Maharashtra Board HSC Physics Question Paper: March 2022 Solutions
Complete, step-by-step solutions for the Class 12 Physics Board Exam 2022.
SECTION − A
Q.1. Select and write the correct answers for the following multiple choice type of questions: [10]
(i) The first law of thermodynamics is concerned with the conservation of _______.
Explanation: The first law of thermodynamics is essentially a restatement of the law of conservation of energy for thermodynamic systems ($Q = \Delta U + W$).
(ii) The average value of alternating current over a full cycle is always _______. [I0 = Peak value of current]
Explanation: An alternating current flows in one direction for the first half cycle and in the opposite direction for the second half cycle symmetrically. Thus, the average value over a complete cycle is zero.
(iii) The angle at which maximum torque is exerted by the external uniform electric field on the electric dipole is _______.
Explanation: Torque \( \tau = pE \sin \theta \). The torque is maximum when \( \sin \theta \) is maximum, i.e., \( \theta = 90^\circ \).
(iv) The property of light which does not change, when it travels from one medium to another is _______.
Explanation: Frequency is a characteristic of the source of the wave and does not change when light passes between media. Velocity and wavelength change.
(v) The root mean square speed of the molecules of a gas is proportional to _______. [T = Absolute temperature of gas]
Explanation: \( v_{rms} = \sqrt{\frac{3RT}{M}} \). Therefore, \( v_{rms} \propto \sqrt{T} \).
(vi) The unit Wbm–2 is equal to _______.
Explanation: Weber per square meter (Wb/m²) is the unit of magnetic flux density or magnetic induction, which is known as Tesla (T).
(vii) When the bob performs a vertical circular motion and the string rotates in a vertical plane, the difference in the tension in the string at horizontal position and uppermost position is _______.
Explanation:
Tension at top \( T_{top} = \frac{mv_t^2}{r} - mg \).
Tension at horizontal \( T_{mid} = \frac{mv_m^2}{r} \).
Using energy conservation: \( \frac{1}{2}mv_m^2 = \frac{1}{2}mv_t^2 + mgr \Rightarrow v_m^2 = v_t^2 + 2gr \).
\( T_{mid} = \frac{m(v_t^2 + 2gr)}{r} = \frac{mv_t^2}{r} + 2mg \).
Difference = \( T_{mid} - T_{top} = (\frac{mv_t^2}{r} + 2mg) - (\frac{mv_t^2}{r} - mg) = 3mg \).
(viii) A liquid rises in glass capillary tube upto a height of 2.5 cm at room temperature. If another glass capillary tube having radius half that of the earlier tube is immersed in the same liquid, the rise of liquid in it will be _______.
Explanation: Capillary rise \( h = \frac{2T \cos \theta}{r \rho g} \). Thus, \( h \propto \frac{1}{r} \).
If the radius is halved (\( r_2 = r_1/2 \)), the height will double.
\( h_2 = 2 \times h_1 = 2 \times 2.5 = 5 \) cm.
(ix) In young’s double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16:1, then the ratio of amplitudes of the two source will be _______.
Explanation:
\( \frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \frac{16}{1} \)
Taking square root: \( \frac{a_1 + a_2}{a_1 - a_2} = \frac{4}{1} \)
\( a_1 + a_2 = 4a_1 - 4a_2 \Rightarrow 3a_1 = 5a_2 \Rightarrow \frac{a_1}{a_2} = \frac{5}{3} \).
(x) The equation of a simple harmonic progressive wave travelling on a string is y = 8 sin (0.02 x – 4t) cm. The speed of the wave is _______.
Explanation: Comparing with standard equation \( y = A \sin(kx - \omega t) \):
\( k = 0.02 \) and \( \omega = 4 \).
Wave speed \( v = \frac{\omega}{k} = \frac{4}{0.02} = \frac{400}{2} = 200 \) cm/s.
HSC Physics Board Papers with Solution
- Physics - March 2025 - English Medium View Answer Key
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- 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 2022 - English Medium View Answer Key
- Physics - July 2022 - English Medium View Answer Key
- Physics - July 2021 - English Medium View Answer Key
- Physics - February 2020 - English Medium View Answer Key
- Physics - March 2013 View
- Physics - October 2013 View Answer Key
- Physics - March 2014 View Answer Key
- Physics - October 2014 View Answer Key
- Physics - March 2015 View Answer Key
- Physics - October 2015 View Answer Key
- Physics - March 2016 View Answer Key
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- Physics - March 2017 View Answer Key
- Physics - July 2017 View Answer Key
Q.2. Answer the following questions: [8]
(i) Define potential gradient of the potentiometer wire.
\( K = \frac{V}{L} \).
(ii) State the formula for critical velocity in terms of Reynold’s number for a flow of a fluid.
(iii) Is it always necessary to use red light to get photoelectric effect?
(iv) Write the Boolean expression for Exclusive – OR (X – OR) gate.
(v) Write the differential equation for angular S.H.M.
(vi) What is the mathematical formula for third postulate of Bohr’s atomic model?
(vii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(viii) Calculate the moment of inertia of a uniform disc of mass 10 kg and radius 60 cm about an axis perpendicular to its length and passing through its centre.
SECTION − B
Attempt any EIGHT questions of the following: [16]
Q.3. Define moment of inertia of a rotating rigid body. State its SI unit and dimensions.
Definition: Moment of inertia of a rigid body about a given axis of rotation is defined as the sum of the products of the mass of each particle of the body and the square of its perpendicular distance from the axis of rotation.
Formula: \( I = \sum_{i=1}^{n} m_i r_i^2 \)
SI Unit: kg m²
Dimensions: [M1 L2 T0]
Q.4. What are polar dielectrics and non polar dielectrics?
- Polar Dielectrics: Substances made of molecules that possess a permanent dipole moment (the center of positive charge and center of negative charge do not coincide). Example: Water (H₂O), HCl.
- Non-polar Dielectrics: Substances made of molecules where the center of positive charge coincides with the center of negative charge, resulting in zero permanent dipole moment. Example: Hydrogen (H₂), Oxygen (O₂), CO₂.
Q.5. What is a thermodynamic process? Give any two types of it.
Definition: A thermodynamic process is a procedure by which the state of a system changes from one equilibrium state to another. It involves changes in state variables like pressure, volume, and temperature.
Types (Any two):
- Isothermal Process: A process occurring at a constant temperature.
- Adiabatic Process: A process where there is no heat exchange with the surroundings.
Q.6. Derive an expression for the radius of the nth Bohr orbit of the electron in hydrogen atom.
Consider an electron of mass \( m \), charge \( e \), moving with velocity \( v_n \) in the \( n^{th} \) orbit of radius \( r_n \) around a nucleus of charge \( +Ze \) (for Hydrogen, Z=1).
1. Electrostatic force provides centripetal force:
$$ \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n^2} = \frac{mv_n^2}{r_n} \implies mv_n^2 = \frac{e^2}{4\pi\epsilon_0 r_n} \quad \dots(1) $$2. Bohr's quantization condition:
$$ mv_n r_n = \frac{nh}{2\pi} \implies v_n = \frac{nh}{2\pi m r_n} \quad \dots(2) $$Substitute eq(2) into eq(1):
$$ m \left( \frac{nh}{2\pi m r_n} \right)^2 = \frac{e^2}{4\pi\epsilon_0 r_n} $$ $$ m \frac{n^2 h^2}{4\pi^2 m^2 r_n^2} = \frac{e^2}{4\pi\epsilon_0 r_n} $$Solving for \( r_n \):
$$ r_n = \frac{\epsilon_0 n^2 h^2}{\pi m e^2} $$This is the required expression. For Hydrogen, \( r_n \propto n^2 \).
Q.7. What are harmonics and overtones (Two points)?
- Harmonics: The fundamental frequency and all its integral multiples are called harmonics. The fundamental frequency is the first harmonic.
- Overtones: The frequencies of vibration higher than the fundamental frequency that are actually present in the emitted sound are called overtones. The first frequency higher than fundamental is the first overtone.
Q.8. Distinguish between potentiometer and voltmeter.
| Potentiometer | Voltmeter |
|---|---|
| It measures the e.m.f of a cell very accurately. | It measures the terminal potential difference, which is slightly less than the actual e.m.f. |
| It works on the null deflection method (draws no current at balance). | It draws some current from the source to deflect the needle. |
Q.9. What are mechanical equilibrium and thermal equilibrium?
- Mechanical Equilibrium: A system is said to be in mechanical equilibrium if there are no unbalanced forces within the system or between the system and its surroundings. The net force and net torque on the system are zero.
- Thermal Equilibrium: A system is in thermal equilibrium if its temperature is uniform throughout and is the same as that of the surroundings. There is no net flow of heat.
Q.10. An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10–11 m with a speed of 3 × 106 m/s. Find the angular momentum of electron.
Given:
Radius \( r = 5.3 \times 10^{-11} \) m
Speed \( v = 3 \times 10^6 \) m/s
Mass of electron \( m = 9.1 \times 10^{-31} \) kg
Formula: Angular momentum \( L = mvr \)
Calculation:
\( L = (9.1 \times 10^{-31}) \times (3 \times 10^6) \times (5.3 \times 10^{-11}) \)
\( L = 9.1 \times 3 \times 5.3 \times 10^{-31+6-11} \)
\( L = 144.69 \times 10^{-36} \)
\( L = 1.4469 \times 10^{-34} \) kg m²/s
Result: Angular momentum is approximately \( 1.45 \times 10^{-34} \) kg m²/s.
Q.11. Plane wavefront of light of wavelength 6000 Å is incident on two slits on a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.
Given:
Wavelength \( \lambda = 6000 \) Å \( = 6 \times 10^{-7} \) m
Screen distance \( D = 2 \) m
Separation of 10 bright fringes means the distance occupied by 10 fringe widths (\( 10 \beta \)) is 2 cm.
\( 10 \beta = 2 \) cm \( = 0.02 \) m.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
\( 10 \left( \frac{\lambda D}{d} \right) = 0.02 \)
\( \frac{10 \times 6 \times 10^{-7} \times 2}{d} = 0.02 \)
\( \frac{120 \times 10^{-7}}{d} = 2 \times 10^{-2} \)
\( d = \frac{1.2 \times 10^{-5}}{2 \times 10^{-2}} \)
\( d = 0.6 \times 10^{-3} \) m = 0.6 mm
Result: The distance between the slits is 0.6 mm.
Q.12. Eight droplets of water each of radius 0.2 mm coalesce into a single drop. Find the decrease in the surface area.
Given:
Number of drops \( n = 8 \)
Radius of small drop \( r = 0.2 \) mm \( = 2 \times 10^{-4} \) m
Step 1: Find radius R of big drop
Volume conservation: \( \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \)
\( R^3 = 8 r^3 \implies R = 2r \)
Step 2: Calculate change in area
Initial Area \( A_1 = n \times 4\pi r^2 = 8 \times 4\pi r^2 = 32\pi r^2 \)
Final Area \( A_2 = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2 \)
Decrease \( \Delta A = A_1 - A_2 = 16\pi r^2 \)
Calculation:
\( \Delta A = 16 \times 3.142 \times (2 \times 10^{-4})^2 \)
\( \Delta A = 16 \times 3.142 \times 4 \times 10^{-8} \)
\( \Delta A = 64 \times 3.142 \times 10^{-8} \approx 201.1 \times 10^{-8} \)
\( \Delta A \approx 2.01 \times 10^{-6} \) m²
Q.13. A 0.1 H inductor, a 25 × 10–6 F capacitor and a 15 Ω resistor are connected in series to a 120 V, 50 Hz AC source. Calculate the resonant frequency.
Given:
\( L = 0.1 \) H
\( C = 25 \times 10^{-6} \) F
Formula: Resonant frequency \( f_r = \frac{1}{2\pi\sqrt{LC}} \)
Calculation:
\( LC = 0.1 \times 25 \times 10^{-6} = 2.5 \times 10^{-6} \)
\( \sqrt{LC} = \sqrt{2.5} \times 10^{-3} \approx 1.581 \times 10^{-3} \)
\( f_r = \frac{1}{2 \times 3.142 \times 1.581 \times 10^{-3}} \)
\( f_r = \frac{1000}{6.284 \times 1.581} = \frac{1000}{9.935} \approx 100.65 \) Hz
Note: The problem mentions a "50 Hz AC source" but asks to calculate the "resonant frequency". The resonant frequency depends only on L and C, not the source frequency. The resonant frequency is approximately 100.65 Hz.
Q.14. The difference between the two molar specific heats of a gas is 9000 J/kg K. If the ratio of the two specific heats is 1.5, calculate the two molar specific heats.
Correction Note: The question states "molar specific heats" but gives the unit "J/kg K", which corresponds to specific heat per unit mass (principal specific heat), denoted by \( c_p \) and \( c_v \). Molar specific heat differences are typically around 8.314 J/mol K. The value 9000 suggests we are dealing with specific heats per unit mass (likely Hydrogen). We will solve for \( c_p \) and \( c_v \).
Given:
Difference: \( c_p - c_v = 9000 \) J/kg K
Ratio: \( \gamma = \frac{c_p}{c_v} = 1.5 \implies c_p = 1.5 c_v \)
Calculation:
Substitute \( c_p \) in the difference equation:
\( 1.5 c_v - c_v = 9000 \)
\( 0.5 c_v = 9000 \)
\( c_v = \frac{9000}{0.5} = 18000 \) J/kg K
Now find \( c_p \):
\( c_p = 1.5 \times 18000 = 27000 \) J/kg K
Result:
\( c_v = 18000 \) J/kg K
\( c_p = 27000 \) J/kg K
SECTION − C
Attempt any EIGHT questions of the following: [24]
Q.15. With the help of a neat diagram, explain the reflection of light on a plane reflecting surface.
Explanation based on Wave Theory (Huygens' Principle):
- Consider a plane wavefront AB incident obliquely on a plane reflecting surface MN.
- Let the wavefront touch the surface at point A at time t=0. At this instant, point B is at a distance 'ct' from the surface (where c is the speed of light).
- According to Huygens' principle, point A acts as a secondary source and emits spherical secondary wavelets in the medium. In time 't', the wavelet from A travels a distance 'ct' (radius).
- Meanwhile, the point B of the incident wavefront moves forward and touches the surface at point C after time t (Distance BC = ct).
- If we draw a tangent from C to the secondary wavelet originating from A, we get the plane CD, which represents the reflected wavefront.
- By geometry of congruent triangles (Triangle ABC and Triangle ADC), it can be proved that the angle of incidence (i) equals the angle of reflection (r).
Q.16. What is magnetization, magnetic intensity and magnetic susceptibility?
- Magnetization (M): It is defined as the net magnetic dipole moment per unit volume of a material. \( M = \frac{m_{net}}{V} \). Unit: A/m.
- Magnetic Intensity (H): It is a quantity representing the strength of the external magnetic field that induces magnetism in a material. \( H = \frac{B}{\mu} \) (in linear media). Unit: A/m.
- Magnetic Susceptibility (\( \chi \)): It is the measure of how easily a substance can be magnetized. It is defined as the ratio of magnetization (M) to the magnetic intensity (H). \( \chi = \frac{M}{H} \). It is a dimensionless quantity.
Q.17. Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.
Let two sound waves with frequencies \( n_1 \) and \( n_2 \) (where \( n_1 > n_2 \)) and same amplitude \( A \) be represented as:
$$ y_1 = A \sin(2\pi n_1 t) $$ $$ y_2 = A \sin(2\pi n_2 t) $$By superposition principle, resultant displacement \( y = y_1 + y_2 \):
$$ y = A [\sin(2\pi n_1 t) + \sin(2\pi n_2 t)] $$Using trigonometric identity \( \sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2}) \):
$$ y = 2A \cos\left(2\pi \frac{n_1 - n_2}{2} t\right) \sin\left(2\pi \frac{n_1 + n_2}{2} t\right) $$This represents a wave with resultant amplitude \( R = 2A \cos(2\pi \frac{n_1 - n_2}{2} t) \).
Intensity is maximum when \( \cos(...) = \pm 1 \). This occurs when the phase is \( k\pi \).
Time interval between two successive maxima is \( T_{beat} = \frac{1}{n_1 - n_2} \).
Beat frequency \( f_{beat} = \frac{1}{T_{beat}} = n_1 - n_2 \).
Q.18. Define:
(a) Inductive reactance
(b) Capacitive reactance
(c) Impedance
- (a) Inductive Reactance (\( X_L \)): The opposition offered by an inductor to the flow of alternating current. \( X_L = \omega L = 2\pi f L \). Unit: Ohm.
- (b) Capacitive Reactance (\( X_C \)): The opposition offered by a capacitor to the flow of alternating current. \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \). Unit: Ohm.
- (c) Impedance (Z): The effective opposition offered by an AC circuit (containing combinations of resistor, inductor, and capacitor) to the flow of current. \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Unit: Ohm.
Q.19. Derive an expression for the kinetic energy of a body rotating with a uniform angular speed.
Consider a rigid body rotating about an axis with uniform angular speed \( \omega \).
Let the body consist of \( n \) particles of masses \( m_1, m_2, \dots, m_n \) at distances \( r_1, r_2, \dots, r_n \) from the axis of rotation.
The linear velocity of the \( i^{th} \) particle is \( v_i = r_i \omega \).
The kinetic energy of the \( i^{th} \) particle is \( E_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2 \).
Total Rotational Kinetic Energy \( E_{rot} \) is the sum of K.E. of all particles:
$$ E_{rot} = \sum_{i=1}^{n} \frac{1}{2} m_i r_i^2 \omega^2 $$ $$ E_{rot} = \frac{1}{2} \omega^2 \left( \sum_{i=1}^{n} m_i r_i^2 \right) $$Since \( \sum m_i r_i^2 = I \) (Moment of Inertia):
$$ E_{rot} = \frac{1}{2} I \omega^2 $$Q.20. Derive an expression for emf (e) generated in a conductor of length (l) moving in uniform magnetic field (B) with uniform velocity (v) along x-axis.
Consider a straight conductor of length \( l \) moving with velocity \( v \) in a uniform magnetic field \( B \). Let \( B \), \( l \), and \( v \) be mutually perpendicular.
Lorentz Force Method:
A free charge \( q \) inside the conductor experiences a magnetic force \( F_m = q(v \times B) \). Magnitude \( F_m = qvB \).
This force pushes electrons to one end, creating a potential difference and an electric field \( E \).
At equilibrium, Electric Force = Magnetic Force.
\( qE = qvB \implies E = vB \).
The induced EMF \( e \) is the potential difference, which is \( e = E \times l \).
Thus, \( e = Blv \).
Q.21. Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.
Consider a sphere of radius \( r \) and density \( \rho \) falling through a fluid of density \( \sigma \) and viscosity \( \eta \).
Forces acting on the sphere at terminal velocity \( v \):
- Weight acting downwards: \( W = \frac{4}{3}\pi r^3 \rho g \)
- Upthrust (Buoyancy) acting upwards: \( F_u = \frac{4}{3}\pi r^3 \sigma g \)
- Viscous drag force acting upwards (Stokes' Law): \( F_v = 6\pi \eta r v \)
At equilibrium (terminal velocity): Downward Force = Upward Forces
$$ W = F_u + F_v $$ $$ \frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v $$ $$ 6\pi \eta r v = \frac{4}{3}\pi r^3 g (\rho - \sigma) $$ $$ v = \frac{4\pi r^3 g (\rho - \sigma)}{3 \times 6\pi \eta r} $$ $$ v = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta} $$Q.22. Determine the shortest wavelengths of Balmer and Paschen series. Given the limit for Lyman series is 912 Å.
Given:
Limit for Lyman series (shortest wavelength, \( n_1=1, n_2=\infty \)):
\( \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty} \right) = R \).
\( \lambda_L = \frac{1}{R} = 912 \) Å.
1. Shortest wavelength of Balmer Series (\( n_1=2, n_2=\infty \)):
\( \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \right) = \frac{R}{4} \)
\( \lambda_B = \frac{4}{R} = 4 \times 912 = 3648 \) Å.
2. Shortest wavelength of Paschen Series (\( n_1=3, n_2=\infty \)):
\( \frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{\infty} \right) = \frac{R}{9} \)
\( \lambda_P = \frac{9}{R} = 9 \times 912 = 8208 \) Å.
Q.23. Calculate the value of magnetic field at a distance of 3 cm from a very long, straight wire carrying a current of 6A.
Given:
Current \( I = 6 \) A
Distance \( r = 3 \) cm \( = 0.03 \) m
\( \mu_0 = 4\pi \times 10^{-7} \) T m/A
Formula: \( B = \frac{\mu_0 I}{2\pi r} \)
Calculation:
\( B = \frac{4\pi \times 10^{-7} \times 6}{2\pi \times 0.03} \)
\( B = \frac{2 \times 10^{-7} \times 6}{0.03} \)
\( B = \frac{12 \times 10^{-7}}{3 \times 10^{-2}} \)
\( B = 4 \times 10^{-5} \) T
Q.24. A parallel plate capacitor filled with air has an area of 6 cm2 and plate separation of a 3 mm. Calculate its capacitance.
Given:
Area \( A = 6 \text{ cm}^2 = 6 \times 10^{-4} \text{ m}^2 \)
Separation \( d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m} \)
\( \epsilon_0 = 8.85 \times 10^{-12} \) F/m
Formula: \( C = \frac{\epsilon_0 A}{d} \)
Calculation:
\( C = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-4}}{3 \times 10^{-3}} \)
\( C = 8.85 \times 10^{-12} \times 2 \times 10^{-1} \)
\( C = 17.7 \times 10^{-13} \) F
\( C = 1.77 \times 10^{-12} \) F = 1.77 pF
Q.25. An emf of 91 mV is induced in the windings of a coil, when the current in a nearby coil is increasing at the rate of 1.3 A/s, what is the mutual inductance (M) of the two coils in mH?
Given:
Induced EMF \( |e| = 91 \) mV \( = 91 \times 10^{-3} \) V
Rate of change of current \( \frac{di}{dt} = 1.3 \) A/s
Formula: \( |e| = M \frac{di}{dt} \)
Calculation:
\( 91 \times 10^{-3} = M \times 1.3 \)
\( M = \frac{91 \times 10^{-3}}{1.3} \)
\( M = \frac{91}{1.3} \times 10^{-3} = 70 \times 10^{-3} \) H
\( M = 70 \) mH
Q.26. Two cells of emf 4V and 2V having respective internal resistance of 1 Ω and 2 Ω are connected in parallel, so as to send current in the same direction through an external resistance of 5 Ω. Find the current through the external resistance.
Given:
Cell 1: \( E_1 = 4 \) V, \( r_1 = 1 \, \Omega \)
Cell 2: \( E_2 = 2 \) V, \( r_2 = 2 \, \Omega \)
External Resistance: \( R = 5 \, \Omega \)
Equivalent EMF and Resistance (Parallel Combination):
\( E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{4}{1} + \frac{2}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{4 + 1}{1 + 0.5} = \frac{5}{1.5} = \frac{10}{3} \) V.
\( r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \, \Omega \).
Total Current (I):
\( I = \frac{E_{eq}}{R + r_{eq}} \)
\( I = \frac{10/3}{5 + 2/3} = \frac{10/3}{17/3} \)
\( I = \frac{10}{17} \) A \( \approx 0.588 \) A.
SECTION − D
Attempt any THREE questions of the following: [12]
Q.27. Derive an expression for a pressure exerted by a gas on the basis of kinetic theory of gases.
Consider an ideal gas enclosed in a cube of side \( L \). Volume \( V = L^3 \). Let there be \( N \) molecules, each of mass \( m \).
Consider a molecule moving with velocity \( v_1 \) having components \( v_{x1}, v_{y1}, v_{z1} \). It hits the wall perpendicular to x-axis.
Momentum before collision: \( mv_{x1} \).
Momentum after collision (elastic): \( -mv_{x1} \).
Change in momentum: \( \Delta p = -2mv_{x1} \). Momentum transferred to wall = \( 2mv_{x1} \).
Time between successive collisions with same wall: \( \Delta t = \frac{2L}{v_{x1}} \).
Force by single molecule: \( f = \frac{\Delta p}{\Delta t} = \frac{2mv_{x1}}{2L/v_{x1}} = \frac{mv_{x1}^2}{L} \).
Total force by N molecules: \( F_x = \frac{m}{L} \sum v_{xi}^2 \).
Pressure \( P = \frac{F}{A} = \frac{F}{L^2} = \frac{m}{L^3} \sum v_{xi}^2 = \frac{m}{V} \sum v_{xi}^2 \).
Since gas is isotropic, average velocity components are equal: \( \overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3} \overline{v^2} \).
Also \( \sum v_{xi}^2 = N \overline{v_x^2} = \frac{N}{3} \overline{v^2} \) (where \( \overline{v^2} \) is mean square speed).
$$ P = \frac{1}{3} \frac{Nm}{V} \overline{v^2} $$Or \( P = \frac{1}{3} \rho v_{rms}^2 \).
Q.28. What is a rectifier? With the help of a neat circuit diagram, explain the working of a half wave rectifier.
Rectifier: A device that converts alternating current (AC) into direct current (DC).
Half Wave Rectifier Working:
- Circuit: Consists of a transformer, a p-n junction diode, and a load resistor \( R_L \) connected in series.
- Positive Half Cycle: When the input AC voltage is positive, the diode is forward biased and conducts current. Voltage appears across \( R_L \).
- Negative Half Cycle: When the input is negative, the diode is reverse biased and does not conduct. No current flows through \( R_L \).
- Output: The output is a pulsating DC voltage present only during positive half cycles.
Q.29. Draw a neat, labelled diagram of a suspended coil type moving coil galvanometer.
The initial pressure and volume of a gas enclosed in a cylinder are 2 × 105 N/m2 and 6 × 10–3 m3 respectively. If the work done in compressing the gas at constant pressure is 150 J, find the final volume of the gas.
Part 1: [Diagram Required: Suspension wire, Mirror, Rectangular Coil, Horseshoe Magnet, Soft Iron Core, Spring].
Part 2: Calculation
Given:
Pressure \( P = 2 \times 10^5 \) N/m²
Initial Volume \( V_1 = 6 \times 10^{-3} \) m³
Work done in compressing \( W_{ext} = 150 \) J. (This implies work done on the gas is +150 J, or work done by the gas is -150 J).
Formula: Work done by gas \( W = P(V_2 - V_1) \)
Calculation:
\( -150 = 2 \times 10^5 (V_2 - 6 \times 10^{-3}) \)
\( V_2 - 6 \times 10^{-3} = \frac{-150}{2 \times 10^5} \)
\( V_2 - 6 \times 10^{-3} = -75 \times 10^{-5} \)
\( V_2 - 6 \times 10^{-3} = -0.75 \times 10^{-3} \)
\( V_2 = 6 \times 10^{-3} - 0.75 \times 10^{-3} \)
\( V_2 = 5.25 \times 10^{-3} \) m³
Q.30. Define second’s pendulum. Derive a formula for the length of second’s pendulum.
A particle performing linear S.H.M. has maximum velocity 25 cm/s and maximum acceleration 100 cm/s2. Find period of oscillations.
Part 1: Second's Pendulum
Definition: A simple pendulum whose time period is exactly 2 seconds.
Derivation: Time period of simple pendulum \( T = 2\pi \sqrt{\frac{L}{g}} \).
For second's pendulum, \( T = 2 \).
\( 2 = 2\pi \sqrt{\frac{L_s}{g}} \implies 1 = \pi \sqrt{\frac{L_s}{g}} \)
\( 1 = \pi^2 \frac{L_s}{g} \implies L_s = \frac{g}{\pi^2} \).
Part 2: Problem
Given: \( v_{max} = 25 \) cm/s, \( a_{max} = 100 \) cm/s².
Formulas: \( v_{max} = A\omega \) and \( a_{max} = A\omega^2 \).
\( \frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega \)
\( \omega = \frac{100}{25} = 4 \) rad/s.
Period \( T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57 \) s.
Q.31. Explain de Broglie wavelength. Obtain an expression for de Broglie wavelength of wave associated with material particles.
The photoelectric work function for a metal is 4.2 eV. Find the threshold wavelength.
Part 1: de Broglie Wavelength
Explanation: According to de Broglie, every moving material particle is associated with a wave, called matter wave or de Broglie wave.
Expression: From Planck's quantum theory, energy of photon \( E = h\nu = \frac{hc}{\lambda} \).
From Einstein's mass-energy relation, \( E = mc^2 \).
Equating both: \( \frac{hc}{\lambda} = mc^2 \implies \lambda = \frac{h}{mc} \).
For a particle of mass m moving with velocity v, momentum \( p = mv \).
Thus, \( \lambda = \frac{h}{p} = \frac{h}{mv} \).
Part 2: Problem
Given: Work function \( W_0 = 4.2 \) eV.
Conversion: \( W_0 = 4.2 \times 1.6 \times 10^{-19} \) J.
Formula: \( W_0 = \frac{hc}{\lambda_0} \implies \lambda_0 = \frac{hc}{W_0} \).
\( h = 6.63 \times 10^{-34} \), \( c = 3 \times 10^8 \).
Calculation:
\( \lambda_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.2 \times 1.6 \times 10^{-19}} \)
\( \lambda_0 = \frac{19.89 \times 10^{-26}}{6.72 \times 10^{-19}} \)
\( \lambda_0 \approx 2.96 \times 10^{-7} \) m \( = 2960 \) Å.