Physics Paper 2 This Paper
Physics Paper 3
Physics Paper 4
Model Question Paper Set - II: Physics
Time: 3 Hours | Total Marks: 70
SECTION – I
i. On being churned the butter separates out of milk due to _______.
ii. If the earth stops rotating, the value of ‘g’ at the equator will _______.
Reason: At the equator, effective gravity is \( g' = g - R\omega^2 \). If rotation stops (\( \omega = 0 \)), \( g' \) becomes \( g \), which is an increase.
iii. If a person, sitting on a rotating table, with his arms outstretched and holding heavy dumb bells in each hand, suddenly lowers his hands, then _______.
Reason: When arms are lowered, mass distribution shifts closer to the axis of rotation, decreasing the moment of inertia \( I \). (Note: By conservation of angular momentum \( L = I\omega \), since \( I \) decreases, \( \omega \) increases, but option D is the direct physical change).
iv. The average kinetic energy of a gas molecule is _______.
Reason: \( K.E. \propto T \).
v. The time period of a spring of force constant k loaded with mass m is _______.
Reason: \( T = 2\pi\sqrt{\frac{m}{k}} \).
vi. Speed of sound in air is 300 m/s. The distance between two successive nodes of a stationary wave of frequency 1000 Hz is _______.
Calculation: \( \lambda = \frac{v}{n} = \frac{300}{1000} = 0.3 \, \text{m} \). Distance between successive nodes = \( \frac{\lambda}{2} = \frac{0.3}{2} = 0.15 \, \text{m} = 15 \, \text{cm} \).
vii. If a wave enters from air to water, then what remains unchanged?
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 2022 - English Medium View
- 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
- Physics - July 2016 View Answer Key
- Physics - March 2017 View
- Physics - July 2017 View
i. What is banking of road? Explain the necessity of banking of the road.
Necessity:
- When a vehicle moves along a curved path, the necessary centripetal force is provided by the friction between the tyres and the road.
- However, friction is not reliable; it decreases when the road is wet or oily.
- To provide a reliable centripetal force independent of friction, roads are banked. This allows the horizontal component of the normal reaction to provide the necessary centripetal force.
ii. A satellite orbits around the earth at a height equal to R of the earth. Find its period. [R = 6.4 × 106 m, g = 9.8 m/s2]
Height \( h = R \)
Orbital radius \( r = R + h = R + R = 2R \)
Formula for period: \( T = 2\pi \sqrt{\frac{r^3}{GM}} \)
Since \( GM = gR^2 \), we have:
\( T = 2\pi \sqrt{\frac{(2R)^3}{gR^2}} = 2\pi \sqrt{\frac{8R^3}{gR^2}} = 2\pi \sqrt{\frac{8R}{g}} \)
Substituting values:
\( T = 2 \times 3.142 \times \sqrt{\frac{8 \times 6.4 \times 10^6}{9.8}} \)
\( T = 6.284 \times \sqrt{\frac{51.2}{9.8} \times 10^6} \)
\( T = 6.284 \times \sqrt{5.224} \times 10^3 \)
\( T = 6.284 \times 2.285 \times 1000 \approx 14360 \, \text{seconds} \)
\( T \approx 4 \, \text{hours} \).
iii. Four particles of masses 4 kg, 2 kg, 2 kg and 4 kg are placed at corners A, B, C and D of a square ABCD of side 2 m. Find M.I. and radius of gyration of system about AD as axis of rotation.
Particles at A (4 kg) and D (4 kg) lie on the axis, so their distance \( r = 0 \).
Particles at B (2 kg) and C (2 kg) are at a distance equal to the side of the square \( r = 2 \, \text{m} \).
Moment of Inertia (I):
\( I = \sum mr^2 = m_A(0)^2 + m_D(0)^2 + m_B(2)^2 + m_C(2)^2 \)
\( I = 0 + 0 + 2(4) + 2(4) = 8 + 8 = 16 \, \text{kg m}^2 \)
Radius of Gyration (K):
Total Mass \( M = 4 + 2 + 2 + 4 = 12 \, \text{kg} \)
\( K = \sqrt{\frac{I}{M}} = \sqrt{\frac{16}{12}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \, \text{m} \approx 1.155 \, \text{m} \).
iv. Define the following terms: a. Phase of S.H.M b. Epoch of S.H.M
b. Epoch of S.H.M: The phase of the particle performing S.H.M. at the start (at time t = 0) is called the Epoch or initial phase constant \( (\phi) \).
v. Find the greatest length of a metal wire that can hang vertically without breaking. [Breaking stress for metal = 6 × 108 N/m2 and density of metal = 9 × 103 kg/m3]
The weight of the wire acts as the force causing stress at the top point.
Force \( F = mg = (\text{Volume} \times \rho)g = (A \times L \times \rho)g \)
Stress \( = \frac{F}{A} = \frac{A L \rho g}{A} = L \rho g \)
For breaking condition:
\( L \rho g = \text{Breaking Stress} \)
\( L = \frac{\text{Breaking Stress}}{\rho g} \)
\( L = \frac{6 \times 10^8}{9 \times 10^3 \times 9.8} \)
\( L = \frac{60000}{88.2} \approx 680.27 \, \text{m} \).
vi. A capillary tube 0.12 mm in diameter has its lower end immersed in a liquid of surface tension 0.054 N/m. If the density of liquid is 860 kg/m3, find the height to which the liquid rises in the tube. [Angle of contact = 30°, g = 9.8 m/s2]
\( d = 0.12 \, \text{mm} \Rightarrow r = 0.06 \, \text{mm} = 6 \times 10^{-5} \, \text{m} \)
\( T = 0.054 \, \text{N/m} \), \( \rho = 860 \, \text{kg/m}^3 \), \( \theta = 30^\circ \)
Formula: \( h = \frac{2T \cos \theta}{r \rho g} \)
\( h = \frac{2 \times 0.054 \times \cos 30^\circ}{6 \times 10^{-5} \times 860 \times 9.8} \)
\( h = \frac{0.108 \times 0.866}{0.50568} \)
\( h = \frac{0.0935}{0.50568} \approx 0.1849 \, \text{m} \).
vii. Explain the reflection of transverse waves from a denser medium.
- The particle at the boundary is not free to vibrate.
- The incident wave exerts a force on the boundary, and by Newton's third law, the boundary exerts an equal and opposite reaction force on the medium.
- This causes a reversal in the phase of vibration.
- A crest is reflected as a trough, and a trough is reflected as a crest.
- There is a phase change of \( \pi \) radians (or 180°) between the incident and reflected waves.
viii. State and derive law of linear density for a vibrating string.
Derivation:
The fundamental frequency of a vibrating string is given by:
\( n = \frac{1}{2l} \sqrt{\frac{T}{m}} \)
where \( n \) = frequency, \( l \) = vibrating length, \( T \) = tension, \( m \) = linear density.
If \( l \) and \( T \) are constant, then \( \frac{1}{2l}\sqrt{T} \) is a constant \( C \).
Therefore, \( n = C \times \frac{1}{\sqrt{m}} \)
\( \Rightarrow n \propto \frac{1}{\sqrt{m}} \).
i. A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 4 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.2. Will the cyclist slip while taking the turn? Calculate maximum safety speed? Will it be constant always?
Speed of cyclist \( v = 18 \, \text{km/h} = 18 \times \frac{5}{18} = 5 \, \text{m/s} \).
Radius \( r = 4 \, \text{m} \).
Coefficient of friction \( \mu = 0.2 \).
Maximum safety speed calculation:
\( v_{max} = \sqrt{\mu rg} \)
\( v_{max} = \sqrt{0.2 \times 4 \times 9.8} = \sqrt{7.84} = 2.8 \, \text{m/s} \).
Will the cyclist slip?
Since the cyclist's speed (5 m/s) is greater than the maximum safety speed (2.8 m/s), the cyclist will slip.
Will it be constant always?
No, the maximum safety speed depends on the coefficient of friction (\( \mu \)), which can change depending on road conditions (wet, dry, oily) and tyre conditions.
ii. State and prove law of conservation of angular momentum. Give 2 examples.
Proof:
Angular momentum \( \vec{L} = \vec{r} \times \vec{p} \).
Differentiating with respect to time:
\( \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) \)
\( \frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} + \frac{d\vec{r}}{dt} \times \vec{p} \)
Since \( \vec{v} = \frac{d\vec{r}}{dt} \), \( \vec{p} = m\vec{v} \), and \( \vec{F} = \frac{d\vec{p}}{dt} \):
\( \frac{d\vec{L}}{dt} = \vec{r} \times \vec{F} + \vec{v} \times (m\vec{v}) \)
Since \( \vec{v} \times \vec{v} = 0 \), and Torque \( \vec{\tau} = \vec{r} \times \vec{F} \):
\( \frac{d\vec{L}}{dt} = \vec{\tau} \)
If external torque \( \vec{\tau} = 0 \), then \( \frac{d\vec{L}}{dt} = 0 \), implying \( \vec{L} = \text{constant} \).
Examples:
1. A ballet dancer brings her arms close to her body to increase her angular speed of rotation.
2. Divers curl their bodies to decrease moment of inertia and increase spin rate while diving.
iii. A brass wire of radius 2 mm is loaded by a mass of 32.8 kg. What would be the decrease in its radius? [Y = 9 × 1010 N/m2, Poisson’s ratio σ = 0.36]
\( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
\( M = 32.8 \, \text{kg} \)
\( Y = 9 \times 10^{10} \, \text{N/m}^2 \)
\( \sigma = 0.36 \)
Longitudinal Stress \( = \frac{F}{A} = \frac{Mg}{\pi r^2} \)
Longitudinal Strain \( l = \frac{\text{Stress}}{Y} = \frac{Mg}{\pi r^2 Y} \)
Lateral Strain \( = \frac{\Delta r}{r} \)
Poisson's ratio \( \sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} \)
\( \text{Lateral Strain} = \sigma \times \frac{Mg}{\pi r^2 Y} \)
Change in radius \( \Delta r = r \times \text{Lateral Strain} = r \times \sigma \times \frac{Mg}{\pi r^2 Y} = \frac{\sigma Mg}{\pi r Y} \)
Substituting values:
\( \Delta r = \frac{0.36 \times 32.8 \times 9.8}{3.142 \times 2 \times 10^{-3} \times 9 \times 10^{10}} \)
\( \Delta r = \frac{115.7184}{56.556 \times 10^7} \)
\( \Delta r \approx 2.046 \times 10^{-7} \, \text{m} \).
iv. What is a heat engine? Explain working and efficiency of heat engine.
Working:
- It consists of a source at high temperature \( T_1 \) and a sink at low temperature \( T_2 \).
- The working substance absorbs heat \( Q_1 \) from the source.
- It performs mechanical work \( W \).
- It rejects the remaining heat \( Q_2 \) to the sink.
- The net work done is \( W = Q_1 - Q_2 \).
It is the ratio of work done by the engine to the heat absorbed from the source.
\( \eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} \).
A. Two organ pipes, open at both ends, are sounded together and 5 beats are heard per second. The length of shorter pipe is 0.24 m, find the length of the other pipe. (Given: velocity of sound in air = 350 m/s, end correction at one end = 0.02 m same for both pipes) [4]
End correction per end \( e = 0.02 \, \text{m} \). Total correction \( 2e = 0.04 \, \text{m} \).
Corrected length \( l_1 = L_1 + 2e = 0.24 + 0.04 = 0.28 \, \text{m} \).
Frequency \( n_1 = \frac{v}{2l_1} = \frac{350}{2 \times 0.28} = \frac{350}{0.56} = 625 \, \text{Hz} \).
Beats \( = |n_1 - n_2| = 5 \).
Since \( L_1 \) is the shorter pipe, \( n_1 > n_2 \).
\( n_2 = n_1 - 5 = 625 - 5 = 620 \, \text{Hz} \).
For pipe 2:
\( n_2 = \frac{v}{2l_2} \)
\( 620 = \frac{350}{2(L_2 + 2e)} \)
\( L_2 + 0.04 = \frac{350}{2 \times 620} = \frac{350}{1240} \approx 0.2822 \, \text{m} \).
\( L_2 = 0.2822 - 0.04 = 0.2422 \, \text{m} \).
Length of the other pipe is approx 0.242 m.
B. Explain surface tension on the basis of molecular theory. [3]
- Molecules deep inside liquid: A molecule well inside the liquid is surrounded by other molecules on all sides. The cohesive forces acting on it are balanced, so the net force is zero.
- Molecules on surface: A molecule in the surface film (thickness equal to molecular range) is surrounded by liquid molecules below and vapor/air molecules above.
- Net Downward Force: The density of liquid is much higher than air/vapor. Thus, the number of liquid molecules in the lower half of the sphere of influence is much greater than air molecules in the upper half. This results in a strong net downward cohesive force.
- Potential Energy: To take a molecule from the interior to the surface, work must be done against this downward force. This work is stored as potential energy.
- Minimum Energy State: Systems tend to minimize potential energy. To do so, the liquid tries to minimize the number of molecules on the surface, thereby minimizing the surface area. This tendency of the surface to contract behaves like a stretched elastic membrane, known as surface tension.
Q.4. A. Discuss analytically the composition of two S.H.M’s of same period and parallel to each other. Obtain their resultant amplitude. Also find the resultant amplitude when phase difference of two S.H.M’s is a. 0 b. π [4]
\( x_1 = A_1 \sin(\omega t + \phi_1) \)
\( x_2 = A_2 \sin(\omega t + \phi_2) \)
Resultant displacement \( x = x_1 + x_2 \)
Expanding and combining using trigonometry, the resultant amplitude \( R \) is given by:
\( R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\phi_1 - \phi_2)} \)
Special Cases:
Let phase difference \( \delta = \phi_1 - \phi_2 \).
a. When \( \delta = 0 \):
\( \cos 0 = 1 \).
\( R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2} = \sqrt{(A_1 + A_2)^2} = A_1 + A_2 \). (Maximum Amplitude)
b. When \( \delta = \pi \):
\( \cos \pi = -1 \).
\( R = \sqrt{A_1^2 + A_2^2 - 2A_1 A_2} = \sqrt{(A_1 - A_2)^2} = |A_1 - A_2| \). (Minimum Amplitude).
B. Find the height of a geostationary satellite (communication satellite) from the surface of the earth. [Mass of the earth = 6 × 1024 kg, radius of the earth = 6400 km, G = 6.67 × 10−11 N.m2/kg2] [3]
From Kepler's law: \( T^2 = \frac{4\pi^2}{GM} r^3 \)
\( r^3 = \frac{GMT^2}{4\pi^2} \)
\( r^3 = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times (86400)^2}{4 \times (3.142)^2} \)
\( r^3 \approx 7.49 \times 10^{22} \)
\( r \approx 4.217 \times 10^7 \, \text{m} = 42170 \, \text{km} \).
Height from surface \( h = r - R \)
\( h = 42170 - 6400 = 35770 \, \text{km} \).
(Approx 36,000 km).
SECTION – II
i. The plane of vibration and the plane of polarisation of a beam of light _______.
ii. To obtain pronounced diffraction with single slit illuminated by light, the slit width should be of the order of _______.
Reason: Slit width must be comparable to wavelength of light (\( \sim 10^{-7} \) m).
iii. Dimensions of electric flux is _______.
Reason: \( \Phi = EA = \frac{F}{q} A = \frac{MLT^{-2}}{AT} L^2 = M L^3 T^{-3} A^{-1} \).
iv. The accuracy of M.C.G can be increased by _______.
Reason: Percentage error is minimized when the deflection reading is large.
v. For a given transistor \( \beta_{dc} \) is 99, then the corresponding value of \( \alpha_{dc} \) for same transistor is _______.
Calculation: \( \alpha = \frac{\beta}{1+\beta} = \frac{99}{100} = 0.99 \).
vi. The energy equivalent of mass 1 u is _______.
vii. A modem acts as a _______ in receiving mode.
i. Explain why microscopes of high magnifying power have oil-immersion objectives.
- The resolving power of a microscope is given by \( R.P. = \frac{2\mu \sin \theta}{\lambda} \), where \( \mu \) is the refractive index of the medium between the object and objective.
- Oil immersion objectives use an oil of high refractive index (\( \mu \approx 1.5 \)) instead of air (\( \mu = 1 \)).
- This increases the numerical aperture (\( \mu \sin \theta \)), thereby increasing the resolving power and brightness of the image at high magnifications.
ii. Write any two points of difference between amplitude modulation and frequency modulation.
| Amplitude Modulation (AM) | Frequency Modulation (FM) |
|---|---|
| Amplitude of carrier wave changes according to modulating signal. Frequency remains constant. | Frequency of carrier wave changes according to modulating signal. Amplitude remains constant. |
| More susceptible to noise. | Less susceptible to noise (Better quality). |
iii. Four resistances 5Ω, 5Ω, 5Ω and 15Ω form a Wheatstone’s network. Find the resistance which when connected across the 15Ω resistance, will balance the network.
For balance condition: \( \frac{P}{Q} = \frac{R}{S_{new}} \).
\( \frac{5}{5} = \frac{5}{S_{new}} \Rightarrow 1 = \frac{5}{S_{new}} \Rightarrow S_{new} = 5 \, \Omega \).
Currently we have \( S = 15 \, \Omega \). To make it \( 5 \, \Omega \), we must connect a resistor \( X \) in parallel with 15.
\( \frac{1}{S_{new}} = \frac{1}{15} + \frac{1}{X} \)
\( \frac{1}{5} = \frac{1}{15} + \frac{1}{X} \)
\( \frac{1}{X} = \frac{1}{5} - \frac{1}{15} = \frac{3-1}{15} = \frac{2}{15} \)
\( X = 7.5 \, \Omega \).
iv. A solenoid of 100 turns per unit length and cross-sectional area 2 × 10−4 m2 carries a current of 6 A. It is placed in horizontal axis at 30° with direction of uniform magnetic field of 0.3 T. Calculate magnetic moment of solenoid and torque experienced by solenoid due to the field.
Note: Since total length is not given, we calculate quantities per unit length.
Magnetic Moment per unit length (\( m \)):
\( m = nIA = 100 \times 6 \times 2 \times 10^{-4} = 0.12 \, \text{A m}^2 \).
Torque per unit length (\( \tau \)):
\( \tau = mB \sin \theta \)
\( \tau = 0.12 \times 0.3 \times \sin 30^\circ \)
\( \tau = 0.036 \times 0.5 = 0.018 \, \text{Nm} \).
v. An alternating emf E = 250 sinωt V is connected to a 1250 Ω resistor. Calculate the rms current through the resistor and the average power dissipated in one cycle.
Peak Current \( I_0 = \frac{E_0}{R} = \frac{250}{1250} = 0.2 \, \text{A} \).
RMS Current \( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.2}{1.414} \approx 0.1414 \, \text{A} \).
Average Power \( P = I_{rms}^2 R \) or \( \frac{E_0 I_0}{2} \).
\( P = \frac{250 \times 0.2}{2} = \frac{50}{2} = 25 \, \text{W} \).
vi. Define cut off potential. Show graphically variation of photoelectric current with collector plate potential for different intensity of incident radiation.
Graph:
(Imagine a graph where X-axis is Potential and Y-axis is Current).
- Curves for intensities \( I_1, I_2, I_3 \) (where \( I_3 > I_2 > I_1 \)) start from the same negative point on X-axis (Stopping Potential \( V_0 \)).
- Current increases with potential and saturates at different levels (Saturation currents \( i_3 > i_2 > i_1 \)).
vii. Calculate the de Broglie wavelength of proton if it is moving with speed of 8 × 106 m/s. [mp = 1.67 × 10−27 kg].
\( m = 1.67 \times 10^{-27} \, \text{kg} \).
\( h = 6.63 \times 10^{-34} \, \text{Js} \).
\( \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 8 \times 10^6} \)
\( \lambda = \frac{6.63}{13.36} \times 10^{-34+27-6} \)
\( \lambda \approx 0.496 \times 10^{-13} \, \text{m} \).
viii. Explain I – V characteristics of zener diode with suitable graph.
Reverse Bias:
- Ideally, current is negligible (leakage current) until breakdown voltage (\( V_z \)) is reached.
- At \( V_z \), current increases sharply with very small change in voltage.
- This breakdown region allows Zener diode to act as a voltage regulator.
i. State and prove Gauss’ theorem of electrostatics.
Proof: Consider a charge +q at center O of a sphere of radius r.
Electric field at surface \( E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \).
Flux through small area \( dS \): \( d\Phi = \vec{E} \cdot d\vec{S} = E dS \cos 0 = E dS \).
Total flux \( \Phi = \oint E dS = E \oint dS = E (4\pi r^2) \).
\( \Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} (4\pi r^2) = \frac{q}{\varepsilon_0} \).
ii. State the principle of working of cyclotron. Also calculate how rapidly should the electric field between the Dees of cyclotron be reversed, if magnetic field of 1.8 Wb/m2 is used to accelerate protons. [charge on proton = 1.6 × 10−19 C, mass of proton = 1.67 × 10−27 kg]
Calculation:
Reversal frequency is the cyclotron frequency \( f \).
\( f = \frac{Bq}{2\pi m} \)
\( f = \frac{1.8 \times 1.6 \times 10^{-19}}{2 \times 3.142 \times 1.67 \times 10^{-27}} \)
\( f = \frac{2.88}{10.494} \times 10^8 \)
\( f \approx 0.274 \times 10^8 = 27.4 \times 10^6 \, \text{Hz} = 27.4 \, \text{MHz} \).
iii. Explain ferromagnetism on the basis of domain theory.
- Ferromagnetic materials are made up of small regions called domains.
- In each domain, atomic magnetic dipole moments are aligned in the same direction due to exchange interaction.
- In an unmagnetized state, domains are randomly oriented, canceling out net magnetic moment.
- When an external magnetic field is applied:
- Domains aligned with the field grow in size (Domain Wall Displacement).
- Other domains rotate towards the field direction (Domain Rotation).
- This results in strong magnetization.
iv. The work function of caesium is 2.22 eV. Find the threshold frequency and photocurrent wavelength of incident light if photocurrent is brought to zero by stopping potential of 0.64 V.
Stopping Potential \( V_s = 0.64 \, \text{V} \).
Threshold Frequency (\( \nu_0 \)):
\( \phi_0 = h\nu_0 \Rightarrow \nu_0 = \frac{\phi_0}{h} \)
\( \nu_0 = \frac{2.22 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = \frac{3.552}{6.63} \times 10^{15} \approx 5.36 \times 10^{14} \, \text{Hz} \).
Wavelength of Incident Light:
Energy of incident photon \( E = \phi_0 + K.E._{max} = \phi_0 + eV_s \)
\( E = 2.22 \, \text{eV} + 0.64 \, \text{eV} = 2.86 \, \text{eV} \).
\( \lambda = \frac{hc}{E} \)
Using \( hc \approx 12400 \, \text{eV Å} \):
\( \lambda = \frac{12400}{2.86} \approx 4335 \, \text{Å} \).
A. A battery of emf 12 V and internal resistance 1Ω is connected in parallel with another battery of emf 10 V and internal resistance 1Ω. The combination is used to send current through an external resistance of 10 Ω. Calculate the current through external resistance. [4]
\( E_1 = 12, r_1 = 1 \).
\( E_2 = 10, r_2 = 1 \).
Equivalent EMF \( E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{12}{1} + \frac{10}{1}}{\frac{1}{1} + \frac{1}{1}} = \frac{22}{2} = 11 \, \text{V} \).
Equivalent Internal Resistance \( r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1 \times 1}{1 + 1} = 0.5 \, \Omega \).
External Resistance \( R = 10 \, \Omega \).
Current \( I = \frac{E_{eq}}{R + r_{eq}} = \frac{11}{10 + 0.5} = \frac{11}{10.5} \approx 1.048 \, \text{A} \).
B. State Faraday’s second law of E.M.I. A conducting loop of area 7 m2 is placed normal to uniform magnetic induction 4.8 Wb/m2. If the magnetic induction is uniformly reduced to 2 Wb/m2 in a time of 2 second. Calculate the induced e.m.f. produced in the loop. [3]
Calculation:
Initial Flux \( \Phi_1 = B_1 A = 4.8 \times 7 = 33.6 \, \text{Wb} \).
Final Flux \( \Phi_2 = B_2 A = 2 \times 7 = 14 \, \text{Wb} \).
Change in flux \( d\Phi = 14 - 33.6 = -19.6 \, \text{Wb} \).
Time \( dt = 2 \, \text{s} \).
Induced emf \( e = -\frac{d\Phi}{dt} = -\frac{-19.6}{2} = 9.8 \, \text{V} \).
Q.8. A. Define half life period. Derive expression for it [3]
Derivation:
Radioactive decay law: \( N = N_0 e^{-\lambda t} \).
At \( t = T_{1/2} \), \( N = \frac{N_0}{2} \).
\( \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \)
\( \frac{1}{2} = e^{-\lambda T_{1/2}} \)
\( 2 = e^{\lambda T_{1/2}} \)
Taking natural log on both sides:
\( \ln 2 = \lambda T_{1/2} \)
\( T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \).
B. i. Explain what is a polaroid. Give its two uses. [2]
Uses: 1. Used in sunglasses to reduce glare. 2. Used in 3D movie cameras and glasses.
B. ii. In a biprism experiment, light of wavelength 5400 Å is used to get interference pattern on screen. The fringe-width changes by 1.8 mm when the screen is moved towards biprism by 50 cm. Find the distance between two virtual images of the slit. [2]
Change in fringe width \( \Delta X = \frac{\lambda \Delta D}{d} \).
Given: \( \lambda = 5400 \, \text{Å} = 5.4 \times 10^{-7} \, \text{m} \).
\( \Delta X = 1.8 \, \text{mm} = 1.8 \times 10^{-3} \, \text{m} \).
\( \Delta D = 50 \, \text{cm} = 0.5 \, \text{m} \).
Find \( d \) (distance between virtual images).
\( d = \frac{\lambda \Delta D}{\Delta X} \)
\( d = \frac{5.4 \times 10^{-7} \times 0.5}{1.8 \times 10^{-3}} \)
\( d = \frac{2.7 \times 10^{-7}}{1.8 \times 10^{-3}} = 1.5 \times 10^{-4} \, \text{m} = 0.15 \, \text{mm} \).