Class 12 NEB Physics Question Solution | 2079 (2022) Question Paper Solution | Physics

NEB Grade XII 2079 (2022) Physics Question Solution

New Course

Group 'A'

Rewrite the correct options of each question in your answer sheet.

1. If L represents momentum, I represents moment of inertia, then \(\frac{L^2}{2I}\) represents
1. Rotational kinetic energy
2. Torque
3. Power
4. Potential energy
Solution: Option (a)
Explanation: Since, the rotational kinetic energy is commonly expressed as, \[\text{K.E.}=\frac{1}{2}mv^2\] Also, \[L=I\omega\] Then, \[\begin{align*} \text{K.E.}&=\frac{1}{2} I \times \frac{L^2}{I^2}\\ &=\frac{L^2}{2I}\\ \end{align*}\]


2. Starting from mean position, a particle in simple harmonic motion takes time \(T_1\) and \(T_2\) to cover first half and next half displacement in moving from mean to extreme position, then
1. \(T_1\)=\(T_2\)
2. \(T_2=2T_1\)
3. \(T_1=2T_2\)
4. \(T_1 \gt T_2\)
Solution: Option (b)
Explanation: Let's chit chat on this question, Ok! T is the time period. If A is the amplitude of the vibrating particle in a simple harmonic motion, \(T_1\) is the time taken by the particle to cover first half displacement i.e., 0 to \(\frac{A}{2}\) and \(T_2\) is the time taken by the particle to cover next half displacement i.e., \(\frac{A}{2}\) to A.
The displacement equation of the simple harmonic equation is, \[y=Asin\omega t\] For the first half displacement, \[\begin{align*} y&=A sin\omega T_1\\ \frac{A}{2}&=A sin\omega T_1\\ \frac{1}{2}&=sin\omega T_1\\ sin\frac{\pi}{6}&=sin\omega T_1\\ \omega T_1&=\frac{\pi}{6}\\ \frac{2\pi}{T} T_1&=\frac{\pi}{6}\\ T_1&=\frac{T}{12}\\ \end{align*}\] For the displacement from the mean position to the extreme position i.e., 0 to A (T' be the half the time period)
\[\begin{align*} y&=A sin\omega T'\\ A&=A sin\omega T'\\ 1&=sin\omega T'\\ T'&=\frac{T}{4}\\ \end{align*}\] So, for the next half displacement, we can write the time as, \[T_2=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}\] So the relation of \(T_1\) and \(T_2\) is thus \[T_2=2T_1\].


3. Water is flowing at 12 m/s in a horizontal pipe. If the pipe widens to twice its original diameter, the flow speed in the wider section is
1. 6 m/s
2. 9 m/s
3. 2 m/s
4. 3 m/s
Solution: Option (d)
Explanation: Use equation of continuity \(av=\text{constant}\) to solve this question.
Hints! Let, initial velocity (v₁)= 12 m/s, initial diameter (d₁)= d (let), final diameter (d₂)= 2d, final velocity (v₂)= ? Do the rest.



4. An ideal gas of ratio of heat capacities = 5/3 at 72\(^\circ\) C is expanded adiabatically to eight times to its original volume. Approximate rise in temperature of the gas is
1. 86 K
2. 186 K
3. 259 K
4. 273 K
This question does not seem to be correct. In adiabatic expansion, the gas cools since the gas does work in expense of its own internal energy. However, the question here asks the rise in temperature, which seems to be incorrect.


5. A carnot cycle includes
1. Two isothermal and two adiabatic processes
2. Two isothermal and two isobaric processes
3. Two isothermal processes
4. Two adiabatic processes
Solution: Option (a)


6. The intensity of sound 'I' and amplitude of vibration 'a' are related as,
1. I\(\propto\) a
2. I\(\propto\) a\(^2\)
3. I \(\propto\) \(\frac{1}{a}\)
4. I \(\propto\) \(\frac{1}{a^2}\)
Solution: Option (b)
Explanation: Don't get confused here! Intensity of sound is directly proportional to the square of its amplitude but inversely proportional to the square of the distance from the source of a sound.


7. A diffraction grating has 400 lines/mm and is illuminated by a monochromatic light of wavelength 6000\(A^0\). The maximum number of diffraction maxima obtained will be
1. 4.16
2. 4
3. 5
4. 2.96
Solution: Option (b)
Explanation: We know, \[\begin{align*} a+b&=\frac{1}{N}\\ &=\frac{1}{400\text{lines}/\text{mm}}\\ &=\frac{1 \text{mm}}{400}\\ &=2.5\times 10^{-6} \hspace{0.1cm} \text{m}\\ \end{align*}\] wavelength (\(\lambda\))= 6000\(\times\)10\(^{-10}\) m
maximum number of diffraction maxima, n=?
For the diffraction maxima, \[(a+b)sin\theta=n\lambda\] For the maximum number of diffraction maxima, \(\theta\) should be 90\(^\circ\) for the value of sin\(\theta\) to be maximum i.e., 1. Then,
\[\begin{align*} (a+b)&=n\lambda\\ 2.5 \times 10^{-6} &=n \times 6000\times 10^{-10}\\ n&=4.167\\ \therefore n&\approx 4\\ \end{align*}\] Please do not get confused here. 'n' should always be an integer. Do not rush in to tick the option (a): it is not an integer.


8. Above curie temperature
1. A ferromagnetic substance becomes paramagnetic
2. A ferromagnetic substance becomes diamagnetic
3. A paramagnetic substance becomes diamagnetic
4. A paramagnetic substance becomes ferromagnetic
Solution: Option (a)


9. If the emf of a thermocouple, one junction of which is kept at 0\(^\circ\) C is given by \(E=a\theta + \frac{1}{2} b\theta^2\). Then, the neutral temperature will be
1. \(\frac{a}{b}\)
2. \(\frac{-a}{b}\)
3. \(\frac{1}{ab}\)
4. \(\frac{-1}{ab}\)
Solution: Option (b)
Explanation: At neutral temperature \(\theta_n\): \[\begin{align*} \left(\frac{dE}{d\theta}\right)_{\theta=\theta_n}&=0\\ \left(\frac{d(a\theta+\frac{1}{2}\theta^2)}{d\theta}\right)_{\theta=\theta_n}&=0\\ a+\frac{1}{2} b\theta_n&=0\\ \therefore \theta_n&=-\frac{a}{b}\\ \end{align*}\]


10. If power factor of an ac circuit is 0.5, then the phase difference between voltage and current in the circuit is
1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{2}\)
4. \(\pi\)
Solution: Option (a)
Explanation: The power factor is given by, \[\begin{align*} cos\phi&=0.5\\ \phi&=cos^{-1}(0.5)\\ \therefore \phi&=\frac{\pi}{3}\\ \end{align*}\]


11. In photoelectric cell, the relation between cut-off voltage (\(V_0\)) and frequency (f) of incident light is represented by

Solution: Option (d)
Explanation: Since, the photoelectric effect is observed only when the frequency of the incident light exceeds the threshold frequency. Here, the threshold frequency is given by the horizontal intercept (i.e., x-intercept) in the graph.



Group 'B'

1. What do you mean by moment of inertia? [1]
The property of a body to resist the change in its rotational state is called the moment of inertia. Mathematically, the moment of inertia of the rigid body is, \[I=\sum_{i=1}^{n} m_i r_i^2\] where, \(m_i\) is the mass of the \(i^{th}\) particle of rigid body, and \(r_i\) is the perpendicular distance of that particle from the axis of rotation.


2. State principle of conservation of angular momentum. [1]
If no external torque (τ) acts on a system, the angular momentum of the system remains constant, no matter what changes takes place within the system.
i.e., if for \(\tau\) = 0, \(\frac{dL}{dt} = 0 \implies\) L=constant.
L = Iω = constant where, I is the moment of inertia of a body and \(\omega\) is the angular velocity of the rotating body.


3. A disc of moment of inertia 5\(\times\)10\(^{-4}\) kgm\(^2\) is rotating freely about the axis through the center at 40 rpm. Calculate the new revolution per minute if some wax of mass 0.02 kg is dropped gently on to the disc 0.08 m from the axis. [3]
Solution:
moment of inertia about center of disc, \(I\) = 5\(\times\)10\(^{-4}\) kgm\(^2\)
initial frequency of rotation of disc, \(f_1\) = 40 rpm = \(\frac{40}{60}\) rps = \(\frac{2}{3}\) rps
final frequency of rotation of disc, \(f_2\) = ?
mass of wax dropped in the disc, \(m\) = 0.02 kg
distance of wax from the axis of rotation, \(r\) = 0.08 m
The moment of inertia of the system constituting the disc and the wax is, \[I'=I+mr^2\] From the conservation of the angular momentum, \[\begin{align*} I \omega_1&=I'\omega_2\\ If_1&I'f_2\\ I \times \frac{2}{3}&=(I+mr^2) f_2\\ 5\times 10^{-4} \times \frac{2}{3}&=(5\times 10^{-4} + 0.02 \times (0.08)^2) f_2\\ f_2&=\frac{5\times 10^{-4} \times \frac{2}{3}}{(5\times 10^{-4} + 0.02 \times (0.08)^2)}\\ &=0.53 \hspace{0.1cm}\text{rps}\\ \therefore f_2&=31.80 \hspace{0.1cm}\text{rpm} \end{align*}\] Thus, the new revolution per minute is 31.80 rpm. Dear all! Here, I could have used rpm without changing it to rps to obtain the final result. But, I do have habit of doing calculations in SI unit. Do as per your comfort.
OR
1. What is meant by simple harmonic motion? [1]
If a particle moves to and fro about a mean position in a straight line such that the acceleration is directed towards the mean position and is directly proportional to the displacement from that position is called simple harmonic motion .
If y is the displacement produced and a be the accleration of a body in a simple harmonic motion then, \[a=-ky\] Here, k is a constant .


2. Show that motion of a simple pendulum is simple harmonic and hence calculate its time period. [3]

A simple pendulum is a heavy point mass object suspended by an inextensible, weightless and flexible string from a rigid support which is free to oscillate in a vertical plane.



Figure: Simple pendulum
Consider a simple pendulum as in fig. in which a bob of mass m is suspended from the fixed support O through a light string of length l . The system can stay in equilibrium if the string is vertical. This is the mean position. If the bob is pulled aside and released, it oscillates in a circular arc with the centre at the point of suspension O.
The position of the bob at any time can be described by the angle \(\theta\) between the string and the vertical. Let us see whether the motion of simple pendulum is simple harmonic or not and find out its time period of oscillation.
When the bob is displaced from its mean position, it oscillates along the path CAB in the vertical plane and at an instant, let the angular displacement be \(\theta\). The force acting on the bob at B are,

• Weight mg of the bob acting vertically downward,
• Tension T in the string along its length towards the point of suspension.
mg resolves into two components : \(mgcos\theta\) and \(mgsin\theta\) as in figure. \(mgcos\theta\) balances the tension in the string and the component \(mgsin\theta\) acts as restoring force and produce acceleration in the bob in the direction of the arrow.
\[\therefore F=-mgsin\theta ... (i) \] Here, negative sign is due to the fact that the force is restoring in nature; it acts opposite to the displacement of the body.
If a be the acceleration of the bob, then \[F=ma=-mgsin\theta\] From this, \[a=-gsin\theta ... (ii)\] For small angular displacement \(\theta\), \(sin\theta \approx \theta=\frac{y}{l}\) where y is the displacement of bob from the mean position.
With this, equation (ii) becomes,
\[a=-g\left(\frac{y}{l}\right)=-\left(\frac{g}{l}\right) y ... (iii)\] For a given pendulum at a given place \(\frac{g}{l}\) is a constant and thus \(a\propto y\). Since, acceleration is directly proportional to the displacement from the mean position, it is the characteristic of SHM. So, simple pendulum executes simple harmonic motion .

Time period of a simple pendulum

Comparing equation (iii) with the standard equation i.e., \(a=-{\omega}^2y\), we get, \[-{\omega}^2=-\frac{g}{l}\] So, \[\omega=\frac{g}{l}\] So, the period of oscillation , \(T=\frac{2\pi}{\omega}\), so \[\begin{align*} \frac{2\pi}{T}&=\sqrt{\frac{g}{l}}\\ T&=2\pi \sqrt{\frac{l}{g}} ... (iv) \end{align*} \]
3. On what factors does the time period of simple pendulum depend? [1]
As can be seen from equation (iv), the time period of the simple pendulum depends on the following:
1. effective length of the pendulum
2. acceleration due to gravity of a place
Interestingly, the time period of simple pendulum does not depend on mass of the bob of the pendulum. 😎



1. State Stoke's law. [1]
Stoke's law states that the viscous force F , acting on a spherical body of radius r moving with terminal velocity v in a fluid of coefficient of viscosity η is given by, \[F=6\pi \eta r v\]
2. Describe a method to determine terminal velocity of a spherical body falling through a viscous liquid using Stoke's law. [2]

Figure: Experimental arrangement for the determination of viscosity of a liquid


Fill a tall glass vessel with the liquid and drop a small steel ball P gently into the liquid so that it falls through the liquid as in figure. Towards the middle of the liquid P reaches a constant velocity i.e., terminal velocity v , which is measured by timing its fall through a distance AB or BC.
The downward force acting on the body is weight W and the upward forces are upthrust U and viscous force F . When P attains the terminal velocity, \[F+U=W ... (i)\] From stoke's law, \[F=6\pi \eta r v \] And, \[\begin{align*} W&=mg\\ &=\rho V g\\ &=\rho \frac{4}{3} \pi r^3 g\\ &=\frac{4}{3} \pi r^3 \rho g \end{align*}\] Here, V is the volume of P ,\(\rho\) is the density of P and r is its radius.

Also, \[\begin{align*} U&=m_f g\\ &=\sigma V g\\ &=\sigma \frac{4}{3}\pi r^3 g\\ &=\frac{4}{3}\pi r^3 \sigma g \end{align*}\] Here, \(m_f\) is the mass of the fluid displaced and \(\sigma\) is the density of the liquid.

From all these expressions, equation (i) becomes, \[\begin{align*} 6\pi \eta r v + \frac{4}{3}\pi r^3 \sigma g&=\frac{4}{3} \pi r^3 \rho g\\ 6\pi \eta r v&=\frac{4}{3} \pi r^3 \rho g-\frac{4}{3}\pi r^3 \sigma g\\ &=\frac{4}{3}\pi r^3 (\rho-\sigma)g\\ \eta&=\frac{2r^2(\rho-\sigma)g}{9v} \end{align*}\] In this way, the coefficient of viscosity of the liquid or fluid is determined.




3. Two spherical raindrops of equal size are falling through air with terminal velocity 10 cm/s. If these two drops were to coalesce to form a single drop, what would be the new terminal velocity? [2]

Figure: Force diagram for the small drops and a bigger drop
Let, \(v\) be the terminal velocity of the small sized drops. \(V\) be the terminal velocity of the bigger drop formed as a result of coalescion of smaller drops. \(r\) be the radius of the small drops, and \(R\) be the radius of the bigger drop.
Since, \[\begin{align*} \text{ volume of bigger drop}& = 2 × \text{volume of smaller drop}\\ \frac{4}{3} \pi R^3&=2 \times \frac{4}{3} \pi r^3\\ R^3&=2r^3\\ R&=2^{\frac{1}{3}}\\ \end{align*}\] For smaller drop, at terminal velocity, \[\begin{align*} F&=W\\ 6\pi \eta r v&=mg\\ 6\pi eta r v&=\frac{4}{3} \pi r^3 \rho g \hspace{0.1cm} \text{... (i)}\\ \end{align*}\] Similarly for bigger drop, \[\begin{align*} 6\pi R V&=\frac{4}{3} \pi R^3 \rho g \hspace{0.1cm} \text{... (ii)}\\ \end{align*}\] Dividing equation (ii) by equation (i), \[\begin{align*} \frac{6\pi \eta R V}{6\pi \eta r v}&=\frac{R^3}{r^3}\\ \frac{RV}{rv}&=\frac{R^3}{r^3}\\ \frac{V}{v}&=\frac{R^2}{r^2}\\ V&=\left(\frac{R}{r}\right)^2 \times v\\ &=\left(\frac{2^{\frac{1}{3}}r}{r}\right)^2 \times v\\ &=2^{\frac{2}{3}}\times 0.1\\ \therefore V&=0.16 \hspace{0.1cm} \text{m/s}\\ \end{align*}\] So, the required terminal velocity is 16 cm/s.


1. What happens to the energy added to an ideal gas when it is heated at (i) constant volume (ii) constant pressure? [2]
From the first law of thermodynamics, \[dQ=dU+pdV \hspace{0.1cm} \text{...(i)}\] where, dQ is the energy added/released to/from the system, dU is the change in internal energy of the system, dV is the change in the volume of the system of gas, and P is the pressure of the gas.
At constant volume, \(dV=0\). Thus, equation (i) reduces to, \[dQ=dU\] So, as can be logically deduced from this equation, it can be seen that the energy added to the gas is spent on increasing the internal energy of the ideal gas.
At constant pressure, \(dP=0\), the equation (i) is unchanged. So, in this case, the energy added to the gas is spent in increasing the internal energy of the gas and also expand the gas (i.e., increase its volume).


2. A carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle? [3]
temperature of the source, T₁ = 600 K
temperature of the sink, T₂ = 300 K
output work, W = 800 J
amount of heat energy supplied to the engine from source, Q₁ = ?
We have, \[\begin{align*} \frac{Q_1}{Q_2}&=\frac{T_1}{T_2}\\ &=\frac{600}{300}\\ Q_1&=2Q_2 \hspace{0.1cm} \text{... (i)}\\ \end{align*}\] Also, \[\begin{align*} W&=Q_1-Q_2\\ 800&=2Q_2 -Q_2\\ 800&=Q_2\\ \end{align*}\] Thus, from equation (i), \[\begin{align*} Q_1&=2 \times 800\\ \therefore Q_1&=1600 \hspace{0.1cm} \text{J}\\ \end{align*}\] So, the required amount of heat energy is 1600 J.


1. What is stationary waves? [1]
The waves formed due to the superposition of two progressive waves of equal frequency and amplitude travelling in opposite directions is called stationary waves. Stationary waves are also called 'standing waves' as they do not transfer the energy in the medium.

2. How stationary waves are formed in an open organ pipe? [1]
Stationary waves are formed in an open organ pipe as a result of superposition of incident waves from one open end and the reflected waves from the other open end. In open end, the antinodes are formed, and several nodes and antinodes are formed within the pipe in different harmonics.

3. Sketch the wave pattern and calculate the frequencies of fundamental vibration and the first overtone for the open organ pipe of length 'L'. [3]


Figure: Modes of vibration in open organ pipe
Let v be the velocity of sound wave of wavelength λ. If L be the length of the open organ pipe, \[L=\frac{\lambda}{2} \hspace{0.1cm}\text{...(i)}\] If f₁ be the fundamental frequency of the vibration, \[f_1=\frac{v}{\lambda}=\frac{v}{2L} \hspace{0.1cm}\text{...(ii)}\] If f₂ be the frequency of the first overtone, \[\begin{align*} f_2&=\frac{v}{\lambda}\\ &=\frac{v}{L}\\ &=2\times \frac{v}{2L}\\ \therefore f_2&=2f_1\\ \end{align*}\]
OR
1. What do you mean by Doppler's effect? [1]
The apparent alteration in frequency due to the relative motion of the source of sound and observer is known as the Doppler effect. For example, the horn of a bus or a car appears to increase in pitch as it approaches a stationary observer; as the moving object passes the observer, the pitch changes and becomes lower.

2. How is it that one can recognize a friend from his voice without seeing him? [2]
One can recognize a friend from his voice without seeing him due to the quality of his sound; one of the three fundamental characteristics of the sound. Quality of a sound depends upon waveform of the sound. It enables us to distinguish between two sounds of the same pitch and loudness produced by two different sources.

3. Write down the factors on which the speed of sound in air depends. [2]
The factors on which the speed of the sound in air depends are:
1. temperature
2. density of the medium
3. moisture or humidity
4. wind
1. What is the principle of potentiometer? [1]
Provided the uniform cross-section of the potentiometer wire, the principle of potentiometer is that the potential drop across the potentiometer wire is directly proportional to the length of the potentiometer wire.i.e., \[V\propto l\] where, V is the voltage drop across the wire, and l is the length of the potentiometer wire.

2. Why is the potentiometer preferred to a voltmeter to measure the emf of a cell? [1]
In a voltmeter, a resistor having a high resistance R is connected in series with the galvanometer having resistance G. When the voltmeter is used in a circuit, its resistance \[R_v=R+G\] is connected in parallel to some element in the circuit. This changes the overall current in the circuit and hence affects the potential difference across that element. Due to this reason, though having large resistance, some current is drawn by the voltmeter. However, potentiometer draws no current through the circuit and still measures the potential difference. So, potentiometer is preferred over voltmeter to measure the given emf of a cell.

3. The emf of a battery is balanced by a length 75 cm on a potentiometer wire and emf of a standard cell of 1.02 volt is balanced by a length 50 cm of it. Calculate the emf of the battery. [3]
Solution: For unknown emf,
emf, \(E_1\) (let)
balancing length, \(l_1\) = 75 cm = 0.75 m
For standard cell,
emf, \(E_2\) = 1.02 V
balancing length, \(l_2\) = 50 cm =0.50 m
We know, \[\begin{align*} \frac{E_1}{E_2}&=\frac{l_1}{l_2}\\ \frac{E_1}{1.02}&=\frac{0.75}{0.50}\\ E_1&=\frac{\frac{0.75}{0.50}}{1.02}\\ \therefore E_1&=1.47 \hspace{0.1cm} \text{V}\\ \end{align*}\] So, the required emf of the battery is 1.47 V. Hold on! Most of you may feel as if this answer may not be sufficient for 3 marks. Don't worry, I used to think the same in my school days. Don't waste your time in deriving the relation used in this solution. This is sufficient. Believe me. 😇
1. Define magnetic flux. [1]

Figure: Magnetic flux through the given surface
Magnetic flux (\(\phi\)) through any surface is defined as the number of magnetic lines of force crossing through the surface. The magnetic flux is given by, \[\phi=BA cos\theta\] where, B is the magnetic field intensity, A is the area of the surface and θ is the angle between normal to the surface (i.e., area vector) and B.

2. State Faraday's laws of electromagnetic induction. [1]
Faraday's law of electromagnetic induction can be stated in following two points:
1. Faraday's first law: When the magnetic flux linking in a conductor or a coil changes, an emf is induced in it.
2. Faraday's second law: The magnitude of induced emf in a conductor or a coil is directly proportional to the rate of change of flux linkages. \[\mathcal{E}=N\frac{d\phi}{dt}\] where, N is the number of turns of the coil.
3. A coil has 5000 turns. What is the emf produced in the coil when the magnetic flux cutting the coil changes by 8\(\times\)10\(^{-4}\) Wb in 0.1 s? [1]
Solution: number of turns of the coil, N = 5000 turns
\(\mathcal{E}\) = ?
change in magnetic flux, \(d\phi\) = 8\(\times\)10\(^{-4}\) Wb
time interval, \(dt\) = 0.1 s
From Faraday's law of electromagnetic induction, \[\begin{align*} \mathcal{E}&=N\frac{d\phi}{dt}\\ &=5000 \times \frac{8 \times 10^{-4}}{0.1}\\ \therefore \mathcal{E}&=40 \hspace{0.1cm} \text{V}\\ \end{align*}\] So, the emf produced in the coil is 40 V.

4. A light aluminium ring is suspended from a long thread as shown in figure. When a strong magnet is moved away from it, the ring follows the magnet, why? What would happen if the magnet were moved towards the ring? [2]

Lenz's law states that the direction of induced current is such that it opposes the cause which produces it. When the strong magnet is moved away from the thread, the current is induced in clockwise direction and thus induce the magnetic field with the coil acting as a magnetic dipole with its south pole towards the north pole of the source magnet. Hence, in this case, the ring is attracted towards the magnet. However, when the magnet is moved towards the ring, the current is induced in a anticlockwise direction and thus the magnetic field is induced in the coil with the coil acting as the magnetic dipole with its north end towards the north pole of the source magnet. Hence, in this case, the ring will be repelled.

1. Why is X-rays radiation process called inverse photoelectric effect? [2]
In photoelectric effect, the electrons are ejected from the metal surface when the radiation of sufficient frequency strikes on it. However, in X-ray production, the electrons are bombarded to the metal target to produce the X-ry radiation. Hence, X-rays radiation process is called the inverse photoelectric effect.

2. State and explain Bragg's law of X-ray diffraction. [3]

Figure: Bragg's diffraction (This image is adapted from

Britannica

accessed on June 12, 2022.)
Bragg's law states that when the X-ray is incident on a crystal surface, its angle of incidence θ will reflect back with the same angle of scattering, θ. And, when the path difference between the reflected waves is equal to the integral multiple of wavelength of X-ray, the constructive interference or the maximum intensity is obtained.
Consider a set of parallel atomic planes separated by a distance 'd' as in figure. A narrow monochromatic X-ray beam of wavelength λ is impinged (striked) on the crystal lattice at a glancing angle θ. The reflected (glancing angle) θ, as shown by the experiment, is equal to the incident angle θ. Consider waves 1 and 2 which are reflected by two atoms A and B as in figure. The path difference between these two waves is, \[\begin{align*} CBD&=CB+BD\\ &=dsin\theta +dsin\theta\\ &=2d sin\theta\\ \end{align*}\] For maximum intensity, the path difference between the reflected waves should be integral multiple of λ. \[2dsin\theta=n\lambda\] where n is an integer i.e., n = 1,2,3,...
This equation is known as Bragg's law of X-ray diffraction. 'n' gives the order of the scattered/reflected beam.
(Better to include!) The Bragg's law is useful for measuring wavelengths and for determining the lattice spacings of crystal.

1. What is pn junction diode? [1]
When a semiconducting material such as silicon or germanium is doped with impurity in such a way that one side has a larger number of acceptor impurities (i.e., p-type) and the other side has a larger number of donor impurities (i.e., n-type), a p-n junction is obtained. The device constructed in such a way is called a p-n junction diode.

2. Describe junction diode as a full wave rectifier. [3]

Figure: Centre tapped full-wave rectifier
A full wave rectifier is a device which converts full cycle of alternating current input into the direct current output. The centre tapped full-wave rectifier is explained along with the figure below.
The input a.c. is applied across the primary of transformer. The secondary coils of the transformer is connected to the n-type end of both the diodes D₁ and D₂. The central tapping of the secondary is connected to the load resistance R_L.
During the positive half-cycle of secondary voltage, one end of the secondary, A, becomes positive and the other end, B, becomes negative. So, the diode D₁ is forward biased and diode D₂ is reverse biased. Thus, the current flows through the load resistor through the path AD₁DR_LCA as the diode D₂ does not conduct.
During the negative half cycle of secondary voltage, end A becomes negative and end B becomes positive. D₂ thus conducts(i.e., forward biased) but D₁ does not conduct. Thus, the current flows through R_L through the path BD₂DR_LCB.
In both half cycles, the unidirectional current flows through the load resistor R_L. Thus, the combination of D₁ and D₂ works as a full-wave rectifier. The input and output voltage of centre tapped full-wave rectifier are shown in figures. The input and output voltage of the centre tapped full-wave rectifier is shown below in figure (a) and (b) respectively.


Figure: (a) Input voltage (b) Output voltage

3. Write truth table related to AND gate. [1]
The truth table of AND gate is,

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

Group 'C'

20. The diagram represents the experimental arrangement used to produce interference fringes in Young's double slits experiment.

1. What are coherent sources of light? [1]
Coherent sources of light are those sources of light that emit continuous light waves of the same wavelength, same frequency, and are in the same phase or have a constant phase difference. Coherent sources of light waves is required for interference of light.

2. What do you mean by interference of light? [1]
The phenomenon of non-uniform distribution of light energy in a medium due to superposition of light waves from two coherent sources is called interference of light.

3. In the above experiment, if the slits S₁ and S₂ are illuminated by a monochromatic source of light of wavelength λ, show that teh width of bright fringe is equal to width is equal to width of dark fringe as given by \(\beta=\frac{\lambda D}{d}\). [4]

Figure: Young's double slit experiment
The experimental arrangement of Young's double slit experiment is as in figure (in simplified form). P be the arbitrary point on the screen at a distance x from the point M. Path difference between the waves reaching point P from S₁ and S₂ is, \[\text{Path difference}=S_2 P - S_1 P\] From the geometry in figure, \[PQ=x-\frac{d}{2}\] and, \[PN=x+\frac{d}{2}\] Also using the pythagoras relation, we can deduce the expression of \((S_2P)^2-(S_1P)^2\) as follows, \[\begin{align*} S_2P^2 - S_1P^2 &=\left[D^2 + \left(x+\frac{d}{2}\right)^2\right]-\left[D^2 + \left(x-\frac{d}{2}\right)^2\right]\\ &=D^2+\left(x+\frac{d}{2}\right)^2-D^2+\left(x+\frac{d}{2}\right)^2\\ &=2xd\\ (S_2P-S_1P)(S_2P+S_1P)&=2xd \hspace{0.1cm} \text{...(i)}\\ \end{align*}\] In practice, P lies very close to M, \[S_2P \approx S_1P \approx D\] So, equation (i) reduces to, \[\begin{align*} (S_2P-S_1P)\times 2D&=2xd\\ S_P-S_1P&=\frac{xd}{D}\\ \end{align*}\] For constructive interference (i.e., bright fringe):
Path difference is the integral multiple of wavelength of light for the constructive interference.
\[\begin{align*} \frac{xd}{D}&=n\lambda\\ x&=\frac{n\lambda D}{d} \hspace{0.1cm} \text{...(ii)}\\ \end{align*}\] where n = 0,1,2,3,...
For n = 0 (central bright fringe),
\[x_0=0\] For n = 1 (first bright fringe),
\[x_1=\frac{\lambda D}{d}\] For n = 2 (second bright fringe),
\[x_2 = \frac{2\lambda D}{d}\] For n = n (n^th bright fringe),
\[x_n = \frac{n\lambda D}{d}\] The distance between two consecutive bright fringes is thus, \[\beta=\frac{n\lambda D}{d} \hspace{0.1cm}\text{... (iii)}\] For destructive interference (i.e., dark fringe):
Path difference is the odd integral multiple of half of the wavelength of light for the destructive interference.
\[\begin{align*} \frac{xd}{D}&=(2n-1)\frac{\lambda}{2}\\ x&=(2n-1) \frac{\lambda D}{2d} \hspace{0.1cm}\text{...(iv)}\\ \end{align*}\] where, n = 1, 2, 3, ... For n = 1 (first dark fringe), \[x_1=\frac{\lambda D}{2d}\] For n = 2 (second dark fringe), \[x_2 = \frac{3\lambda D}{2d}\] For n = n (n^th dark fringe), \[x_n=(2n-1)\frac{\lambda D}{2d}\] The distance between two consecutive dark fringe is thus, \[\beta=\frac{n\lambda D}{d}\hspace{0.1cm}\text{...(v)}\] From equation (iii) and (v), it is seen that the width of bright fringe is equal to the width of dark fringe is given by, \[\beta=\frac{\lambda D}{d}\]
4. If the distance between slits and the screen is doubled and slits separation is halved, what will be the effect on fringe width? [1]
The fringe width is given by,\[\beta=\frac{\lambda D}{d}\] When the distance between the slits and screen is doubled (i.e., 2D), and the slits separation is halved (i.e., d/2), the fringe width will become, \[\begin{align*} \beta^{'}&=\frac{\lambda 2D}{\frac{d}{2}}\\ &=4\frac{\lambda D}{d}\\ \beta^{'}&=4\times \beta\\ \end{align*}\] Thus, the fringe width will increase by 4 times.

5. What happened to fringe width if whole apparatus is immersed in water? [1]
The fringe width is given by, \[\beta=\frac{\lambda D}{d}\] When the whole apparatus is immersed in water, D and d remains unchanged, but the wavelength, λ changes. So, \(\beta \propto \lambda\).Since, the refractive index of water is greater than the refractive index of air i.e., \(\mu_w \gt \mu_a\). And as, \(\mu \propto \frac{1}{\lambda}\), it can be seen that the wavelength of light is less in water than in air, and thus the fringe width decreases when whole apparatus is immersed in water.

1. State and explain Biot-Savart law. [2]

Figure: Current carrying wire
Consider a wire XY carrying current I from X to Y. Let dl be the small length element in XY. Biot-Savart's law states that the magnetic field produced by the length element dl at a distance r (i.e., at point P) from the wire is given by,
1. directly proportional to the magnitude of current passed, \(dB \propto I \hspace{0.1cm}\text{...(i)}\)
2. directly proportional to the length of the element dl, \(dB \propto dl \hspace{0.1cm}\text{...(ii)}\)
3. directly proportional to the sine of the angle between the current carrying wire and the line joining r, \(dB \propto sin\theta \hspace{0.1cm}\text{...(iii)}\)
4. inversely proportional to the square of the line joining the centre of the element and the point P, \(dB \propto \frac{1}{r^2}\hspace{0.1cm} \text{...(iv)}\)
Combining (i) to (iv), we get, \[\begin{align*} dB \propto&\frac{Idlsin\theta}{r^2}\\ &=k\frac{Idlsin\theta}{r^2}\\ dB&=\frac{\mu_0}{4\pi} \frac{Idlsin\theta}{r^2}\\ \end{align*}\] Here, k is the proportionality constant with value \(k=\frac{\mu_0}{4\pi}\) with \(\mu_0=4\pi \times 10^{-7}\) Henary per meter as the permeability of free space.

2. Use this law to find the magnetic field at the center of current carrying circular coil. [3]
Let us consider a circular coil of radius r with centre at O carrying current I. dl be the small length element of the circular coil.

Figure: Current carrying coil
The magnetic field produced at the centre of the coil i.e., O due the current flowing through it is given by Biot-Savart's law as: \[\begin{align*} dB&=\frac{\mu_0}{4\pi} \frac{Idlsin\theta}{r^2} &=\frac{\mu_0}{4\pi}\frac{Idl}{r^2} \hspace{0.1cm} \because \theta=90^\circ\\ \end{align*}\] To find the magnetic field produced by the whole coil, interpreting equation (i) with limits 0 and 2\(\pi r\);
\[\begin{align*} B&=\int_{0}^{2\pi r} \frac{\mu_0}{4\pi} \frac{Idl}{r^2}\\ &=\frac{\mu_0}{4\pi}\int_{0}^{2\pi r} \frac{Idl}{r^2}\\ &=\frac{\mu_0I}{4\pi r^2}\int_{0}^{2\pi r}dl\\ &=\frac{\mu_0I}{4\pi r^2} [l]_{0}^{2\pi r}\\ &=\frac{\mu_0 I}{2r}\\ \end{align*}\] If the coil has 'n' turns, \[B=\frac{\mu_0 n I}{2r}\]

3. A circular coil of 100 turns has a radius of 10 cm and carries a current of 5 A. Determine the magnetic field at a point on the axis of the coil at a distance of 5 cm from the center of the coil. [3]
Solution:
number of turns, n = 100
radius of circular loop / coil, a = 10 cm = 0.10 m
current, I = 5 A
magnetic field at x = 0.05 m, B = ?
μ₀ = 4π×10^-7 T/mA

Now, \[\begin{align*} B&=\frac{\mu_0 n I a^2}{2 (a^2 + x^2)^{3/2}}\\ &=\frac{ 4 \pi \times 10^{-7} \times 100 \times 5 \times 0.10^2}{2(0.10^2 + 0.05^2)^{3/2}}\\ \therefore B&= 2.2 \times 10^{-3} \hspace{0.1cm} T\\ \end{align*}\]
The required magnetic field is 2.2×10^-3 T.
OR An inductor, a capacitor and a resistor are connected in series along with a source of alternating voltage.
1. Derive a relationship between the current and voltage. [3]


Figure (a) shows an inductor, a capacitor, and a resistor connected in series with an a.c. source of emf, E(rms value).(b) shows the phasor diagram for the case when voltage across the inductor is greater than the voltage across the capacitor. I be the rms value of the current flowing through all the circuit elements.
Potential difference across inductor, L, is;
\[V_L=IX_L\] Potential difference across capacitor, C, is;
\[V_C=IX_C\] Potential difference across resistor, R, is;
\[V_R=IX_R\] The resultant of \(V_L\) and \(V_C\) for \(V_L \gt V_C\) is, \[V_L-V_C=IX_L-IX_C=I\left(\omega L-\frac{1}{\omega C}\right)\] in the direction of the positive Y-axis.
The resultant of \(V_L-V_C\) and \(V_R\) is, \[\begin{align*} E&=\sqrt{(V_L-V_C)^2+V_R^2}\\ &=\sqrt{I^2\left(\omega L-\frac{1}{\omega C}\right)^2 +I^2R^2}\\ \therefore E&=I\sqrt{\left(\omega L-\frac{1}{\omega C}\right)^2 +R^2}\hspace{0.1cm}\text{...(i)}\\ \end{align*}\] Equation (i) is the required relation between the voltage and current for \(V_L\gt V_C\).
Similarly for \(V_L \lt V_C\), \[E=I\sqrt{\left(\frac{1}{\omega C}-\omega L\right)^2 +R^2}\hspace{0.1cm}\text{...(ii)}\]
2. Draw phasor diagram to show the lagging or leading relationship of voltage and current in the circuit connection. [2]
The phase angle is related to the voltage is, \[\begin{align*} tan\phi&=\frac{V_L-V_C}{V_R}\\ &=\frac{X_L-X_C}{R}\\ \end{align*}\] Case (i) : When \(X_L \gt X_C\), tanΦ>0; the voltage leads the current (circuit is more inductive in nature).


Case (ii) : When \(X_L \lt X_C\), tanΦ<0; the voltage lags the current (circuit is more capacitive in nature).


Figure (a) represents the phasor diagram for the case \(X_L \gt X_C\) and (b) represents the phasor diagram for the case \(X_C \gt X_L\). Since, the same current flows through the series LCR circuit, current, I, is taken as the reference phasor.

3. In the series LCR circuit, R=25 Ω, L=30 mH and C=10 µ F and these elements are connected to 240 volts, 50 Hz ac source. Calculate the current in the circuit and voltmeter reading across the capacitor. [3]
Solution;
resistance, R = 25 Ω
inductance, L = 30 mH = 30×10^-3H
capacitance, C = 10 µ F = 10×10^-6 F
rms voltage, E = 240 V
frequency, f = 50 Hz
current, I = ?
voltage across capacitor, V = ?
Now, the inductive reactance is, \[\begin{align*} X_L&=\omega L\\ &=2\pi f L\\ &= 2\pi \times 50 \times 30 \times 10^{-3}\\ &= 3 \pi \\ \end{align*}\] Similarly, the capacitive reactance is, \[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2\pi f C}\\ &= \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}}\\ &=\frac{1000}{\pi}\\ \end{align*}\] Since \(X_C \gt X_L\), the impedance of the LCR circuit is, \[\begin{align*} Z&=\sqrt{(X_C-X_L)^2 + R^2}\\ &=\sqrt{(\frac{1000}{\pi}-3\pi)^2+(25)^2}\\ \therefore Z&=309.89 \hspace{0.1cm}\Omega\\ \end{align*}\] The current in the circuit is, \[\begin{align*} I&=\frac{E}{Z}\\ &=\frac{240}{309.89}\\ \therefore I&=0.77 \hspace{0.1cm} \text{A}\\ \end{align*}\] Then, the voltage acros the capacitor is, \[\begin{align*} V_C&=IX_C\\ &=0.77 \times \frac{1000}{\pi}\\ &=245.09 \hspace{0.1cm}\text{V}\\ \end{align*}\] Thus, the current in the circuit is 0.77 A and the voltage reading across the capacitor is 245.09 V.
22. The work function of metal is 6.4\(\times\)10\(^{-19}\) J.
1. Explain what is meant by the term 'work function'. [2]
The maximum energy that must be given to an electron to take it out of the metal is called the work function of the metal . Work function is denoted by the symbol φ and is different for different metals.

2. Light with frequency of 1.2\(\times\)10\(^{15}\) Hz is shone onto the metal surface. Find whether or not the photons of this light will cause the photoelectric effect to take place. (h=6.62\(\times\)10\(^{-34}\) Js) [2]
Solution:
For the photons of the given light to cause the photoelectric effect, the its energy should be equal or greater than the work function of the metal. Let's try to see that.
work function of the metal, φ = 6.4\(\times\)10\(^{-19}\) J
frequency of the light, f = 1.2\(\times\)10\(^{15}\) Hz
So, the energy of the photon is given by, \[\begin{align*} E&=hf\\ &=6.62\times 10^{-34}\times 1.2 \times 10^{15}\\ &=7.9 \times 10^{-19} \hspace{0.1cm}\text{J}\\ \end{align*}\] Since the energy of the photons of the light is greater than the work function of the metal, it will cause the photoelectric effect.

3. The light source is now replaced with a light source of frequency 1.5\(\times\)10\(^{15}\) Hz.
1. Write down Einstein's photoelectric equation. [1]
Einstein's photoelectric equation can be written as, \[(K.E.)_{max}=E-\phi\] Furthermore, it can be simplified to, \[(K.E.)_{max}=h(f-f_0)\] where, h is the Planck's constant, f is the incident frequency, and f₀ is the threshold frequency.
where, E is the energy of the incident photons of the light, \(\phi\) is the work function of the metal, and (K.E.)_max is the maximum kinetic energy of the photoelectrons.

2. The photons from the source contain more energy then is required to release the electrons. How much extra energy is available after the electron has been released? [2]
As can be seen from the above relation, the kinetic energy in the range 0 to K.E. _max is available for the photoelectrons from the remaining energy of the incident light after releasing the photoelectrons from the metal surface.

3. Photons come up from three lamps that emit red, green and blue light. Which of these lamps produces photons with the highest energy? [1]
The energy of the photons is given by the relation, \[E=hf=h\frac{c}{\lambda}\] Since the red light has minimum wavelength and maximum frequency, the photons of red light will have highest energy. Hence, the lamp emitting red light will produce photons with the highest energy.

OR
1. State radioactive decay law. [2]
Radioactive decay law states that the rate of disintegration of radioactive substance is directly proportional to the number of atoms present at that instant. If N be the number of atoms left after time t, decay law deduce to, \[\begin{align*} \frac{dN}{dt}\propto& N\\ \frac{dN}{dt}&=-\lambda N\\ \end{align*}\] where λ is a proportionality constant called the decay constant and negative sign indicates that the number of atom N decreases with time.
2. Derive the relation \(N=N_0e^{-\lambda t}\). [3]
Let \(N_0\) be the initial number of atoms present in the radioactive sample at time, t=0 and \(N\) be the number of atoms left after time t.
From the decay law, \[\frac{dN}{dt} \propto N\] \[\frac{dN}{dt}=-\lambda N\] where \(\lambda\) is the decay constant. This can be expressed as, \[\frac{dN}{N}=-\lambda dt\] Now, taking integration on both sides, \[\begin{align*} \int_{N_0}^{N} \frac{dN}{N}&=-\lambda \int_{0}^{t} dt\\ [ln N]_{N_0}^{N}&=-\lambda[t]_{0}^{t}\\ lnN-lnN_0&=-\lambda t\\ ln{\frac{N}{N_0}}=-\lambda t\\ \therefore N&=N_0 e^{-\lambda t}\\ \end{align*}\]
3. A sample of radioactive isotopes contains 50% of its original number in 2 year. Then
1. What is half-life? [1]
Solution:
Let N₀ be the initial number of atoms and N be the number of atoms present after time t.
Here, \(N=50\% of N_0 = \frac{N_0}{2}\)
time , t= 2 years Since, t is the time taken by the radioactive atom to deduce to half of its original number of atoms, t itself is the half-life. \[\therefore T_{\frac{1}{2}}=2 \hspace{0.1cm}\text{years}\]
2. If there are 10⁶ such nuclei remaining after 8 years, how many numbers are there in the beginning? [2]
Solution:
Given,
N = 10⁶
t = 8 years
N₀ = ?
From the decay equation, \[\begin{align*} N&=N_0 e^{-\lambda t}\\ 10^6&=N_0 e^{-\frac{0.693}{T_{\frac{1}{2}} \times 8}}\\ 10^6&=N_0 e^{-2.772}\\ N_0&=\frac{10^6}{e^{-2.772}}\\ \therefore N_0&=1.59 \times 10^{7}\\ \end{align*}\] Hence, the original number of atoms is 1.59×10⁷.