BOARD QUESTION PAPER : MARCH 2017
Note: Use of only logarithmic table is allowed.
All symbols have their usual meaning unless otherwise stated.
Note:
- All questions are compulsory.
- Neat diagrams must be drawn wherever necessary.
- Figures to the right indicate full marks.
- Use of only logarithmic table is allowed.
- All symbols have their usual meaning unless otherwise stated.
- Answers to both sections must be written in the same answerbook.
- Answer to every question must be written on a new page.
SECTION – I
Q.1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
i. If the pressure of an ideal gas decreases by 10% isothermally, then its volume will _______.
For an isothermal process, \(PV = \text{constant}\). Let initial pressure be \(P_1\) and volume \(V_1\).
Given pressure decreases by 10%, so \(P_2 = P_1 - 0.10P_1 = 0.9P_1\).
Since \(P_1 V_1 = P_2 V_2\):
\(V_2 = \frac{P_1 V_1}{P_2} = \frac{P_1 V_1}{0.9 P_1} = \frac{1}{0.9} V_1 \approx 1.1111 V_1\)
Percentage increase in volume = \(\frac{V_2 - V_1}{V_1} \times 100 = (1.1111 - 1) \times 100 = 11.11\%\).
Note: The closest specific option provided in the board paper is 11.4%, though exact calculation yields 11.1%. Some textbooks approximate small changes as \(\frac{\Delta V}{V} \approx -\frac{\Delta P}{P}\) which would give 10% (Option C), but for a 10% change, this approximation has significant error. Board keys often point to Option D.
ii. Stretching of a rubber band results in _______.
Work is done against the restoring elastic forces during stretching. This work is stored in the rubber band as elastic potential energy.
iii. When the angular acceleration of a rotating body is zero, which physical quantity will be equal to zero?
According to the rotational analog of Newton's second law, Torque \(\tau = I\alpha\). If angular acceleration \(\alpha = 0\), then \(\tau = 0\) (assuming Moment of Inertia \(I \neq 0\)).
iv. In a damped harmonic oscillator, periodic oscillations have _______ amplitude.
In damped oscillations, energy is dissipated due to resistive forces, causing the amplitude to decay exponentially (gradually) over time.
v. A sine wave of wavelength ‘\(\lambda\)’ is travelling in a medium. What is the minimum distance between two particles of the medium which always have the same speed?
Particle speed in a wave is given by \(v(t) = \omega A \cos(kx - \omega t)\).
For two particles to always have the same speed (magnitude of velocity), i.e., \(|v_1| = |v_2|\), their phase difference \(\Delta \phi\) must be such that \(\cos(\theta) = \pm \cos(\theta + \Delta \phi)\).
This occurs when \(\Delta \phi = \pi\).
Since phase difference \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\), we have \(\pi = \frac{2\pi}{\lambda} \Delta x \Rightarrow \Delta x = \frac{\lambda}{2}\).
vi. Velocity of transverse wave along a stretched string is proportional to _______.
The velocity of a transverse wave on a stretched string is given by \(v = \sqrt{\frac{T}{m}}\), where \(T\) is tension and \(m\) is linear mass density. Thus, \(v \propto \sqrt{T}\).
vii. Find the wavelength at which a black body radiates maximum energy, if its remperature is 427°C. (Wein’s constant \(b = 2.898 \times 10^{-3}\) mK)
Using Wien's Displacement Law: \(\lambda_{max} T = b\).
Temperature \(T = 427^\circ\text{C} = 427 + 273 = 700 \text{ K}\).
\(\lambda_{max} = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{700}\)
\(\lambda_{max} \approx 0.00414 \times 10^{-3} \text{ m} = 4.14 \times 10^{-6} \text{ m}\).
HSC Physics Board Papers with Solution
- Physics - March 2025 - English Medium View Answer Key
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- Physics - July 2021 - English Medium View Answer Key
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- Physics - July 2017 View
Q.2. Attempt any SIX : [12]
i. Explain the concept of centripetal force.
Centripetal Force:
1. It is the force acting on a particle performing uniform circular motion, which is directed along the radius towards the centre of the circle.
2. It is necessary for maintaining circular motion.
3. Magnitude: \(F_{cp} = \frac{mv^2}{r} = mr\omega^2\), where \(m\) is mass, \(v\) is linear velocity, \(r\) is radius, and \(\omega\) is angular velocity.
4. It is a real force arising from interactions (e.g., tension in a string, gravitational force, friction).
ii. Prove that root mean square velocity of gas molecule is directly proportional to the square root of its absolute temperature.
According to the kinetic theory of gases, the pressure \(P\) exerted by an ideal gas is given by:
\(P = \frac{1}{3} \rho \overline{c^2}\), where \(\rho\) is density and \(\overline{c^2}\) is the mean square velocity.
Since \(\rho = \frac{M}{V}\) (Mass/Volume), we have:
\(PV = \frac{1}{3} M \overline{c^2}\)
For one mole of gas, ideal gas equation is \(PV = RT\).
\(\therefore RT = \frac{1}{3} M \overline{c^2}\)
\(\overline{c^2} = \frac{3RT}{M}\)
Taking square root on both sides to find RMS velocity (\(c_{rms}\)):
\(c_{rms} = \sqrt{\overline{c^2}} = \sqrt{\frac{3RT}{M}}\)
Since \(R\) and \(M\) are constants for a given gas:
\(c_{rms} \propto \sqrt{T}\)
Thus, RMS velocity is directly proportional to the square root of absolute temperature.
iii. Obtain the differential equation of linear simple harmonic motion.
In linear Simple Harmonic Motion (S.H.M.), the restoring force \(F\) is directly proportional to the displacement \(x\) from the mean position and acts opposite to it.
\(F = -kx\) (where \(k\) is the force constant).
By Newton's second law, \(F = ma = m \frac{d^2x}{dt^2}\).
Equating the forces:
\(m \frac{d^2x}{dt^2} = -kx\)
\(\frac{d^2x}{dt^2} + \frac{k}{m}x = 0\)
Let \(\frac{k}{m} = \omega^2\) (where \(\omega\) is angular frequency).
\(\therefore \frac{d^2x}{dt^2} + \omega^2 x = 0\)
This is the differential equation of linear S.H.M.
iv. Draw a neat, labelled diagram for a liquid surface in contact with a solid, when the angle of contact is acute.
Diagram Description: A concave meniscus is formed. The solid is partially immersed. Vectors shown:
- \(T_1\): Force due to surface tension at solid-liquid interface (downwards along solid).
- \(T_2\): Force due to surface tension at solid-air interface (upwards along solid).
- \(T_3\): Force due to surface tension at liquid-air interface (tangent to meniscus).
- Angle of contact \(\theta\) is less than 90° (inside the liquid).
v. A hole is drilled half way to the centre of the Earth. A body is dropped into the hole. How much will it weigh at the bottom of the hole if the weight of the body on the Earth’s surface is 350 N?
Given: Weight on surface \(W = mg = 350\) N.
Depth \(d = \frac{R}{2}\) (halfway to center).
Acceleration due to gravity at depth \(d\) is given by \(g_d = g(1 - \frac{d}{R})\).
\(g_d = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}\).
Weight at depth \(d\) is \(W_d = m g_d = m (\frac{g}{2}) = \frac{mg}{2} = \frac{W}{2}\).
\(W_d = \frac{350}{2} = 175\) N.
Answer: The body will weigh 175 N.
vi. A solid sphere of mass 1 kg rolls on a table with linear speed 2 m/s, find its total kinetic energy.
Given: Mass \(M = 1\) kg, Velocity \(v = 2\) m/s.
For a solid sphere, Moment of Inertia \(I = \frac{2}{5}MR^2\).
Total Kinetic Energy \(E = \text{Translational K.E.} + \text{Rotational K.E.}\)
\(E = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\)
Since rolling without slipping, \(v = R\omega\), so \(\omega = v/R\).
\(E = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2\)
\(E = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = (\frac{1}{2} + \frac{1}{5})Mv^2 = \frac{7}{10}Mv^2\)
Calculation: \(E = \frac{7}{10} \times 1 \times (2)^2 = 0.7 \times 4 = 2.8\) J.
Answer: Total kinetic energy is 2.8 J.
vii. A transverse wave is produced on a stretched string 0.9 m long and fixed at its ends. Find the speed of the transverse wave, when the string vibrates while emitting second overtone of frequency 324 Hz.
Given: Length \(L = 0.9\) m, Frequency \(f = 324\) Hz.
Mode: Second overtone. For a string fixed at both ends, the second overtone corresponds to the 3rd harmonic (\(n=3\), 3 loops).
Frequency formula: \(f_n = \frac{n v}{2L}\)
\(324 = \frac{3 \times v}{2 \times 0.9}\)
\(324 = \frac{3v}{1.8} = \frac{v}{0.6}\)
\(v = 324 \times 0.6 = 194.4\) m/s.
Answer: The speed of the transverse wave is 194.4 m/s.
viii. A body cools at the rate of 0.5°C / minute when it is 25° C above the surroundings. Calculate the rate of cooling when it is 15°C above the same surroundings.
According to Newton's Law of Cooling, rate of cooling \(\frac{d\theta}{dt} = k(\theta - \theta_0)\), where \((\theta - \theta_0)\) is the excess temperature.
Case 1:
Rate \((R_1) = 0.5^\circ\)C/min.
Excess temp \((\theta_1 - \theta_0) = 25^\circ\)C.
\(0.5 = k(25) \Rightarrow k = \frac{0.5}{25} = 0.02 \text{ min}^{-1}\).
Case 2:
Excess temp \((\theta_2 - \theta_0) = 15^\circ\)C.
Rate \((R_2) = k(15) = 0.02 \times 15\).
\(R_2 = 0.3^\circ\)C/min.
Answer: The rate of cooling is 0.3°C/minute.
Q.3. Attempt any THREE [9]
i. Show that period of a satellite revolving around the Earth depends upon mass of the Earth.
Consider a satellite of mass \(m\) revolving around the Earth (mass \(M\)) in a circular orbit of radius \(r\) with velocity \(v_c\).
The necessary centripetal force is provided by the gravitational force.
\(\frac{mv_c^2}{r} = \frac{GMm}{r^2}\)
\(\therefore v_c = \sqrt{\frac{GM}{r}}\)
The time period \(T\) is given by \(T = \frac{\text{Circumference}}{\text{velocity}} = \frac{2\pi r}{v_c}\).
Substitute \(v_c\):
\(T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} = 2\pi \sqrt{\frac{r^3}{GM}}\)
From this expression, it is clear that \(T\) depends on \(M\) (Mass of Earth) inversely under the square root. Thus shown.
ii. Obtain an expression for torque acting on a rotating body with constant angular acceleration. Hence state the dimensions and SI unit of torque.
Consider a rigid body rotating about an axis with constant angular acceleration \(\alpha\).
The body consists of \(n\) particles of masses \(m_1, m_2, \dots, m_n\) at distances \(r_1, r_2, \dots, r_n\) from the axis.
Linear acceleration of particle \(i\) is \(a_i = r_i \alpha\).
Force on particle \(i\) is \(f_i = m_i a_i = m_i r_i \alpha\).
Torque on particle \(i\) about the axis is \(\tau_i = f_i r_i = (m_i r_i \alpha) r_i = m_i r_i^2 \alpha\).
Total torque \(\tau = \sum \tau_i = \sum m_i r_i^2 \alpha = (\sum m_i r_i^2) \alpha\).
Since \(I = \sum m_i r_i^2\) (Moment of Inertia),
\(\tau = I \alpha\)
SI Unit: Newton-meter (N m)
Dimensions: \([L^2 M^1 T^{-2}]\) (Energy dimensions)
iii. The total energy of free surface of a liquid drop is \(2\pi\) times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in SI unit).
Let \(T\) be the surface tension and \(r\) be the radius of the drop.
Surface energy \(E\) of a liquid drop (having 1 free surface) is given by \(E = T \times \text{Area} = T(4\pi r^2)\).
Given condition: \(E = 2\pi T\).
Equating the two:
\(T(4\pi r^2) = 2\pi T\)
Dividing both sides by \(2\pi T\):
\(2r^2 = 1 \Rightarrow r^2 = 0.5 \Rightarrow r = \frac{1}{\sqrt{2}} \text{ m}\).
Diameter \(D = 2r = \frac{2}{\sqrt{2}} = \sqrt{2}\) m.
Answer: The diameter of the drop is \(\sqrt{2}\) m (approx 1.414 m).
iv. A vehicle is moving on a circular track whose surface is inclined towards the horizon at an angle of 10°. The maximum velocity with which it can move safely is 36 km / hr. Calculate the length of the circular track. [\(\pi = 3.142\)]
Given: Banking angle \(\theta = 10^\circ\).
Velocity \(v = 36 \text{ km/hr} = 36 \times \frac{5}{18} = 10 \text{ m/s}\).
Assuming this is the optimum/safe speed for banking (ignoring friction as \(\mu\) isn't given), formula is \(v = \sqrt{rg \tan \theta}\).
Square both sides: \(v^2 = rg \tan \theta\).
\(r = \frac{v^2}{g \tan \theta}\).
Using \(g = 9.8 \text{ m/s}^2\) and \(\tan(10^\circ) \approx 0.1763\).
\(r = \frac{100}{9.8 \times 0.1763} \approx \frac{100}{1.7277} \approx 57.88 \text{ m}\).
Length of track = Circumference = \(2\pi r\).
\(L = 2 \times 3.142 \times 57.88 \approx 363.7 \text{ m}\).
Answer: The length of the circular track is approximately 363.7 m.
Q.4. A. Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy w. r. t. instantaneous displacement.
Proof:
Consider a particle of mass \(m\) performing S.H.M. with amplitude \(A\) and angular frequency \(\omega\).
Displacement \(x\). Velocity \(v = \omega \sqrt{A^2 - x^2}\).
1. Potential Energy (P.E.): \(E_p = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2\).
2. Kinetic Energy (K.E.): \(E_k = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)\).
3. Total Energy (T.E.): \(E = E_p + E_k\)
\(E = \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2 x^2\)
\(E = \frac{1}{2}m\omega^2 A^2\).
Since \(m, \omega, A\) are constants, Total Energy is conserved at any position \(x\).
Graphical Representation:
- P.E. is a parabola opening upwards (\(U\)-shape) with minimum at \(x=0\).
- K.E. is a parabola opening downwards (inverted \(U\)) with maximum at \(x=0\).
- T.E. is a straight horizontal line.
B. Two sound notes have wavelengths \(\frac{83}{170}\) m and \(\frac{83}{172}\) m in the air. These notes when sounded together produce 8 beats per second. Calculate the velocity of sound in the air and frequencies of the two notes. [7]
Given:
\(\lambda_1 = \frac{83}{170}\) m, \(\lambda_2 = \frac{83}{172}\) m.
Since \(\lambda_1 > \lambda_2\) (denominator is smaller), \(f_1 < f_2\).
Beat frequency \(N = f_2 - f_1 = 8\) Hz.
Using \(v = f\lambda \Rightarrow f = v/\lambda\).
\(f_1 = \frac{v}{83/170} = \frac{170v}{83}\)
\(f_2 = \frac{v}{83/172} = \frac{172v}{83}\)
Substitute into beat equation:
\(\frac{172v}{83} - \frac{170v}{83} = 8\)
\(\frac{2v}{83} = 8\)
\(2v = 8 \times 83\)
\(v = 4 \times 83 = 332\) m/s.
Frequencies:
\(f_1 = \frac{170 \times 332}{83} = 170 \times 4 = 680\) Hz.
\(f_2 = \frac{172 \times 332}{83} = 172 \times 4 = 688\) Hz.
Answer: Velocity of sound = 332 m/s; Frequencies are 680 Hz and 688 Hz.
A. Explain the formation of stationary waves by analytical method. Show the formation of stationary wave diagramatically.
Analytical Method:
Consider two identical progressive waves travelling in opposite directions:
\(y_1 = A \sin(\omega t - kx)\) (positive x-direction)
\(y_2 = A \sin(\omega t + kx)\) (negative x-direction)
By superposition principle, \(y = y_1 + y_2\).
Using \(\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})\):
\(y = 2A \sin(\omega t) \cos(-kx) = 2A \cos(kx) \sin(\omega t)\).
This represents a stationary wave where amplitude \(R = 2A \cos(kx)\) varies with position \(x\).
- Nodes: Points where amplitude is zero (\(\cos kx = 0\)). \(x = (2n+1)\lambda/4\).
- Antinodes: Points where amplitude is max (\(\cos kx = \pm 1\)). \(x = n\lambda/2\).
Diagram: Shows loops with Nodes (N) and Antinodes (A).
B. A mass of 1 kg is hung from a steel wire if radius 0.5 mm and length 4 m. Calculate the extension produced. What should be the area of cross-section of the wire so that elastic limit is not exceeded? Change in radius is negligible.
(Given : \(g = 9.8 \text{ m/s}^2\); Elastic limit of steel is \(2.4 \times 10^8 \text{ N/m}^2\); \(Y\) for steel \((Y_{steel}) = 20 \times 10^{10} \text{ N/m}^2\); \(\pi = 3.142\)) [7]
Part 1: Extension Calculation
\(m = 1\) kg, \(L = 4\) m, \(r = 0.5 \text{ mm} = 5 \times 10^{-4}\) m.
Area \(A = \pi r^2 = 3.142 \times (5 \times 10^{-4})^2 = 3.142 \times 25 \times 10^{-8} = 7.855 \times 10^{-7} \text{ m}^2\).
Force \(F = mg = 1 \times 9.8 = 9.8\) N.
Using Young's Modulus formula: \(l = \frac{FL}{AY}\).
\(l = \frac{9.8 \times 4}{7.855 \times 10^{-7} \times 20 \times 10^{10}}\)
\(l = \frac{39.2}{157.1 \times 10^3} = \frac{39.2}{1.571 \times 10^5} \approx 2.49 \times 10^{-4} \text{ m}\).
Extension \(\approx 0.25\) mm.
Part 2: Area for Elastic Limit
Stress \(\le\) Elastic Limit.
\(\frac{F}{A'} \le 2.4 \times 10^8\)
\(A' \ge \frac{F}{2.4 \times 10^8} = \frac{9.8}{2.4 \times 10^8}\)
\(A' \ge 4.083 \times 10^{-8} \text{ m}^2\).
Answer: Extension is 0.25 mm. Required area is \(4.08 \times 10^{-8} \text{ m}^2\).
SECTION – II
Q.5. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
i. If A.C. voltage is applied to a pure capacitor, then voltage across the capacitor ______.
In a purely capacitive circuit, Current (I) leads Voltage (V) by 90° (\(\pi/2\)). Conversely, Voltage lags behind Current by \(\pi/2\).
ii. In Doppler effect of light, the term “red shift” is used for ______.
Red shift implies the light spectrum shifts towards red (longer wavelength). Since \(c = f\lambda\), an increase in wavelength corresponds to a decrease in frequency.
iii. If a watch-glass containing a small quantity of water is placed on two dissimilar magnetic poles, then water ______.
Water is diamagnetic. Diamagnetic substances move from stronger to weaker parts of the magnetic field. Between dissimilar poles (N-S), the field is strongest in the center, so water is pushed away to the sides, causing a depression.
iv. Any device that converts one form of energy into another is termed as ______.
A transducer is defined as a device that converts a signal in one form of energy to another form of energy.
v. When a p-n-p transistor is operated in saturation region, then its ______.
For a transistor to work in saturation (closed switch mode), both the emitter-base and collector-base junctions must be forward biased.
vi. In a photon-electron collision ______.
In fundamental particle collisions like photon-electron (Compton effect), the fundamental conservation laws of total energy and total linear momentum always hold true.
vii. If the charge on the condenser of 10 \(\mu\)F is doubled, then the energy stored in it becomes ______.
Energy \(E = \frac{Q^2}{2C}\).
If charge \(Q\) becomes \(2Q\), new energy \(E' = \frac{(2Q)^2}{2C} = \frac{4Q^2}{2C} = 4E\).
Q.6. Attempt any SIX: [12]
i. Distinguish between the phenomenon of interference and diffraction of light.
Difference:
1. Origin: Interference is due to superposition of waves from two distinct coherent sources. Diffraction is due to superposition of secondary wavelets from different parts of the same wavefront.
2. Fringe Width: In interference, all bright fringes are usually of equal intensity and equal width. In diffraction, the central maximum is widest and brightest; secondary maxima decrease in intensity and width.
3. Minima: In interference, minima can be perfectly dark. In diffraction, minima are not perfectly dark.
ii. Explain how moving coil galvanometer is converted into a voltmeter. Derive the necessary formula.
A Moving Coil Galvanometer (MCG) is converted into a voltmeter by connecting a high resistance (\(R_s\)) in series with it.
Let \(G\) be the resistance of the galvanometer and \(I_g\) be the current for full-scale deflection.
To measure a potential difference \(V\):
Total resistance of the circuit = \(G + R_s\).
By Ohm's law, \(V = I_g (G + R_s)\).
\(G + R_s = \frac{V}{I_g}\)
\(R_s = \frac{V}{I_g} - G\).
This is the required value of series resistance.
iii. State the advantages of potentiometer over voltmeter.
- A potentiometer measures the EMF of a cell in an open circuit (null point method), drawing no current from the cell, thus giving the accurate EMF. A voltmeter draws current, measuring terminal voltage (which is less than EMF).
- It can measure the internal resistance of a cell.
- It is more sensitive than a voltmeter.
- It can be used to compare EMFs of two cells.
iv. Draw a neat, labelled block diagram of a receiver for the detection of amplitude modulated wave.
Block Diagram Sequence:
Receiving Antenna \(\rightarrow\) Amplifier/Tuner \(\rightarrow\) IF (Intermediate Frequency) Stage \(\rightarrow\) Detector (Demodulator) \(\rightarrow\) Audio Amplifier \(\rightarrow\) Speaker (Output).
v. A rectangular coil of a moving coil galvanometer contains 100 turns, each having area 15 cm\(^2\). It is suspended in the radial magnetic field 0.03 T. The twist constant of suspension fibre is \(15 \times 10^{-10}\) N-m/degree. Calculate the sensitivity of the moving coil galvanometer.
Given: \(N = 100\), \(A = 15 \text{ cm}^2 = 15 \times 10^{-4} \text{ m}^2\), \(B = 0.03\) T.
\(C\) (twist constant) \(= 15 \times 10^{-10}\) N-m/degree.
Current Sensitivity \(S_i = \frac{\theta}{I} = \frac{NBA}{C}\).
\(S_i = \frac{100 \times 0.03 \times 15 \times 10^{-4}}{15 \times 10^{-10}}\)
\(S_i = \frac{3 \times 15 \times 10^{-4}}{15 \times 10^{-10}} = 3 \times 10^{-4} \times 10^{10} = 3 \times 10^6\).
Answer: Sensitivity is \(3 \times 10^6\) degree/Ampere.
vi. The magnetic flux through a loop is varying according to a relation \(\phi = 6t^2 + 7t + 1\) where \(\phi\) is in milliweber and \(t\) is in second. What is the e.m.f. induced in the loop at \(t = 2\) second?
Induced EMF \(e = \left| \frac{d\phi}{dt} \right|\).
\(\phi = (6t^2 + 7t + 1) \times 10^{-3}\) Wb.
\(\frac{d\phi}{dt} = (12t + 7) \times 10^{-3}\) V.
At \(t = 2\) s:
\(e = (12(2) + 7) \times 10^{-3} = (24 + 7) \times 10^{-3} = 31 \times 10^{-3}\) V.
Answer: Induced EMF is 31 mV (millivolts).
vii. An unknown resistance is placed in the left gap and resistance of 50 ohm is placed in the right gap of a meter bridge. The null point is obtained at 40 cm from the left end. Determine the unknown resistance.
Left gap \(X\) (unknown), Right gap \(R = 50 \, \Omega\).
Balancing length from left, \(l_x = 40\) cm.
Length for R, \(l_R = 100 - 40 = 60\) cm.
Principle of Meter Bridge: \(\frac{X}{R} = \frac{l_x}{l_R}\)
\(\frac{X}{50} = \frac{40}{60} = \frac{2}{3}\)
\(X = 50 \times \frac{2}{3} = \frac{100}{3} = 33.33 \, \Omega\).
Answer: Unknown resistance is 33.33 \(\Omega\).
viii. Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is \(2.14 \times 10^{-10}\) m and \(1.09 \times 10^6\) m/s respectively. [\(\pi = 3.142\)]
Frequency \(f = \frac{v}{2\pi r}\).
Given \(v = 1.09 \times 10^6\) m/s, \(r = 2.14 \times 10^{-10}\) m.
\(f = \frac{1.09 \times 10^6}{2 \times 3.142 \times 2.14 \times 10^{-10}}\)
\(f = \frac{1.09 \times 10^{16}}{13.447} \approx 0.08105 \times 10^{16}\) Hz.
\(f \approx 8.1 \times 10^{14}\) Hz.
Answer: Frequency is \(8.1 \times 10^{14}\) Hz.
Q.7. Attempt any THREE: [9]
i. Explain with a neat diagram, how a p-n junction diode is used as a half wave rectifier.
Principle: A PN junction diode conducts only when forward biased.
Circuit: AC source connected to primary of transformer. Secondary connected to Diode (D) and Load Resistance (\(R_L\)) in series.
Operation:
- During positive half cycle of AC, diode is forward biased and conducts current. Output appears across \(R_L\).
- During negative half cycle, diode is reverse biased and does not conduct. Output is zero.
- Result is pulsating DC.
ii. Explain self induction and mutual induction.
Self Induction: The phenomenon of production of induced EMF in a coil due to the change of current in the same coil is called self induction. (\(\phi = LI\), \(e = -L \frac{dI}{dt}\)).
Mutual Induction: The phenomenon of production of induced EMF in a secondary coil due to the change of current in a primary (neighboring) coil is called mutual induction. (\(\phi_s = M I_p\), \(e_s = -M \frac{dI_p}{dt}\)).
iii. A cube of marble having each side 1 cm is kept in an electric field of intensity 300 V/m. Determine the energy contained in the cube of dielectric constant 8. [Given : \(\in_0 = 8.85 \times 10^{-12} \text{ C}^2\text{/Nm}^2\)]
Volume of cube \(V = (1 \text{ cm})^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3\).
Electric field \(E = 300\) V/m. Dielectric constant \(K = 8\).
Energy density \(u = \frac{1}{2} K \in_0 E^2\).
Total Energy \(U = u \times V = \frac{1}{2} K \in_0 E^2 V\).
\(U = \frac{1}{2} \times 8 \times 8.85 \times 10^{-12} \times (300)^2 \times 10^{-6}\)
\(U = 4 \times 8.85 \times 10^{-18} \times 9 \times 10^4\)
\(U = 35.4 \times 9 \times 10^{-14} = 318.6 \times 10^{-14} \text{ J}\)
\(U = 3.186 \times 10^{-12}\) J.
Answer: Energy contained is \(3.19 \times 10^{-12}\) J.
iv. An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. If the frequency of revolution of an electron is \(9 \times 10^9\) MHz, calculate the orbital angular momentum.
[Given : Charge on an electron = \(1.6 \times 10^{-19}\) C; Gyromagnetic ratio = \(8.8 \times 10^{10}\) C/kg; \(\pi = 3.142\)]
Frequency \(f = 9 \times 10^9 \text{ MHz} = 9 \times 10^{15} \text{ Hz}\).
Radius \(r = 0.53 \text{ Å} = 0.53 \times 10^{-10} \text{ m}\).
Angular momentum \(L\) is related to magnetic moment \(M_{orb}\).
Also, \(L = mvr = m(2\pi r f)r = 2\pi m f r^2\).
Mass of electron \(m\) can be found from Gyromagnetic ratio (\(\gamma = \frac{e}{2m}\)).
\(m = \frac{e}{2\gamma}\).
Substitute \(m\) in \(L\):
\(L = 2\pi (\frac{e}{2\gamma}) f r^2 = \frac{\pi e f r^2}{\gamma}\).
\(L = \frac{3.142 \times 1.6 \times 10^{-19} \times 9 \times 10^{15} \times (0.53 \times 10^{-10})^2}{8.8 \times 10^{10}}\)
\(L = \frac{45.24 \times 10^{-4} \times 0.2809 \times 10^{-20}}{8.8 \times 10^{10}}\)
\(L \approx \frac{12.7 \times 10^{-24}}{8.8 \times 10^{10}} \approx 1.44 \times 10^{-34} \text{ J s}\).
Answer: Orbital angular momentum is \(1.44 \times 10^{-34}\) J s.
Q.8. A. Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram.
Description:
Fresnel's biprism consists of two acute angle prisms joined base to base. It creates two coherent virtual sources (\(S_1, S_2\)) from a single slit source \(S\) via refraction.
Procedure:
1. Measure fringe width \(X\) (or \(\beta\)) using a micrometer eyepiece.
2. Measure distance \(D\) between slit and eyepiece.
3. Measure distance \(d\) between virtual sources using conjugate foci method (convex lens inserted between biprism and eyepiece). \(d = \sqrt{d_1 d_2}\).
Formula: \(\lambda = \frac{X d}{D}\).
Diagram: Shows Slit, Biprism, Virtual Sources \(S_1, S_2\), Screen/Eyepiece, and interference region.
B. The width of plane incident wavefront is found to be doubled on refraction in denser medium. If it makes an angle of 65° with the normal, calculate the refractive index for the denser medium. [7]
Let width of incident wavefront be \(W_1\) and refracted wavefront be \(W_2\).
Given \(W_2 = 2 W_1\).
From geometry of wavefront refraction: \(\frac{W_2}{W_1} = \frac{\cos r}{\cos i}\).
Note: Usually for denser medium, \(r < i\), so \(\cos r > \cos i\), and width increases. This matches the problem statement.
\(2 = \frac{\cos r}{\cos 65^\circ}\)
\(\cos r = 2 \cos 65^\circ = 2 \times 0.4226 = 0.8452\).
\(r = \cos^{-1}(0.8452) \approx 32^\circ 18' \approx 32.3^\circ\).
Refractive index \(\mu = \frac{\sin i}{\sin r}\).
\(\mu = \frac{\sin 65^\circ}{\sin 32.3^\circ} \approx \frac{0.9063}{0.5344} \approx 1.696\).
Answer: The refractive index is approximately 1.70.
A. Draw a neat, labelled energy level diagram for H atom showing the transitions. Explain the series of spectral lines for H atom, whose fixed inner orbit numbers are 3 and 4 respectively.
Diagram: Energy levels \(n=1, 2, 3, 4, \dots, \infty\). Arrows showing transitions down to specific levels (Lyman to 1, Balmer to 2, Paschen to 3, etc.).
Explanation:
1. Paschen Series (Inner orbit \(n=3\)): Transitions from higher orbits \(m = 4, 5, 6, \dots\) to \(n=3\). Lies in the near infrared region. Formula: \(\frac{1}{\lambda} = R (\frac{1}{3^2} - \frac{1}{m^2})\).
2. Brackett Series (Inner orbit \(n=4\)): Transitions from higher orbits \(m = 5, 6, 7, \dots\) to \(n=4\). Lies in the far infrared region. Formula: \(\frac{1}{\lambda} = R (\frac{1}{4^2} - \frac{1}{m^2})\).
B. The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Is the photoelectric effect possible for either of them if the incident wavelength is 5180 Å?
[Given : Planck’s constant = \(6.63 \times 10^{-34}\) J.s.; Velocity of light = \(3 \times 10^8\) m/s; 1 eV = \(1.6 \times 10^{-19}\) J] [7]
Incident Wavelength \(\lambda = 5180 \text{ Å} = 5.18 \times 10^{-7}\) m.
Energy of incident photon \(E = \frac{hc}{\lambda}\).
\(E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5.18 \times 10^{-7}}\)
\(E = \frac{19.89 \times 10^{-26}}{5.18 \times 10^{-7}} = 3.84 \times 10^{-19} \text{ J}\).
Convert to eV:
\(E (\text{eV}) = \frac{3.84 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.40 \text{ eV}\).
Condition for Photoelectric Effect: Incident Energy \(>\) Work Function (\(\phi\)).
1. Potassium: \(\phi = 2.25\) eV. Since \(2.40 > 2.25\), Photoelectric effect is possible.
2. Caesium: \(\phi = 2.14\) eV. Since \(2.40 > 2.14\), Photoelectric effect is possible.
Answer: Yes, photoelectric effect is possible for both potassium and caesium.