Physics (54) Board Question Paper Solution
HSC 2025 Maharashtra Board Exam - Fully Solved J-287
SECTION – A
Q. 1. Select and write the correct answers for the following multiple choice type of questions: [10]
(i) “If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.” This statement refers to :
Answer: (a) zeroth law of thermodynamics
(ii) In Bernoulli’s theorem, which of the following is constant?
Answer: (d) Energy
(iii) Which of the following materials belongs to diathermanous substance?
Answer: (c) glass
(iv) Electric potential ‘V’ at a distance ‘r’ from a point charge is directly proportional to _____.
Answer: (c) \( \frac{1}{r} \)
(v) Which of the following equations gives correct expression for the internal resistance of a cell by using potentiometer?
Answer: (d) \( r = R\left(\frac{E}{V} - 1\right) \)
(vi) An electron, a proton, an \(\alpha\)-particle and a hydrogen atom are moving with the same kinetic energy. The associated de-Broglie wavelength will be longest for –
Answer: (b) electron
(vii) The gate which produces high output, when its both inputs are high is –
Answer: (b) AND gate
(viii) The power rating of a ceiling fan rotating with a constant torque of 2 Nm with an angular speed of 2\(\pi\) rad/s will be _____.
Answer: (d) 4\(\pi\) W
Explanation: \( P = \tau \omega = 2 \times 2\pi = 4\pi \) W.
Explanation: \( P = \tau \omega = 2 \times 2\pi = 4\pi \) W.
(ix) A string of length 2 m is vibrating with 2 loops. The distance between its node and adjacent antinode is _____.
Answer: (a) 0.5 m
Explanation: Length of 2 loops = \( L = \lambda \). So \( \lambda = 2 \)m. Distance N to A = \( \lambda/4 = 0.5 \)m.
Explanation: Length of 2 loops = \( L = \lambda \). So \( \lambda = 2 \)m. Distance N to A = \( \lambda/4 = 0.5 \)m.
(x) A transformer increases an alternating e.m.f. from 220V to 880V. If primary coil has 1000 turns, the number of turns in the secondary coil are _____.
Answer: (d) 4000
Explanation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow \frac{880}{220} = \frac{N_s}{1000} \Rightarrow N_s = 4000 \).
Explanation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow \frac{880}{220} = \frac{N_s}{1000} \Rightarrow N_s = 4000 \).
HSC Physics Board Papers with Solution
- Physics - March 2025 - English Medium View Answer Key
- Physics - March 2025 - Marathi Medium View Answer Key
- Physics - March 2025 - Hindi Medium View Answer Key
- Physics - March 2024 - English Medium View Answer Key Answer Key
- Physics - March 2024 - Marathi Medium View Answer Key
- Physics - March 2024 - Hindi Medium View Answer Key
- Physics - March 2023 - English Medium View Answer Key
- Physics - July 2023 - English Medium View Answer Key
- Physics - March 2013 View
- Physics - October 2013 View
- Physics - March 2014 View
- Physics - October 2014 View
- Physics - March 2015 View
- Physics - July 2015 View
- Physics - March 2016 View
- Physics - July 2016 View
- Physics - March 2017 View
- Physics - July 2017 View
Q. 2. Answer the following questions : [8]
(i) At what temperature the surface tension of a liquid becomes zero?
At the critical temperature.
(ii) Define self inductance.
Self inductance is the phenomenon of production of induced e.m.f. in a coil due to the change in current in the same coil.
(iii) What is the work done by an external uniform magnetic field perpendicular to the velocity of a moving charge?
The work done is zero. (Since the magnetic force is always perpendicular to the direction of motion/displacement).
(iv) What do you mean by a thermodynamic system?
A thermodynamic system is a collection of a large number of molecules (matter or radiation) confined within a definite boundary, separated from the surroundings.
(v) What is value of B called, when H = 0 is in the hysteresis loop?
It is called Retentivity or Remanence.
(vi) State the formula for angle of banking.
The formula is: \( \tan \theta = \frac{v^2}{rg} \) or \( \theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \).
(vii) Calculate the electric field intensity at a point just near the surface of a charged plane sheet, measured from its mid-point. [ \(\sigma = 8.85 \mu C/m^2\) ]
Formula: \( E = \frac{\sigma}{2\epsilon_0} \)
Given: \( \sigma = 8.85 \times 10^{-6} \, \text{C/m}^2 \), \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \)
Calculation: \( E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = \frac{10^6}{2} = 5 \times 10^5 \, \text{N/C} \) (or V/m).
Given: \( \sigma = 8.85 \times 10^{-6} \, \text{C/m}^2 \), \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \)
Calculation: \( E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = \frac{10^6}{2} = 5 \times 10^5 \, \text{N/C} \) (or V/m).
(viii) Find kinetic energy of 1 litre of an ideal gas at S.T.P.
Formula: K.E. \( = \frac{3}{2} PV \)
Given: \( P = 1.013 \times 10^5 \, \text{N/m}^2 \) (at STP), \( V = 1 \text{ litre} = 10^{-3} \, \text{m}^3 \).
Calculation: \( E = 1.5 \times 1.013 \times 10^5 \times 10^{-3} \approx 151.95 \) J.
Given: \( P = 1.013 \times 10^5 \, \text{N/m}^2 \) (at STP), \( V = 1 \text{ litre} = 10^{-3} \, \text{m}^3 \).
Calculation: \( E = 1.5 \times 1.013 \times 10^5 \times 10^{-3} \approx 151.95 \) J.
SECTION – B
Attempt any EIGHT questions of the following : [16]
Q. 3. Explain harmonics and overtones.
Harmonics: The term harmonic is used to indicate the fundamental frequency and all its integral multiples. The fundamental frequency is the first harmonic, \(2n\) is the second harmonic, etc.
Overtones: Overtones are the frequencies higher than the fundamental frequency which are actually present in the vibration. The first frequency higher than fundamental is the first overtone, the next is the second overtone, etc.
Overtones: Overtones are the frequencies higher than the fundamental frequency which are actually present in the vibration. The first frequency higher than fundamental is the first overtone, the next is the second overtone, etc.
Q. 4. Using Newton’s law of viscosity for streamline flow, derive an expression for coefficient of viscosity.
According to Newton's law of viscosity, the viscous force \( F \) acting between two layers of a liquid is directly proportional to:
1. The area of the layer \( A \)
2. The velocity gradient \( \frac{dv}{dx} \)
Therefore, \( F \propto A \frac{dv}{dx} \Rightarrow F = \eta A \frac{dv}{dx} \).
Where \( \eta \) is a constant called the coefficient of viscosity.
Expression: \( \eta = \frac{F}{A (dv/dx)} \).
1. The area of the layer \( A \)
2. The velocity gradient \( \frac{dv}{dx} \)
Therefore, \( F \propto A \frac{dv}{dx} \Rightarrow F = \eta A \frac{dv}{dx} \).
Where \( \eta \) is a constant called the coefficient of viscosity.
Expression: \( \eta = \frac{F}{A (dv/dx)} \).
Q. 5. State the formula for magnetic induction produced by a current in a circular arc of a wire. Hence find the magnetic induction at the centre of a current carrying circular loop.
Formula for Arc: \( B = \frac{\mu_0 I}{4\pi r} \theta \), where \( \theta \) is the angle subtended by the arc at the centre in radians.
For a circular loop: The angle subtended \( \theta = 2\pi \) radians.
Substituting this value:
\( B = \frac{\mu_0 I}{4\pi r} (2\pi) = \frac{\mu_0 I}{2r} \).
For a circular loop: The angle subtended \( \theta = 2\pi \) radians.
Substituting this value:
\( B = \frac{\mu_0 I}{4\pi r} (2\pi) = \frac{\mu_0 I}{2r} \).
Q. 6. State and prove the law of conservation of angular momentum.
Statement: Angular momentum of a rotating body remains constant if the resultant external torque acting on the body is zero.
Proof:
We know that torque \( \vec{\tau} = \frac{d\vec{L}}{dt} \).
If \( \vec{\tau}_{ext} = 0 \), then \( \frac{d\vec{L}}{dt} = 0 \).
This implies \( \vec{L} = \text{constant} \).
Since \( L = I\omega \), if \( L \) is constant, \( I\omega = \text{constant} \).
Proof:
We know that torque \( \vec{\tau} = \frac{d\vec{L}}{dt} \).
If \( \vec{\tau}_{ext} = 0 \), then \( \frac{d\vec{L}}{dt} = 0 \).
This implies \( \vec{L} = \text{constant} \).
Since \( L = I\omega \), if \( L \) is constant, \( I\omega = \text{constant} \).
Q. 7. State any four advantages of light emitting diode (LED).
1. Energy efficient (Low power consumption).
2. Long operational life.
3. Fast switching capability (fast response time).
4. Rugged and durable (no filament to break).
5. Environmentally friendly (no mercury).
2. Long operational life.
3. Fast switching capability (fast response time).
4. Rugged and durable (no filament to break).
5. Environmentally friendly (no mercury).
Q. 8. Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm² when it is maintained at 127ºC.
Given:
Area \( A = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2 = 2 \times 10^{-2} \text{ m}^2 \)
Time \( t = 1 \text{ min} = 60 \text{ s} \)
Temp \( T = 127 + 273 = 400 \text{ K} \)
Stefan's Constant \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
Formula: \( Q = \sigma A t T^4 \)
Calculation:
\( Q = 5.67 \times 10^{-8} \times (2 \times 10^{-2}) \times 60 \times (400)^4 \)
\( Q = 5.67 \times 1.2 \times 10^{-7} \times 256 \times 10^8 \)
\( Q = 5.67 \times 1.2 \times 256 \times 10^1 \)
\( Q = 17418.24 \text{ J} \approx 1.74 \times 10^4 \text{ J} \).
Area \( A = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2 = 2 \times 10^{-2} \text{ m}^2 \)
Time \( t = 1 \text{ min} = 60 \text{ s} \)
Temp \( T = 127 + 273 = 400 \text{ K} \)
Stefan's Constant \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \)
Formula: \( Q = \sigma A t T^4 \)
Calculation:
\( Q = 5.67 \times 10^{-8} \times (2 \times 10^{-2}) \times 60 \times (400)^4 \)
\( Q = 5.67 \times 1.2 \times 10^{-7} \times 256 \times 10^8 \)
\( Q = 5.67 \times 1.2 \times 256 \times 10^1 \)
\( Q = 17418.24 \text{ J} \approx 1.74 \times 10^4 \text{ J} \).
Q. 9. Two coils having self inductances 60 mH each, are coupled with each other. If the coefficient of coupling is 0.75, calculate the mutual inductance between them.
Given: \( L_1 = 60 \text{ mH}, L_2 = 60 \text{ mH}, k = 0.75 \)
Formula: \( M = k \sqrt{L_1 L_2} \)
Calculation:
\( M = 0.75 \sqrt{60 \times 60} \)
\( M = 0.75 \times 60 \)
\( M = 45 \text{ mH} \).
Formula: \( M = k \sqrt{L_1 L_2} \)
Calculation:
\( M = 0.75 \sqrt{60 \times 60} \)
\( M = 0.75 \times 60 \)
\( M = 45 \text{ mH} \).
Q. 10. In a series LCR circuit, if resistance, inductive reactance and capacitive reactance are 3\(\Omega\), 8\(\Omega\) and 4\(\Omega\) respectively, calculate phase difference between voltage and current.
Given: \( R = 3\Omega, X_L = 8\Omega, X_C = 4\Omega \)
Formula: \( \tan \phi = \frac{X_L - X_C}{R} \)
Calculation:
\( \tan \phi = \frac{8 - 4}{3} = \frac{4}{3} \)
\( \phi = \tan^{-1}(1.333) \approx 53.13^\circ \) or \( 53^\circ 8' \).
Formula: \( \tan \phi = \frac{X_L - X_C}{R} \)
Calculation:
\( \tan \phi = \frac{8 - 4}{3} = \frac{4}{3} \)
\( \phi = \tan^{-1}(1.333) \approx 53.13^\circ \) or \( 53^\circ 8' \).
Q. 11. State the advantages of a potentiometer over a voltmeter.
1. It measures the emf of a cell very accurately because it draws no current from the cell at the null point (acts as an ideal voltmeter).
2. It can be used to measure the internal resistance of a cell.
3. It is more sensitive than a voltmeter.
4. It can be used to compare emfs of two cells.
2. It can be used to measure the internal resistance of a cell.
3. It is more sensitive than a voltmeter.
4. It can be used to compare emfs of two cells.
Q. 12. Draw a neat and labelled circuit diagram for a full wave rectifier.
(Students should draw a circuit diagram containing: A center-tapped step-down transformer, two diodes D1 and D2 connected to the secondary coil, and a load resistor \( R_L \) connected between the center tap and the common cathode point.)
[Diagram Placeholder: Center-tap Transformer + 2 Diodes + Load Resistor]
Q. 13. A body of mass 0.8 kg performs linear S.H.M. It experiences a restoring force of 0.4N, when its displacement from mean position is 4 cm. Determine Force constant and Period of S.H.M.
Given: \( m = 0.8 \text{ kg} \), \( F = 0.4 \text{ N} \), \( x = 4 \text{ cm} = 0.04 \text{ m} \).
1. Force constant (k):
\( F = kx \Rightarrow k = \frac{F}{x} = \frac{0.4}{0.04} = 10 \text{ N/m} \).
2. Period (T):
\( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.8}{10}} = 2\pi \sqrt{0.08} \)
\( T \approx 2 \times 3.142 \times 0.2828 \approx 1.78 \text{ s} \).
1. Force constant (k):
\( F = kx \Rightarrow k = \frac{F}{x} = \frac{0.4}{0.04} = 10 \text{ N/m} \).
2. Period (T):
\( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.8}{10}} = 2\pi \sqrt{0.08} \)
\( T \approx 2 \times 3.142 \times 0.2828 \approx 1.78 \text{ s} \).
Q. 14. A gas of 0.5 mole at 300 K expands isothermally from an initial volume of 2.0 litre to a final volume of 6.0 litre. What is the work done by gas?
Given: \( n = 0.5 \), \( T = 300 \text{ K} \), \( V_1 = 2 \text{ L}, V_2 = 6 \text{ L} \), \( R = 8.319 \text{ J/mol K} \).
Formula: \( W = nRT \ln\left(\frac{V_2}{V_1}\right) = 2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) \)
Calculation:
\( W = 2.303 \times 0.5 \times 8.319 \times 300 \times \log_{10}(3) \)
\( W = 2.303 \times 1247.85 \times 0.4771 \)
\( W \approx 1371 \text{ J} \).
Formula: \( W = nRT \ln\left(\frac{V_2}{V_1}\right) = 2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) \)
Calculation:
\( W = 2.303 \times 0.5 \times 8.319 \times 300 \times \log_{10}(3) \)
\( W = 2.303 \times 1247.85 \times 0.4771 \)
\( W \approx 1371 \text{ J} \).
SECTION – C
Attempt any EIGHT questions of the following : [24]
Q. 15. Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field, performing S.H.M.
Consider a bar magnet of magnetic dipole moment \( \mu \) and moment of inertia \( I \) suspended in a uniform magnetic field \( B \).
If deflected by angle \( \theta \), restoring torque \( \tau = -\mu B \sin \theta \).
For small \( \theta \), \( \sin \theta \approx \theta \), so \( \tau = -\mu B \theta \).
Also \( \tau = I \alpha = I \frac{d^2\theta}{dt^2} \).
\( I \alpha = -\mu B \theta \Rightarrow \alpha = -\left(\frac{\mu B}{I}\right) \theta \).
This is S.H.M condition with \( \omega^2 = \frac{\mu B}{I} \).
Period \( T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{\mu B}} \).
If deflected by angle \( \theta \), restoring torque \( \tau = -\mu B \sin \theta \).
For small \( \theta \), \( \sin \theta \approx \theta \), so \( \tau = -\mu B \theta \).
Also \( \tau = I \alpha = I \frac{d^2\theta}{dt^2} \).
\( I \alpha = -\mu B \theta \Rightarrow \alpha = -\left(\frac{\mu B}{I}\right) \theta \).
This is S.H.M condition with \( \omega^2 = \frac{\mu B}{I} \).
Period \( T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{\mu B}} \).
Q. 16. In a thermodynamic system, define – (a) Mechanical equilibrium (b) Chemical equilibrium and (c) Thermal equilibrium
(a) Mechanical equilibrium: A system is said to be in mechanical equilibrium if there are no unbalanced forces within the system and between the system and the surroundings (pressure is constant throughout).
(b) Chemical equilibrium: A system is in chemical equilibrium if there are no chemical reactions going on within the system and no transfer of matter from one part of the system to another.
(c) Thermal equilibrium: A system is in thermal equilibrium if its temperature is uniform throughout and does not change with time.
(b) Chemical equilibrium: A system is in chemical equilibrium if there are no chemical reactions going on within the system and no transfer of matter from one part of the system to another.
(c) Thermal equilibrium: A system is in thermal equilibrium if its temperature is uniform throughout and does not change with time.
Q. 17. What is Brewster’s law? Derive the formula for Brewster’s angle.
Brewster's Law: It states that the tangent of the polarizing angle (Brewster's angle) is numerically equal to the refractive index of the medium.
Derivation:
When light is incident at polarizing angle \( i_p \), the reflected and refracted rays are perpendicular.
\( r + 90^\circ + i_p = 180^\circ \Rightarrow r = 90^\circ - i_p \).
By Snell's law: \( \mu = \frac{\sin i_p}{\sin r} = \frac{\sin i_p}{\sin(90^\circ - i_p)} \).
\( \mu = \frac{\sin i_p}{\cos i_p} = \tan i_p \).
Derivation:
When light is incident at polarizing angle \( i_p \), the reflected and refracted rays are perpendicular.
\( r + 90^\circ + i_p = 180^\circ \Rightarrow r = 90^\circ - i_p \).
By Snell's law: \( \mu = \frac{\sin i_p}{\sin r} = \frac{\sin i_p}{\sin(90^\circ - i_p)} \).
\( \mu = \frac{\sin i_p}{\cos i_p} = \tan i_p \).
Q. 18. Derive an expression for law of radioactive decay. Define one becquerel (Bq).
Derivation: Rate of decay is proportional to number of nuclei present.
\( \frac{dN}{dt} \propto -N \Rightarrow \frac{dN}{dt} = -\lambda N \).
\( \frac{dN}{N} = -\lambda dt \).
Integrating: \( \ln N = -\lambda t + C \).
At \( t=0, N=N_0 \), so \( C = \ln N_0 \).
\( \ln N - \ln N_0 = -\lambda t \Rightarrow \ln(N/N_0) = -\lambda t \).
\( N = N_0 e^{-\lambda t} \).
One Becquerel (Bq): It is defined as the activity of a radioactive sample which decays at the rate of one disintegration per second.
\( \frac{dN}{dt} \propto -N \Rightarrow \frac{dN}{dt} = -\lambda N \).
\( \frac{dN}{N} = -\lambda dt \).
Integrating: \( \ln N = -\lambda t + C \).
At \( t=0, N=N_0 \), so \( C = \ln N_0 \).
\( \ln N - \ln N_0 = -\lambda t \Rightarrow \ln(N/N_0) = -\lambda t \).
\( N = N_0 e^{-\lambda t} \).
One Becquerel (Bq): It is defined as the activity of a radioactive sample which decays at the rate of one disintegration per second.
Q. 19. State and prove Kirchhoff’s law of heat radiation.
Statement: At a given temperature, the coefficient of absorption of a body is equal to its coefficient of emission (emissivity). \( a = e \).
Proof: Consider an ordinary body A and a black body B in an enclosure at constant temp T. Let R be emissive power of A, \( R_b \) for B.
Energy absorbed by A = \( a \times \) incident energy.
Energy emitted by A = \( R \).
At equilibrium, Energy emitted = Energy absorbed.
If Q is incident per unit area per sec: \( R = a Q \).
For black body, \( a=1 \), so \( R_b = Q \).
Substituting Q: \( R = a R_b \Rightarrow \frac{R}{R_b} = a \).
But \( \frac{R}{R_b} = e \) (definition of emissivity).
Therefore, \( a = e \).
Proof: Consider an ordinary body A and a black body B in an enclosure at constant temp T. Let R be emissive power of A, \( R_b \) for B.
Energy absorbed by A = \( a \times \) incident energy.
Energy emitted by A = \( R \).
At equilibrium, Energy emitted = Energy absorbed.
If Q is incident per unit area per sec: \( R = a Q \).
For black body, \( a=1 \), so \( R_b = Q \).
Substituting Q: \( R = a R_b \Rightarrow \frac{R}{R_b} = a \).
But \( \frac{R}{R_b} = e \) (definition of emissivity).
Therefore, \( a = e \).
Q. 20. Obtain an expression for practical determination of end correction – (i) for a pipe open at both ends and (ii) for a pipe closed at one end.
(i) Pipe open at both ends:
\( v = 2n_1(l_1 + 2e) \) and \( v = 2n_2(l_2 + 2e) \).
\( n_1 l_1 + 2n_1 e = n_2 l_2 + 2n_2 e \).
\( 2e(n_1 - n_2) = n_2 l_2 - n_1 l_1 \).
\( e = \frac{n_2 l_2 - n_1 l_1}{2(n_1 - n_2)} \).
(ii) Pipe closed at one end:
\( v = 4n_1(l_1 + e) \) and \( v = 4n_2(l_2 + e) \).
\( n_1 l_1 + n_1 e = n_2 l_2 + n_2 e \).
\( e(n_1 - n_2) = n_2 l_2 - n_1 l_1 \).
\( e = \frac{n_2 l_2 - n_1 l_1}{n_1 - n_2} \).
\( v = 2n_1(l_1 + 2e) \) and \( v = 2n_2(l_2 + 2e) \).
\( n_1 l_1 + 2n_1 e = n_2 l_2 + 2n_2 e \).
\( 2e(n_1 - n_2) = n_2 l_2 - n_1 l_1 \).
\( e = \frac{n_2 l_2 - n_1 l_1}{2(n_1 - n_2)} \).
(ii) Pipe closed at one end:
\( v = 4n_1(l_1 + e) \) and \( v = 4n_2(l_2 + e) \).
\( n_1 l_1 + n_1 e = n_2 l_2 + n_2 e \).
\( e(n_1 - n_2) = n_2 l_2 - n_1 l_1 \).
\( e = \frac{n_2 l_2 - n_1 l_1}{n_1 - n_2} \).
Q. 21. A conducting bar is rotating... Obtain an expression for the rotational e.m.f.
Consider a rod of length \( l \) rotating with angular velocity \( \omega \) in field \( B \).
Consider small element \( dr \) at distance \( r \) from pivot. Velocity \( v = r\omega \).
Induced emf in element \( de = B v dr = B (r\omega) dr \).
Total emf \( E = \int_0^l B \omega r \, dr = B\omega \left[\frac{r^2}{2}\right]_0^l \).
\( E = \frac{1}{2} B \omega l^2 \).
Consider small element \( dr \) at distance \( r \) from pivot. Velocity \( v = r\omega \).
Induced emf in element \( de = B v dr = B (r\omega) dr \).
Total emf \( E = \int_0^l B \omega r \, dr = B\omega \left[\frac{r^2}{2}\right]_0^l \).
\( E = \frac{1}{2} B \omega l^2 \).
Q. 22. An electron in hydrogen atom stays in its second orbit for \( 10^{-8} \) s. How many revolutions will it make around the nucleus in that time?
Given: \( n=2, t=10^{-8} \) s.
Velocity in 2nd orbit \( v_2 = \frac{v_1}{2} \approx \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \) m/s.
Radius of 2nd orbit \( r_2 = 4 r_1 \approx 4 \times 0.53 \text{ \AA} = 2.12 \times 10^{-10} \) m.
Frequency \( f = \frac{v_2}{2\pi r_2} = \frac{1.09 \times 10^6}{2\pi \times 2.12 \times 10^{-10}} \approx 8.2 \times 10^{14} \) Hz.
Number of revolutions \( N = f \times t = 8.2 \times 10^{14} \times 10^{-8} = 8.2 \times 10^6 \).
Velocity in 2nd orbit \( v_2 = \frac{v_1}{2} \approx \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \) m/s.
Radius of 2nd orbit \( r_2 = 4 r_1 \approx 4 \times 0.53 \text{ \AA} = 2.12 \times 10^{-10} \) m.
Frequency \( f = \frac{v_2}{2\pi r_2} = \frac{1.09 \times 10^6}{2\pi \times 2.12 \times 10^{-10}} \approx 8.2 \times 10^{14} \) Hz.
Number of revolutions \( N = f \times t = 8.2 \times 10^{14} \times 10^{-8} = 8.2 \times 10^6 \).
Q. 23. A flywheel of a motor has mass 100 kg and radius 1.5 m. The motor develops a constant torque of 2000 Nm. The flywheel starts rotating from rest. Calculate the work done during the first 4 revolutions.
Given: \( \tau = 2000 \) Nm, \( \theta = 4 \) revs \( = 4 \times 2\pi = 8\pi \) rad.
Work Done:
\( W = \tau \theta \)
\( W = 2000 \times 8\pi = 16000\pi \) J.
\( W \approx 16000 \times 3.142 = 50,272 \) J.
Work Done:
\( W = \tau \theta \)
\( W = 2000 \times 8\pi = 16000\pi \) J.
\( W \approx 16000 \times 3.142 = 50,272 \) J.
Q. 24. A galvanometer has a resistance of 50 \(\Omega\) and a current of 2 mA is needed for its full scale deflection. Calculate resistance required to convert it...
Given: \( G = 50\Omega, I_g = 2 \text{ mA} = 0.002 \text{ A} \).
(i) Into Ammeter of 0.5 A range:
\( I = 0.5 \text{ A} \). Shunt \( S = \frac{I_g G}{I - I_g} \).
\( S = \frac{0.002 \times 50}{0.5 - 0.002} = \frac{0.1}{0.498} \approx 0.2008 \, \Omega \).
(ii) Into Voltmeter of 10 V range:
\( V = 10 \text{ V} \). Series Resistance \( R = \frac{V}{I_g} - G \).
\( R = \frac{10}{0.002} - 50 = 5000 - 50 = 4950 \, \Omega \).
(i) Into Ammeter of 0.5 A range:
\( I = 0.5 \text{ A} \). Shunt \( S = \frac{I_g G}{I - I_g} \).
\( S = \frac{0.002 \times 50}{0.5 - 0.002} = \frac{0.1}{0.498} \approx 0.2008 \, \Omega \).
(ii) Into Voltmeter of 10 V range:
\( V = 10 \text{ V} \). Series Resistance \( R = \frac{V}{I_g} - G \).
\( R = \frac{10}{0.002} - 50 = 5000 - 50 = 4950 \, \Omega \).
Q. 25. Diameter of a water drop is 0.6 mm. Calculate the pressure inside a liquid drop. (T = 72 dyne/cm, atm pressure = \(1.013 \times 10^5\) N/m²)
Given: \( r = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m} \), \( T = 0.072 \text{ N/m} \).
Excess Pressure: \( P_{ex} = \frac{2T}{r} = \frac{2 \times 0.072}{3 \times 10^{-4}} = \frac{0.144}{0.0003} = 480 \text{ Pa} \).
Total Pressure: \( P_{in} = P_{atm} + P_{ex} = 101300 + 480 = 101780 \text{ N/m}^2 \).
Excess Pressure: \( P_{ex} = \frac{2T}{r} = \frac{2 \times 0.072}{3 \times 10^{-4}} = \frac{0.144}{0.0003} = 480 \text{ Pa} \).
Total Pressure: \( P_{in} = P_{atm} + P_{ex} = 101300 + 480 = 101780 \text{ N/m}^2 \).
Q. 26. A solenoid of length \(\pi\) m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5A. Calculate the magnetic field at its centre along the axis.
Given: \( L = \pi \text{ m}, N = 1000, I = 5 \text{ A} \).
Turns per unit length \( n = \frac{N}{L} = \frac{1000}{\pi} \).
Formula: \( B = \mu_0 n I \)
Calculation:
\( B = 4\pi \times 10^{-7} \times \frac{1000}{\pi} \times 5 \)
\( B = 4 \times 10^{-7} \times 5000 = 20 \times 10^{-4} = 2 \times 10^{-3} \text{ T} \).
Turns per unit length \( n = \frac{N}{L} = \frac{1000}{\pi} \).
Formula: \( B = \mu_0 n I \)
Calculation:
\( B = 4\pi \times 10^{-7} \times \frac{1000}{\pi} \times 5 \)
\( B = 4 \times 10^{-7} \times 5000 = 20 \times 10^{-4} = 2 \times 10^{-3} \text{ T} \).
SECTION – D
Attempt any THREE questions of the following : [12]
Q. 27. What is Ferromagnetism? Explain it on the basis of domain theory.
Ferromagnetism: The property of substances (like Fe, Co, Ni) to be strongly attracted by magnets and to become permanent magnets.
Domain Theory:
1. Ferromagnetic materials consist of small regions called 'domains' where dipoles are aligned in same direction.
2. In unmagnetized state, domains are randomly oriented, net moment is zero.
3. In external field, domains parallel to field grow in size (wall displacement) or rotate (rotation) to align with field, causing strong magnetization.
Domain Theory:
1. Ferromagnetic materials consist of small regions called 'domains' where dipoles are aligned in same direction.
2. In unmagnetized state, domains are randomly oriented, net moment is zero.
3. In external field, domains parallel to field grow in size (wall displacement) or rotate (rotation) to align with field, causing strong magnetization.
Q. 28. Obtain an expression for average power dissipated in a series LCR circuit.
Instantaneous power \( P = V I \).
Let \( V = V_m \sin\omega t \) and \( I = I_m \sin(\omega t \pm \phi) \).
\( P = V_m I_m \sin\omega t \sin(\omega t \pm \phi) \).
Using trig identity and averaging over one cycle:
Average of \( \cos(2\omega t) \) terms is zero.
\( P_{avg} = \frac{V_m I_m}{2} \cos\phi = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos\phi \).
\( P_{avg} = V_{rms} I_{rms} \cos\phi \).
Let \( V = V_m \sin\omega t \) and \( I = I_m \sin(\omega t \pm \phi) \).
\( P = V_m I_m \sin\omega t \sin(\omega t \pm \phi) \).
Using trig identity and averaging over one cycle:
Average of \( \cos(2\omega t) \) terms is zero.
\( P_{avg} = \frac{V_m I_m}{2} \cos\phi = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos\phi \).
\( P_{avg} = V_{rms} I_{rms} \cos\phi \).
Q. 29. Distinguish between interference and diffraction of light... Calculate angular fringe separation in water.
Distinction: Interference is superposition of waves from two coherent sources. Diffraction is bending of waves around corners from secondary wavelets of same wavefront.
Numerical:
Given: \( \theta_{air} = 0.20^\circ \), \( \mu = 1.33 \).
Fringe separation in air \( \theta_{air} = \frac{\lambda}{d} \).
In water \( \lambda_w = \frac{\lambda}{\mu} \).
New separation \( \theta_{water} = \frac{\lambda_w}{d} = \frac{\lambda}{\mu d} = \frac{\theta_{air}}{\mu} \).
\( \theta_{water} = \frac{0.20}{1.33} \approx 0.15^\circ \).
Numerical:
Given: \( \theta_{air} = 0.20^\circ \), \( \mu = 1.33 \).
Fringe separation in air \( \theta_{air} = \frac{\lambda}{d} \).
In water \( \lambda_w = \frac{\lambda}{\mu} \).
New separation \( \theta_{water} = \frac{\lambda_w}{d} = \frac{\lambda}{\mu d} = \frac{\theta_{air}}{\mu} \).
\( \theta_{water} = \frac{0.20}{1.33} \approx 0.15^\circ \).
Q. 30. State Einstein’s photoelectric equation and mention physical significance... Calculate energy of incident photon.
Equation: \( h\nu = \phi_0 + K_{max} \).
Significance: Energy of photon is used to overcome work function (\(\phi_0\)) and impart kinetic energy (\(K_{max}\)) to the electron.
Numerical:
Given \( \lambda = 4000 \text{ \AA} = 4 \times 10^{-7} \text{ m} \).
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \)
\( E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J} \) (approx 3.1 eV).
Significance: Energy of photon is used to overcome work function (\(\phi_0\)) and impart kinetic energy (\(K_{max}\)) to the electron.
Numerical:
Given \( \lambda = 4000 \text{ \AA} = 4 \times 10^{-7} \text{ m} \).
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \)
\( E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J} \) (approx 3.1 eV).
Q. 31. State any four uses of Van de Graaff generator. Calculate the displacement current.
Uses: 1. To produce high DC voltage. 2. To accelerate charged particles. 3. Nuclear physics experiments. 4. To study collision experiments.
Numerical:
Given: \( \frac{dE}{dt} = 2 \times 10^{11} \text{ V/ms} = 2 \times 10^{11} \text{ V m}^{-1}\text{s}^{-1} \).
Area \( A = 20 \text{ cm}^2 = 2 \times 10^{-3} \text{ m}^2 \).
Formula: \( I_d = \epsilon_0 A \frac{dE}{dt} \).
Calculation: \( I_d = 8.85 \times 10^{-12} \times 2 \times 10^{-3} \times 2 \times 10^{11} \)
\( I_d = 35.4 \times 10^{-4} \text{ A} = 3.54 \text{ mA} \).
Numerical:
Given: \( \frac{dE}{dt} = 2 \times 10^{11} \text{ V/ms} = 2 \times 10^{11} \text{ V m}^{-1}\text{s}^{-1} \).
Area \( A = 20 \text{ cm}^2 = 2 \times 10^{-3} \text{ m}^2 \).
Formula: \( I_d = \epsilon_0 A \frac{dE}{dt} \).
Calculation: \( I_d = 8.85 \times 10^{-12} \times 2 \times 10^{-3} \times 2 \times 10^{11} \)
\( I_d = 35.4 \times 10^{-4} \text{ A} = 3.54 \text{ mA} \).