Solution: If angle of incidence is equal to Brewster’s angle, the transmitted light is slightly polarised and reflected light is plane polarised.

Question 17. For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance.
Solution:
Key concept:
• Resolving power is the ability of an imaging device to separate (i.e., to see as distinct) points of an object that are located at a small angular distance or it is the power of an optical instrument to separate far – away objects, that are close together, into individual images. The term resolution or minimum resolvable distance is the minimum distance between distinguishable objects in an image, although the term is loosely used by many users of microscopes and telescopes to describe resolving power. In scientific analysis, in general, the term “resolution” is used to describe the precision with which any instrument measures Ratio of the least separation,
For electrons accelerated through 100 V, the de-Broglie wavelength, 12.27

Question 18. Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is. half the distance between the slits. Obtain the value of D in terms of X such that the first minima on the screen falls at a distance D from the centre O.

Solution:

Long Answer Type Questions

Question 19.

Figure shown has a two sift arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I₀ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
Solution: The resultant amplitude of wave reaching on screen will be the sum of amplitude of either wave in perpendicular and parallel polarisation.

Question 20.

A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

Solution:

Question 21.

Solution:

Question 22.

Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 8 Electromagnetic Waves

Multiple Choice Questions (MCQs)
Single Correct Answer Type
Question 1. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in
(a) visible region (b) infrared region
(c) ultraviolet region (d) microwave region
Solution:

Question 2.

Solution: (b)
Key concept: When a wave is reflected from a denser medium or perfectly reflecting wall made with optically inactive material, then the type of wave doesn’t change but only its phase changes by 180° or π radian.

Question 3. Light with an energy flux of 20 W/cm² falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm², the total momentum delivered (for complete absorption) during 30 min is
(a) 36 x 10^-5 kg-m/s (b) 36 x 10^-4 kg-m/s
(c) 108 x 10⁴ kg-m/s (d) 1.08 x 10⁷ kg-m/s
Solution:

Question 4. The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

Solution:

Question 5.

Solution: (d)
Key concept: A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. The electric vector is responsible for the optical effects of an EM wave and is called the light vector.

Question 6. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

Solution:


Hence the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.
It means the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is 1:1.

• When the incident EM wave is completely absorbed by a surface, it delivers energy u and momentum u/c to the surface.
• When a wave of energy u is totally reflected from the surface, the momentum delivered to surface is 2u/c.

Question 7. An EM wave radiates outwards from a dipole antenna, with E₀ as the amplitude of its electric field vector. The electric field E₀ which transports significant energy from the source falls off as

Solution: (c) A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector (E₀) which transports significant energy from the source falls intensity inversely as the distance (r) from the antenna,

One or More Than One Correct Answer Type
Question 8.


Solution: (a, d) We are given that the electric field vector of an electromagnetic wave travels in a vacuum along z-direction as,

Question 9.

Solution: (a, b, c)

Question 10. A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B.
(a) E_x, B_y (b) E_y, B_z
(c) B_x,E_y (d) E_z,B_y
Solution: (b, d)

Here in the question electromagnetic wave is propagating along x-direction. So, electro and magnetic field vectors should have either y-direction or 2-direction.

Question 11. A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. The electromagnetic waves produced
(a) will have frequency of 10⁹ Hz
(b) will have frequency of 2 x 10⁹ Hz
(c) will have wavelength of 0.3 m
(d) fall in the region of radio waves
Solution: (a, c, d)
Here we are given the frequency by which the charged particles oscillates about its mean equilibrium position, it is equal to 10⁹ Hz. The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.
So, frequency of electromagnetic waves produced by the charged particle is v= 10⁹ Hz.

Question 12. The source of electromagnetic waves can be a charge
(a) moving with a constant velocity
(b) moving in a circular orbit
(c) at rest
(d) falling in an electric field
Solution: (b, d)
Key concept:

• An electromagnetic wave can be produced by accelerated or oscillating charge.
• An oscillating charge is accelerating continuously, it will radiate electromagnetic waves continuously.
• Electromagnetic waves are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number.

Here, in option (b) charge is moving in a circular orbit.
In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.
In option (d), the charge is falling in electric field. If a charged particle is moving in electric field it experiences a force or we can say it accelerates. We know an accelerating charge particle radiates electromagnetic waves. Hence option (d) is also correct.
Also, we know that a charge starts accelerating when it falls in an electric field.
Important points:

• In an atom an electron is circulating around the nucleus in a stable orbit, although accelerating does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit.
• A simple LC oscillator and energy source can produce waves of desired frequency

Question 13. An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?
(a) Radiation pressure is I/c if the wave is totally absorbed
(b) Radiation pressure is —I/c if the wave is totally reflected
(c) Radiation pressure is 2I/c if the wave is totally reflected
(d) Radiation pressure is in the range I/c < p < 2I/c for real surfaces
Solution: (a, c, d)
Key concept: Radiation pressure (p) is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.
Let us consider a surface exposed to electromagnetic radiation as shown in figure. The radiation is falling normally on the surface. Further, intensity of radiation is I and area of surface exposed to radiation is A.

Very Short Answer Type Questions
Question 14. Why is the orientation of the portable radio with respect to broadcasting station important?
Solution: The electromagnetic waves are plane polarised, so the receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave. So the receiving antenna should be parallel to electric/magnetic part of the wave. That is why the orientation of the portable radio with respect to broadcasting station is important.

Question 15. Why does microwave oven heats up a food item containing water molecules most efficiently?
Solution: The microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.

Question 16. The charge on a parallel plate capacitor varies as q=q₀ cos 2πvt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor.
Solution:

Question 17. A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency?
Solution:

It means the displacement current decreases as the conduction current is equal to the displacement current.

Question 18. The magnetic field of a beam emerging from a fitter facing a flood light is given by B₀= 12 x 10^-8 sin (1.20 x 10⁷z- 3.60 x 10¹⁵ t) T What is the average intensity of the beam?
Solution:

Question 19.

Solution:

Question 20. Professor CV Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.
Solution: The properties of an electromagnetic wave is same as other waves. Like other wave an electromagnetic wave also carries energy and momentum. Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped professor C V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it.

Short Answer Type Questions
Question 21.

Solution:

Question 22. Electromagnetic waves with wavelength
(i) λ₁, is used in satellite communication.
(ii) λ₂, is used to kill germs in water purifier.
(iii) λ₃, is used to detect leakage of oil in underground pipelines.
(iv) λ₄, is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.
Solution: (a) (i) In satellite communications, microwave is widely used. Hence λ₁, is the wavelength of microwave.
(ii) In water purifier, ultraviolet rays are used to kill germs. So, λ₂ is the wavelength of UV rays.
(iii) X-rays are used to detect leakage of oil in underground pipelines. So, λ₃ is the wavelength of X-rays.
(iv) Infrared rays are used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.
(b) Wavelength of X-rays < wavelength of UV < wavelength of infrared < wavelength of microwave.
=> λ₃<λ₂<λ₄<λ₁

Question 23.

Solution:

Question 24. You are given a 2 μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
Solution:

Question 25. Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is I/C.
Solution: Let us consider a surface exposed to electromagnetic radiation. The radiation is falling normally on the surface. Further, intensity of radiation is I and area of surface exposed to radiation is A.

Question 26. What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of room, its intensity essentially remains constant.
What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?
Solution: We know intensity of light from a point source I α 1/r², r is the distance from point source.
As the distance is doubled, so the intensity becomes one-fourth the initial value. But in case of laser it does not spread, so its intensity remain same.
Some geometrical characteristics of LASER beam which are responsible for the constant intensity is
(i) Unidirection (ii) Monochromatic
(iii) Coherent light (iv) Highly collimated
These characteristics are missing in the case of normal light from the bulb.

Question 27. Even though an electric field E exerts a force qE on a charged particle yet electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.
Solution: The electric field of an electromagnetic wave is an oscillation field. It exerts electric force on a charged particle, but this electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure though it transfer the energy. In fact, radiation pressure appears as a result of the action of the magnetic field of the wave on the electric currents induced by the electric field of the same wave.

Long Answer Type Questions
Question 28.


Solution:

Question 29.

Solution:

Question 30.

(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current l^d.
(iii) Compare the conduction current I₀ with the displacement current I₀^d.
Solution:

Question 31.

Solution:

Question 32.

Solution: (i) Total energy carried by electromagnetic wave is due to electric field vector and magnetic field vector. In electromagnetic wave, E and B vary from point to point and from moment to moment.

NCERT Exemplar Problems Class 12 Physics Chapter 7 Alternating Current

Multiple Choice Questions (MCQs)
Single Correct Answer Type
Question 1. If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is

Solution: (b)
Key concept: Equation for i and V: Alternating current or voltage varying as sine function can be written as

Question 2. An alternating current generator has an internal resistance R_gand an internal reactance X_g It is used to supply power to a passive load consisting of a resistance R_g and a reactance X_L. For maximum power to be delivered from the generator to the load, the value of X_L is equal to
(a) zero (b) X_g
(c) -X_g (d) R_g
Solution: (c) For maximum power to be delivered from the generator (or internal reactance X_g) to the load (of reactance, X_L),
=> X_L + X_g = 0 (the total reactance must vanish)
=>X_L=-X_g

Question 3.

Solution:

Question 4. To reduce the resonant frequency in an L-C-R series circuit with a generator,
(a) the generator frequency should be reduced
(b) another capacitor should be added in parallel to the first
(c) the iron core of the inductor should be removed
(d) dielectric in the capacitor should be removed
Solution: (b)

Question 5. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
(a) R = 20 Ω, L = 1.5 H, C = 35μF
(b) R = 25 Ω, L = 2.5 H, C = 45 μF
(c) R=15Ω, L = 3.5H, C = 30 μF
(d) R = 25 Ω, L = 1.5 H, C = 45 μF
Solution: (c)


where R is the resistance, L is the inductance and C is the capacitance of the circuit.
For high Q factor R should be low, L should be high and C should be low. These conditions are best satisfied by the values given in option (c).
Important point: Be careful while writing formula for quality factor, this formula we used in this case is only for series L-C-R circuit.

Question 6. An inductor of reactance 1 Ω and a resistor of 2 Ω are connected in series to the terminals of a 6 V (rms) AC source. The power dissipated in the circuit is
(a) 8 W (b) 12 W
(c) 14.4 W (d) 18 W
Solution: (c) According to the problem, XL = 1 Ω , R = 2 Ω ,

Question 7. The output of a step-down transformer is measured to be 24 V when connected to a 12 W light bulb. The value of the peak current is

Solution: (a)

One or More Than One Correct Answer Type
Question 8. As the frequency of an AC circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
(a) Inductor and capacitor (b) Resistor and inductor
(c) Resistor and capacitor (d) Resistor, inductor and capacitor
Solution: (a, d) Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.
Reactance of an inductor of inductance L is X_L = 2πvL, where v is the frequency of the AC circuit.

Question 9. In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
(a) Only resistor (b) Resistor and an inductor
(c) Resistor and a capacitor (d) Only a capacitor
Solution: (c, d) This is the similar problem as we discussed above. In this problem, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreased as increase in frequency. So, one element that must be connected is capacitor. We can also connect a resistor in series.
For a capacitive circuit,
X_C = 1/ωC =1/2πfC
When frequency increases, X_C decreases. Hence current in the circuit increases.

Question 10. Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
(a) For a given power level, there is a lower current
(b) Lower current implies less power loss
(c) Transmission lines can be made thinner
(d) It is easy to reduce the voltage at the receiving end using step-down transformers
Solution: (a, b, d)

Question 11. For an L-C-R circuit, the power transferred from the driving source to the driven oscillator is P = I² Z cos Ф.
(a) Here, the power factor cos Ф > 0, P > 0
(b) The driving force can give no energy to the oscillator (P = 0) in some cases
(c) The driving force cannot syphon out (P < 0) the energy out of oscillator
(d) The driving force can take away energy out of the oscillator
Solution:

Question 12. When an AC voltage of 220 V is applied to the capacitor C
(a) the maximum voltage between plates is 220 V
(b) the current is in phase with the applied voltage
(c) the charge on the plates is in phase with the applied voltage
(d) the power delivered to the capacitor is zero
Solution: (c, d) If the alternating voltage is applied to the capacitor, the plate connected to the positive terminal of the source will be at higher potential and the plate connected to the negative terminal of source will be at lower potential. So the plates capacitor is charged.

Question 13.

Solution:

Very Short Answer Type Questions
Question 14. If an L-C circuit is considered analogous to a harmonically oscillating spring- block system, which energy of the L-C circuit would be analogous to potential energy and which one analogous to kinetic energy?
Solution: When a charged capacitor C having an initial charge q0 is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant.
The oscillation of the LC circuit are an electromagnetic analog to the mechanical oscillation of a block-spring system.
The total energy of the system remains conserved.

Question 15. Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

Solution:
Key concept: The element with infinite resistance will be considered as open circuit and the element with zero resistance will be considered as short circuited.

Question 16. Study the circuits (a) and (b) shown in figure and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?
(b) Can the rms current in circuit (b) be larger than that in (a)?
Solution:

Question 17. Can the instantaneous power output of an AC source ever be negative? Can the average power output be negative?
Solution:

Question 18. In a series LCR circuit, the plot of I_max versus co is shown in figure. Find the bandwidth and mark in the figure.

Solution:

Question 19. The alternating current in a circuit is described by the graph shown in figure. Show rms current in this graph.

Solution:

Question 20. How does the sign of the phase angle Ф, by which the supply voltage leads the current in an L-C-R series circuit, change as the supply frequency is gradually increased from very low to very high values?
Solution:

Short Answer Type Questions
Question 21. A device ‘X is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in figure.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device X.

Solution: (a) Power is the product of voltage and current (Power = P = VI).
So, the curve of power will be having maximum amplitude, equals to the product of amplitudes of voltage (V) and current (I) curve. Frequencies , of B and C are-equal, therefore they represent V and I curves. So, the curve A represents power.
(b) The full cycle of the graph (as shown by shaded area in the diagram) consists of one positive and one negative symmetrical area.

Hence, average power consumption over a cycle is zero.
(c) Here phase difference between V and I is π /2 therefore, the device ‘X’ may be an inductor (L) or capacitor (C) or the series combination of L and C.

Question 22. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?
Solution: For a Direct Current (DC),
1 ampere = 1 coulomb/sec
Direction of AC changes with the frequency of source with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of AC.
So, r.m.s. value of AC is equal to that value of DC, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time.

Question 23. A coil of 0.01 H inductance and 1 ω resistance is connected to 200 V, 50 Hz AC supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current.
Solution:

Question 24. A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.
Solution:

Question 25. Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
Solution: Capacitor plates get charged and discharged when an AC voltage is applied across the plates. So the current through capacitor is as a result of charging charge. Because the frequency of the capacitive circuit increases, the polarities of the charged plates change more rapidly with time, giving rise to a’larger current. The capacitance reactance (X_C) due to a capacitor C varies
as the inverse of the frequency (f) (as X_C=1/2π fC) and hence approaches zero as v approaches infinity. The current is zero in a DC capacitive circuit, which corresponds to zero proportional and infinite reactance. Also, Since X_C is inversely proportional to frequency, capacitors tend to pass high-frequency current and to block low-frequency currents and DC (just the opposite of inductors).

Question 26. Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
Solution: The inductive reactance is given by X_L = 2πfL, X_L is proportional to the frequency and current is inversely proportional to the reactance. An inductor opposes the flow of current through it by developing a back emf according to Lenz’s law. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice-versa.
Since, the induced emf is proportional to the rate of change of current.

Long Answer Type Questions
Question 27.

Solution:

Question 28.

Solution:

Question 29. Consider the L-C-R circuit shown in figure. Find the net current i and the phase of i. Show that i = V/Z. Find the-impedance Z for this circuit.

Solution: Key concept: In the circuit given above consists of a capacitor (C) and an inductor (L) connected in series and the combination is connected in parallel with a resistance R. Due to this combination there is an oscillation of electromagnetic energy.

Question 30.

Solution:

Question 31.

Solution:

NCERT Exemplar Problems Class 12 Physics Chapter 6 Electromagnetic Induction

Multiple Choice Questions (MCQs)
Single Correct Answer Type
Question 1.

Solution: (c)
Key concept: Magnetic flux is defined as the total number of magnetic lines of force passing normally through an area placed in a magnetic field and is equal to the magnetic flux linked with that area.

Question 2.

Solution:

Question 3. A cylindrical bar magnet is Rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then,
(a) a direct current flows in the ammeter A
(b) no current flows through the ammeter A
(c) an alternating sinusoidal current flows through the ammeter A with a time period 2π /ω
(d) a time varying non-sinusoidal current flows through the ammeter A
Solution: (b)
Key concept: The phenomenon of electromagnetic induction is used in this problem. Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes (or a moving conductor cuts the magnetic flux) an emf is produced in the circuit (or emf induces across the ends of the conductor) is called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A. Hence the ammeter shows no deflection.

Question 4. There are two coils A and B as shown in figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that
(a) there is a constant current in the clockwise direction in A.
(b) there is a varying current in A.
(c) there is no current in A.
(d) there is a constant current in the counter clockwise direction in A.

Solution: (d)
Key concept: Due to variation in the flux linked with coil B an emf will be induced in coil B.
Current in coil B becomes zero when coil A stops moving, it is possible only if the current in coil A is constant. If the current in coil A would be variable, there must be some changing flux and then there must be an induced emf. Hence an induced current will be in coil B even when coil A is not moving.

Question 5. Same as problem 4 except the coil A is made to rotate about a vertical axis (figure). No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counter-clockwise and the coil A is as shown at this instant, t = 0, is
(a) constant current clockwise.
(b) varying current clockwise.
(c) varying current counter clockwise.
(d) constant current counter clockwise.

Solution: (a)
Key concept: In this problem, the Lenz’s law is applicable so let us introduce Lenz’s law first. .
Lenz’s law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon law of conservation of energy.
When the current in coil B (at t= 0) is counter-clockwise and the coil A is considered above it. The counter clockwise flow of the current in coil B is equivalent to north pole of magnet and magnetic field lines are eliminating upward to coil A. When coil A starts rotating at t = 0, the current in A is constant along clockwise direction by Lenz’s rule.

Question 6. The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as
(a) l and A increase (b) l decreases and A increases
(c) l increases and A decreases (d) both l and A decrease
Solution: (b)
Key concept: The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.
L = μ_r μ₀ n² Al
where, n = N/l (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross section: Larger the area of cross section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases μr times if it is wound over an iron core of relative permeability μ_r.
The long solenoid of cross-sectional area A and length l, having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is L = μ_rμ₀ N² A/l
So, the self inductance L of a solenoid increases as l decreases and A increases because L is directly proportional to area and inversely proportional to length.
Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.

One or More Than One Correct Answer Type
Question 7. A metal plate is getting heated. It can be because
(a) a direct current is passing through the plate
(b) it is placed in a time varying magnetic field
(c) it is placed in a space varying magnetic field, but does not vary with time
(d) a current (either direct or alternating) is passing through the plate
Solution: (a, b, c)
Key concept: Eddy Current: When a changing magnetic flux is applied to a bulk piece of conducting material, then circulating currents called eddy currents are induced in the material. Because the resistance of the bulk conductor is usually low, eddy currents often have large magnitudes and heat up the conductor.
(1) These are circulating currents like eddies in water.
(2) Experimental concept is given by Focault, hence also named as “Focault current”.
(3) The production of eddy currents in a metallic block leads to the loss of electric energy in the form of heat.
(4) By lamination, slotting processes, the resistance path for circulation of eddy current increases, resulting into weakening them and also reducing losses causes by them.

A metal plate is getting heated when a DC or AC current is passed through the plate, known as heating effect of current. This current (called eddy current) is induced in the plate when a metal plate is subjected to a time varying magnetic field, i.e., the magnetic flux linked with the plate changes and eddy currents comes into existence which make the plate hot.

Question 8. An emf is produced in a coil, which is not connected to an external voltage source. This can be due to
(a) the coil being in a time varying magnetic field
(b) the coil moving in a time varying magnetic field
(c) the coil moving in a constant magnetic field
(d) the coil is stationary in external spatially varying magnetic field, which does not change with time
Solution: (a, b, c)
Key concept: As we know whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes, an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.

In this problem, magnetic flux linked with the isolated coil changes when the coil is placed in the region of a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field.

Question 9. The mutual inductance M₁₂ of coil 1 with respect to coil 2
(a) increases when they are brought nearer
(b) depends on the current passing through the coils
(c) increases when one of them is rotated about an axis
(d) is the same as M₂₁ of coil 2 with respect to coil 1
Solution: (a, d)
Key concept: Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.

Question 10. A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
(a) the magnetic field is constant
(b) the magnetic field is in the same plane as the circular coil and it may or may not vary
(c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably
(d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction
Solution: (b, c)
Key concept: As we know whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
The induced emf is given by rate of change of magnetic flux linked with the circuit, i.e., e = -dФ/dt
According to the problem there is no electromotive force produced in the coil. Then the various arrangement are to be thought of in such a way that the magnetic flux linked with the coil does not change even if the coil is placed and expanded in magnetic field.
When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or we can say that direction of magnetic field is perpendicular to the direction of area (increasing) so that their dot product is always zero and hence change in magnetic flux is also zero.

Or
The magnetic field has a perpendicular (to the plane of the coil) component whose-magnitude is decreasing suitably in such a way that the dot product of magnetic field and surface area of plane of coil remain constant at every instant.

Very Short Answer Type Questions
Question 11. Consider a magnet surrounded by a wire, with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.

Solution:

Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux. The induced emf is given by rate of change of magnetic flux linked with the circuit i.e, e= – dФ/dt . so flux linked will change when either magnetic field, area or the angle between B and A changes.
If the switch is closed, the circuit will complete. But to induce emf in the circuit, we need:
(i) a changing magnetic field, but the bar magnet is stationary so it is not possible in this situation.
(ii) A changing area, which is also not possible because area is also constant as coil is not expanding or compressed.
(iii) Angle between B bar and A bar changes, which is also not possible in this situation because orientation of bar magnet and coil is fixed.
Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is induced in the coil and hence no current will flow in the circuit.

Question 12. A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.
Solution:
Key concept: Lenz ‘s Law:
This law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon the law of conservation of energy.
(1) When AT-pole of a bar magnet moves towards the coil, the flux associated with the loop increases and an emf is induced in it. Since the circuit of loop is closed, induced current also flows in it.
(2) Cause of this induced current, is approach of north pole and therefore to oppose the cause, i.e., to repel the approaching north pole, the induced current in loop is in such a direction so that the front face of loop behaves as north pole. Therefore induced current as seen by observer O is in anticlockwise direction (figure).

According to the given situation as the coil is stretched so that there are gaps between successive elements of the spiral coil, i.e., the wires are pulled apart which lead to the flux leak through the gaps. According to Lenz’s law, the emf induced in these spirals must oppose this decrease in magnetic flux, which can be done by an increase in current. So, the current will increase.

Question 13. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.
Solution: This problem is based on Lenz’s law and according to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it.

When the iron core is inserted in the current carrying solenoid, the magnetic field increases due to the magnetisation of iron core and hence the flux increases.
So, the emf induced in the coil must oppose this increase in flux, so the current induced in the coil in such a direction that it will oppose the increasing magnetic field which can be done by making decrease in current. So, the current will decrease.

Question 14. Consider a metal ring kept oft the top of a fixed solenoid (say on a cardboard) (figure). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain.

Solution: This problem is based on Lenz’s law and according to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it.
Initially there is no flux linked with the ring or we can say that initially flux through the ring is zero. When the switch is closed current start flowing in the circuit, magnetic flux is linked through the ring. Thus increase in flux takes place. According to Lenz’s law, this increase will be resisted and this can happen if the ring moves away from the solenoid.

This happen because the flux increases will cause an anticlockwise current (as seen from the top in the ring in figure.), i.e., opposite direction to that in the solenoid.
This makes the same sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming same magnetic pole in front of each other. Hence, they will repel each other and the ring will move upward.

Question 15. Consider a metal ring kept (supported by a cardboard) on the top of a fixed solenoid carrying a current I (see figure of Question 14). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?
Solution: This problem is based on Lenz’s law and according to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it.

When the switch is opened, current in the circuit of solenoid stops flowing. Initially there is some magnetic flux linked with the solenoid and now if current in the circuit stops, the magnetic flux falls to zero or we can say that magnetic flux linked through the ring decreases. According to Lenz’s law, this decrease in flux will be opposed and the ring experiences downward force towards the solenoid.
This happen because the current i decrease will cause a clockwise current (as seen from the top in the ring in figure) to increase the decreasing flux. This can be done if the direction of induced magnetic field is same as that of solenoid. This makes the opposite sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming opposite magnetic pole in front of each other.
Hence, they will -attract each other but as ring is placed at the cardboard it could not be able to move downward.

Question 16. Consider a metallic pipe with ah inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, it takes more time to come down than it takes for a similar un-magnetised cylindrical iron bar dropped through the metallic pipe. Explain.
Solution: Key concept: Lem’s Law.
This law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon the law of conservation of energy.
Eddy Current. When a changing magnetic flux is applied to a bulk piece of conducting material, then circulating currents called eddy currents are induced in the material. Because the resistance of the bulk conductor is usually low, eddy currents often have large magnitudes and heat up the conductor.
When a cylindrical bar magnet is dropped through the metallic pipe flux linked with the cylinder changes and consequently eddy currents are produced in the metallic pipe. According to Lenz’s law, these currents will oppose the motion of the magnet, which is the cause of induction.

Therefore, magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an un-magnetised iron bar will not produce eddy currents and will fall with acceleration due to gravity g.
Thus, the magnet will take more time to come down than it takes for a similar un-magnetised cylindrical iron bar dropped through the metallic pipe, so, magnetised magnet takes more time.

Short Answer Type Questions
Question 17. A magnetic field in a certain region is given by B = B₀ cos (ωt) k and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at (a, 0, 0) at

Solution:
Key concept:
(1) First law: When ever the number of magnetic lines of force(magnetic flux) passing through a circuit changes, an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.

Question 18. Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula Ф = B₁dA₁, B₂dA₂ …. Now, if we choose two different surfaces S₁ and S₂ having C as their edge, would we get the same answer for flux. Justify your answer.

Solution: We would get the same answer for magnetic flux. Let us discuss its reason in detail. The magnetic flux linked with the surface can be considered as the number of magnetic field lines passing through the surface. So, let dФ = B.dA represents magnetic lines in an area A to B.
Magnetic field cannot end or start in space, this property of magnetic field lines based upon the concept of continuity. Therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂. Therefore, in both the cases we get the same magnetic flux.

Important point: Magnetic field lines can neither be originated nor be destroyed in space. This property is based on the concept of continuity.

Question 19. Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

Solution:
Key concept: This problem is based upon the motional emf.
Consider a conducting rod of length l moving with a uniform velocity v perpendicular to a uniform magnetic field B bar, directed into the plane of the paper. Let the rod be moving to the right as shown in figure. The conducting electrons also move to the right as they are trapped within the rod.

Question 20. A (current versus time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (u) a maximum. If the back emf at t = 3 s is e, find the back emf at t = 7 s, 15 s and 40 s. OA, AB and BC are straight line segments.

Solution:
Key concept: Whenever the electric current passing through a coil or circuit changes, the magnetic flux linked with it will also change. As a result of this, in accordance with’Faraday’s laws of electromagnetic induction, an emf is induced in the coil or the circuit which opposes the change that causes this induced emf is called back emf, the current so produced in the coil is called induced current. The induced emf is given by

Question 21. There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 10^-2 Wb passes through B (no current through B). If no current passes through A and a current of 1A passes through B, what is the flux through A ?
Solution:

Long Answer Type Questions
Question 22. A magnetic field B = B₀ sin (ωt)k covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

Solution: Key concept: In this problem the emf induced across AB is motional emf due to its motion, and emf induced by change in magnetic flux linked with the loop change due to change of magnetic field.

Question 23. A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B = B(t) k

(i) Write down an equation for the acceleration of the wire XY
(ii) If B is independent of time, obtain v(t), assuming v(0) = u₀
(iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in.
Solution: First we have to analyse the situation as shown in the figure. Let the parallel wires are at y = 0 and y = L and are placed along x-axis. Wire XY is along y-axis.
At t = 0, wire AB starts from x = 0 and moves with a velocity v. Let at time t, wire is at x(t) = vt.
(where, x(t) is the displacement as a function of time).
Let us redraw the diagram as shown below.

Question 24. ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity ω (figure). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor A’BDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

Solution:

Question 25. Consider an infinitely long wire carrying a current I(t), with dI/dt=λ =constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is R (figure).

Solution: To approach these types of problems integration is very useful to find the total magnetic flux linked with the loop.
Let us first consider an elementary strip of length l and width dr at a distance r from an infinite long current carrying wire. The magnetic field at strip due to current carrying wire is given by

Question 26. A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t) = I₀(I – t/T) for 0 ≤ t ≤ T and I(0) = 0 for t >T(figure). Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R.

Solution: To find the charge that passes through the circuit first we have to find the relation between instantaneous current and instantaneous magnetic flux linked with it. The emf induced can be obtained by differentiating the expression of magnetic flux linked w.r.t. t and then applying Ohm’s law, we get

Question 27. A magnetic field B is confined to a region r ≤ a and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b> a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time Δt. Find the angular velocity ω of the ring after the field vanishes.
Solution:
Key concept: According to the law of EMI, when magnetic field changes in the circuit, then magnetic flux linked with the circuit also changes and this changing magnetic flux leads to an induced emf in the circuit. Here, magnetic field decreases which causes induced emf and hence, electric field around the ring. The torque experienced by the ring produces change in angular momentum.
As the magnetic field is brought to zero in time At, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring as discussed above. The induces emf causes the induced electric field E around the ring.

Question 28. A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Solution:
Let us first divide the magnetic field in the components one is along the inclined plane = B sin θ and other component of magnetic field is perpendicular the plane =B cos θ
Now, the conductor moves with speed v perpendicular to B cos θ, component of magnetic field. This causes motional emf across two ends of rod, which is

Question 29. Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Solution: This is the similar problem as we discussed above. Here, a conductor of length d moves with speed v, perpendicular to the magnetic field B as shown in figure. Due to this a motional emf is induced across two ends of rod (e = vBd). Since, switch S is closed at time t = 0, capacitor is charged by this potential difference. Let Q(t) be the charge on the capacitor and current flows from A to B.
Now, the induced current

Question 30. Find the current in the sliding cod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Solution: This is the similar problem as we discussed above. Here, a conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. Due to this an motional emf is induced across two ends of rod (e = vBd). Since, switch S is closed at time t = 0. current start growing in inductor by the potential difference due to motional emf.

Question 31. A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is B_z = B₀ (1+ λz). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, λ and acceleration due to gravity g.
Solution: In this problem a relation is established between induced current, power lost and velocity acquired by freely falling ring.
The magnetic flux linked with the metallic ring of mass m and radius l ring being horizontal falling under gravity in a region having a magnetic field whose z-component of magnetic field is B_z = B₀(1 + λz) is

Question 32. A long solenoid S has n turns per metre, with radius a. At the centre of this coil, we place a smaller coil of N turns and radius b (where b < a). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil. Plot a graph showing nature of variation in emf, if current varies as a function of mt² + C.
Solution:
Key concept: Magnetic field due to a solenoid is given by, B = μ₀ni where signs are as usual.
In this problem the current is varying with time. Due to this an emf is induced in the coil of radius b.