Model Question Paper Set - IV Physics Solutions
SECTION – I
Q. 1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
i. In uniform circular motion, the velocity vector and acceleration vector are _______.
ii. The ratio of kinetic energy of a body orbiting near the earth’s surface and the kinetic energy of the same body escaping the earth’s gravitational field is _______.
Escape Velocity: \(v_e = \sqrt{\frac{2GM}{R}}\)
Ratio of KE: \(\frac{KE_o}{KE_e} = \frac{\frac{1}{2}mv_o^2}{\frac{1}{2}mv_e^2} = \frac{v_o^2}{v_e^2} = \frac{GM/R}{2GM/R} = \frac{1}{2} = 0.5:1\)
iii. Which one of the following substances possesses the highest elasticity?
iv. A circular disc of mass M and radius R is rotating about its axis with uniform speed v. What is its kinetic energy?
Angular velocity \(\omega = \frac{v}{R}\).
Rotational K.E. = \(\frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{4}Mv^2\).
v. A body which absorbs all the radiations incident over it is called a _______.
vi. A liquid does not wet the sides of a solid, if the angle of contact is _______.
vii. When two sound waves are superimposed, beats are produced when they have _______.
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
Q. 2. Attempt any SIX: [12]
i. What is the angular speed of the minute hand of a clock? If the minute hand is 6 cm long. What is the linear speed of its tip?
Solution:
Time period of minute hand \(T = 60 \text{ min} = 60 \times 60 = 3600 \text{ s}\).
Length of hand (radius) \(R = 6 \text{ cm} = 6 \times 10^{-2} \text{ m}\).
Angular speed (\(\omega\)):
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{3600} = \frac{\pi}{1800} \text{ rad/s} \approx 1.745 \times 10^{-3} \text{ rad/s}\).
Linear speed (\(v\)):
\(v = R\omega = 6 \times 10^{-2} \times \frac{\pi}{1800} = \frac{\pi}{300} \times 10^{-2} \text{ m/s} \approx 1.047 \times 10^{-4} \text{ m/s}\).
ii. Assuming the earth to be a homogeneous sphere, determine the density of the earth from following data: g = 9.8 m/s², G = 6.673 × 10⁻¹¹ Nm²/kg², R = 6400 km
Solution:
Formula for acceleration due to gravity: \(g = \frac{GM}{R^2}\)
Mass of earth \(M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \rho\)
Substituting M in g:
\(g = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3}\pi R \rho G\)
Rearranging for density \(\rho\):
\(\rho = \frac{3g}{4\pi R G}\)
Given: \(g=9.8\), \(R=6.4 \times 10^6 \text{ m}\), \(G=6.673 \times 10^{-11}\)
\(\rho = \frac{3 \times 9.8}{4 \times 3.142 \times 6.4 \times 10^6 \times 6.673 \times 10^{-11}}\)
\(\rho = \frac{29.4}{536.56 \times 10^{-5}} \approx 5479 \text{ kg/m}^3\)
iii. Discuss the physical significance of radius of gyration. Also discuss the factors on which radius of gyration depends.
Physical Significance:
The radius of gyration (k) represents the distribution of mass of a body about a given axis of rotation. It allows us to replace a body of complex shape with a point mass 'M' placed at a distance 'k' from the axis, such that the Moment of Inertia remains the same (\(I = Mk^2\)). It is a measure of how far the mass is theoretically concentrated from the axis.
Factors affecting Radius of Gyration:
1. The position and direction of the axis of rotation.
2. The distribution of mass about the axis of rotation.
iv. Particle performing S.H.M starts from extreme position. Plot a graph of displacement against time.
Solution:
When a particle starts from the extreme position (positive), the displacement equation is \(x = A \cos(\omega t)\).
At \(t=0, x=A\); \(t=T/4, x=0\); \(t=T/2, x=-A\); \(t=3T/4, x=0\); \(t=T, x=A\).
The graph is a cosine curve starting at maximum amplitude.
v. What is meant by an angle of contact? State the main characteristics of angle of contact.
Definition: The angle between the tangent drawn to the free surface of the liquid and the surface of the solid at the point of contact, measured within the liquid, is called the angle of contact.
Characteristics:
1. It is constant for a given solid-liquid pair.
2. It depends on the nature of the solid and liquid.
3. It depends on impurities present.
4. It depends on the temperature of the liquid.
vi. A racing car passes stand of race course, travelling at a speed of 180 km/hr. If the noise from the exhaust has a frequency of 300 Hz, determine the frequency heard by audience as the car approaches stand. (Assume speed of sound as 340 m/s)
Solution:
Source velocity \(v_s = 180 \text{ km/hr} = 180 \times \frac{5}{18} = 50 \text{ m/s}\).
Speed of sound \(v = 340 \text{ m/s}\).
Observer is stationary, \(v_o = 0\).
Source frequency \(n = 300 \text{ Hz}\).
As source approaches observer, apparent frequency \(n'\) is given by:
\(n' = n \left( \frac{v}{v - v_s} \right)\)
\(n' = 300 \left( \frac{340}{340 - 50} \right)\)
\(n' = 300 \left( \frac{340}{290} \right) \approx 351.72 \text{ Hz}\).
vii. Distinguish between harmonics and overtones.
Harmonics:
1. Harmonics are integral multiples of the fundamental frequency.
2. All harmonics may or may not be present in a given sound note.
Overtones:
1. Overtones are the frequencies of sound higher than the fundamental frequency that are actually produced by the instrument.
2. The first overtone is the immediate higher frequency after the fundamental, regardless of whether it is an integer multiple.
viii. Calculate the temperature at which the average K.E. of a molecule of a gas will be same as that of an electron accelerated through 4 volt. [kB = 1.4 × 10⁻²³ J/molecule K, e = 1.6 × 10⁻¹⁹C]
Solution:
K.E. of gas molecule = \(\frac{3}{2} k_B T\)
K.E. of electron = \(eV\)
Given: \(V = 4 \text{ V}\), \(e = 1.6 \times 10^{-19} \text{ C}\), \(k_B = 1.4 \times 10^{-23} \text{ J/K}\).
Equating: \(\frac{3}{2} k_B T = eV\)
\(T = \frac{2eV}{3k_B} = \frac{2 \times 1.6 \times 10^{-19} \times 4}{3 \times 1.4 \times 10^{-23}}\)
\(T = \frac{12.8 \times 10^{-19}}{4.2 \times 10^{-23}}\)
\(T = 3.047 \times 10^4 \text{ K} \approx 30476 \text{ K}\).
Q. 3. Attempt any THREE: [9]
i. A stone weighing 2 kg is whirled in a vertical circle at the end of a rope of length 0.5 m. Find the tension at lowest position, mid position, highest position.
Note: The question implies the minimum speed condition for completing the circle, or simply asking for the formulaic difference. Assuming "whirled" implies the critical case (just completing the circle) or standard critical values:
Given: \(m = 2 \text{ kg}\), \(r = 0.5 \text{ m}\), assume \(g = 9.8 \text{ m/s}^2\).
For critical velocity (minimum to complete circle):
Velocity at highest point \(v_H = \sqrt{rg}\)
Velocity at lowest point \(v_L = \sqrt{5rg}\)
Velocity at mid point \(v_M = \sqrt{3rg}\)
Tension Formulas:
Highest (\(T_H\)): \(T_H = \frac{mv_H^2}{r} - mg = 0\) (Minimum case)
Lowest (\(T_L\)): \(T_L = \frac{mv_L^2}{r} + mg = 6mg\)
Midway (\(T_M\)): \(T_M = \frac{mv_M^2}{r} = 3mg\)
Calculations:
\(T_{Highest} = 0 \text{ N}\)
\(T_{Mid} = 3 \times 2 \times 9.8 = 58.8 \text{ N}\)
\(T_{Lowest} = 6 \times 2 \times 9.8 = 117.6 \text{ N}\)
*(If the stone has higher energy than critical, specific velocity must be given. Assuming critical condition based on context).*
ii. Show that under certain conditions, a simple pendulum performs linear S.H.M.
Solution:
Consider a simple pendulum of length 'L' and mass 'm'. Displace it by a small angle \(\theta\).
Restoring force \(F = -mg \sin \theta\).
Condition: If the amplitude is very small, \(\theta\) is small, so \(\sin \theta \approx \theta\).
\(F \approx -mg\theta\).
From geometry, arc length \(x = L\theta \Rightarrow \theta = x/L\).
Substituting \(\theta\): \(F = -mg(x/L) = -(\frac{mg}{L})x\).
Since \(m, g, L\) are constants, \(F = -k x\), where \(k = mg/L\).
Since Force is directly proportional to displacement and opposite in direction, the motion is Linear S.H.M.
iii. Derive an expression for strain energy using calculus method.
Solution:
Let a wire of length \(L\) and area \(A\) be stretched by a force \(F\).
Let \(l\) be the extension produced at any instant. Young's modulus \(Y = \frac{FL}{Al}\) \(\Rightarrow F = \frac{YAl}{L}\).
Work done \(dW\) to extend it further by \(dl\) is \(dW = F \cdot dl\).
Total work done in stretching from \(0\) to \(l\):
\(W = \int_{0}^{l} F \, dl = \int_{0}^{l} \frac{YAl}{L} \, dl\)
\(W = \frac{YA}{L} \int_{0}^{l} l \, dl = \frac{YA}{L} \left[ \frac{l^2}{2} \right]_0^l\)
\(W = \frac{1}{2} \frac{YA l^2}{L}\)
Rearranging: \(W = \frac{1}{2} \times \left( \frac{YAl}{L} \right) \times l = \frac{1}{2} \times \text{Load} \times \text{Extension}\).
This work is stored as Strain Energy.
iv. The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin(2πx/3) cos(120πt) where x and y are in m and t is s. Do all the points on the string oscillate with the same frequency and amplitude? What is the amplitude of a point 0.375 m away from one end?
Solution:
1. Frequency: Yes, in a stationary wave, all points vibrate with the same frequency. \( \omega = 120\pi \Rightarrow f = 60 \text{ Hz}\).
2. Amplitude: No, amplitude varies with position x. Amplitude function \(A(x) = 0.06 \sin(\frac{2\pi x}{3})\).
3. Amplitude at \(x = 0.375 \text{ m}\):
\(A = 0.06 \sin\left(\frac{2\pi (0.375)}{3}\right)\)
\(A = 0.06 \sin\left(\frac{2\pi (3/8)}{3}\right) = 0.06 \sin\left(\frac{2\pi}{8}\right) = 0.06 \sin(\frac{\pi}{4})\)
\(A = 0.06 \times \frac{1}{\sqrt{2}} = 0.06 \times 0.707\)
\(A \approx 0.0424 \text{ m}\).
Q.4. (A) and (B) [Attempt Set 1 OR Set 2]
Set 1
A. Obtain an expression for the binding energy of a satellite... Find total energy and binding energy...
Derivation:
Total Energy (TE) = KE + PE.
\(PE = -\frac{GMm}{r}\). \(KE = \frac{1}{2}mv_c^2 = \frac{GMm}{2r}\).
\(TE = -\frac{GMm}{2r}\).
Binding Energy (BE) is energy required to escape = \(-TE = \frac{GMm}{2r}\).
Problem:
\(h = 1600 \text{ km}\), \(R = 6400 \text{ km}\) \(\Rightarrow r = R+h = 8000 \text{ km} = 8 \times 10^6 \text{ m}\).
\(M_{earth} = 6 \times 10^{24} \text{ kg}\), \(m = 1500 \text{ kg}\).
\(BE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1500}{2 \times 8 \times 10^6}\)
\(BE = \frac{60030 \times 10^{13}}{16 \times 10^6} \approx 3.75 \times 10^{10} \text{ J}\).
Total Energy \(TE = -BE = -3.75 \times 10^{10} \text{ J}\).
B. Define root mean square velocity of gas molecules and derive an expression for it using the expression for the pressure exerted by the gas.
Definition: RMS velocity is the square root of the arithmetic mean of the squares of the velocities of the molecules of the gas.
Derivation:
Pressure \(P = \frac{1}{3} \rho C_{rms}^2\), where \(\rho\) is density.
\(C_{rms}^2 = \frac{3P}{\rho}\)
\(C_{rms} = \sqrt{\frac{3P}{\rho}}\)
Using Ideal Gas Law \(PV = nRT\), \(\rho = M/V\) (where M is molar mass).
\(P/\rho = RT/M\).
\(C_{rms} = \sqrt{\frac{3RT}{M}}\).
Set 2
A. Derive expression for velocity of a ring and solid cylinder having same radii rolling down the smooth inclined plane without slipping.
Solution:
General formula for rolling velocity at bottom of height h: \(v = \sqrt{\frac{2gh}{1 + \frac{I}{MR^2}}}\)
For Ring: \(I = MR^2\).
\(v_{ring} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}\).
For Solid Cylinder: \(I = \frac{1}{2}MR^2\).
\(v_{cyl} = \sqrt{\frac{2gh}{1 + 0.5}} = \sqrt{\frac{2gh}{1.5}} = \sqrt{\frac{4gh}{3}}\).
B. Eight droplets of water, each of radius 0.2 mm, coalesce into a single drop. Find the change in total surface energy. [Surface tension = 0.072 N/m]
Solution:
\(n = 8\), \(r = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}\). \(T = 0.072\).
Volume conservation: \(V_{big} = 8 \times V_{small} \Rightarrow R^3 = 8r^3 \Rightarrow R = 2r\).
Initial Area \(A_1 = 8 \times 4\pi r^2\).
Final Area \(A_2 = 1 \times 4\pi R^2 = 4\pi (2r)^2 = 4 \times 4\pi r^2\).
Change in Area \(\Delta A = A_1 - A_2 = 32\pi r^2 - 16\pi r^2 = 16\pi r^2\).
Change in Energy \(\Delta E = T \times \Delta A\)
\(\Delta E = 0.072 \times 16 \times 3.142 \times (2 \times 10^{-4})^2\)
\(\Delta E = 0.072 \times 50.272 \times 4 \times 10^{-8}\)
\(\Delta E \approx 1.45 \times 10^{-7} \text{ J}\).
SECTION – II
Q. 5. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
i. A wavefront is _______.
ii. When 2 identical capacitors are connected in series then resultant capacitance is _______.
iii. In meter bridge experiment to minimise errors, the wire used must be _______.
iv. µ₀ is analogous to _______.
v. Dimensions of mutual inductance are _______.
vi. The kinetic energy of emitted photoelectrons _______.
vii. The diode used in 7-segment display unit is _______.
Q. 6. Attempt any SIX: [12]
i. Draw a labelled block diagram of elements of communication system.
Answer: The diagram consists of:
Information Source \(\rightarrow\) Transmitter \(\rightarrow\) Channel (with Noise) \(\rightarrow\) Receiver \(\rightarrow\) User of Information.
ii. A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence of 45°. Find ratio of width of the beam in glass to that of air if R. I. of glass is 1.5
Solution:
Relation between widths: \(\frac{w_{glass}}{w_{air}} = \frac{\cos r}{\cos i}\).
Snell's Law: \(\mu = \frac{\sin i}{\sin r} \Rightarrow 1.5 = \frac{\sin 45}{\sin r}\).
\(\sin r = \frac{1/\sqrt{2}}{1.5} = \frac{0.707}{1.5} = 0.471\).
\(\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - 0.222} = \sqrt{0.778} \approx 0.882\).
Ratio: \(\frac{w_g}{w_a} = \frac{0.882}{\cos 45} = \frac{0.882}{0.707} \approx 1.25\).
iii. In a biprism experiment... (Red \(\lambda_1 = 6400 \AA\), 6th bright band) ... (Blue \(\lambda_2\), 8th bright band). Find \(\lambda_2\).
Solution:
The position of the \(n^{th}\) bright band is \(x_n = \frac{n \lambda D}{d}\).
Since the cross-wire is fixed, position is same.
\(x_6 (\text{Red}) = x_8 (\text{Blue})\)
\(6 \lambda_1 = 8 \lambda_2\)
\(\lambda_2 = \frac{6}{8} \lambda_1 = 0.75 \times 6400 \AA = 4800 \AA\).
iv. Two resistances prepared from the wire of the same material having diameters in ratio 2:1 and lengths equal are connected in the left and right gap... Determine the distance of null point from the left end.
Solution:
\(R = \rho \frac{L}{A} \propto \frac{1}{d^2}\) (since L and material are same).
Given diameters \(d_1 : d_2 = 2 : 1\).
Resistance ratio: \(\frac{R_1}{R_2} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
Meter bridge condition: \(\frac{R_1}{R_2} = \frac{l_x}{100 - l_x}\).
\(\frac{1}{4} = \frac{l_x}{100 - l_x}\)
\(100 - l_x = 4l_x \Rightarrow 5l_x = 100 \Rightarrow l_x = 20 \text{ cm}\).
v. An infinite line charge produces a field of 1.8 × 10³ N/C at a distance of 3 cm. Calculate the linear charge density.
Solution:
Formula: \(E = \frac{\lambda}{2\pi \epsilon_0 r}\).
Given \(E = 1.8 \times 10^3\), \(r = 0.03 \text{ m}\).
\(\frac{1}{2\pi \epsilon_0} = 2 \times 9 \times 10^9 = 18 \times 10^9\).
\(1.8 \times 10^3 = (18 \times 10^9) \frac{\lambda}{0.03}\)
\(\lambda = \frac{1.8 \times 10^3 \times 0.03}{18 \times 10^9}\)
\(\lambda = \frac{5.4 \times 10^1}{18 \times 10^9} = 0.3 \times 10^{-8} \text{ C/m} = 3 \times 10^{-9} \text{ C/m}\).
vi. Explain what is a photoelectric cell and how it is used in exposure meter.
Answer: A photoelectric cell is a device that converts light energy into electrical energy. In an exposure meter (used in photography), the light intensity controls the current produced by the cell. A galvanometer connected to the cell measures this current. The deflection indicates the correct exposure time and aperture setting based on the ambient light.
vii. Explain what is a Curie temperature and its importance.
Answer: Curie temperature is the critical temperature above which a ferromagnetic substance becomes paramagnetic. It is important because it marks a phase transition in magnetic materials where magnetic domain structure collapses due to thermal agitation.
viii. Write the Boolean expression for the output and identify the output function.
Analysis of Logic Gate Diagram:
Input A goes through a NOT gate \(\rightarrow \bar{A}\).
Input B goes through a NOT gate \(\rightarrow \bar{B}\).
Both enter an AND gate (D-shape).
Output \(Y = \bar{A} \cdot \bar{B}\).
By De Morgan's Law, \(Y = \overline{A + B}\).
Function: This acts as a NOR gate.
Q. 7. Attempt any THREE: [9]
i. Prove Snell’s law using Huygen’s principle.
Outline:
1. Consider a plane wavefront incident on a boundary.
2. Let \(v_1\) and \(v_2\) be speeds in medium 1 and 2.
3. Draw the secondary wavelets. Time taken 't' for point B to reach C is same as A to reach E.
4. \(BC = v_1 t\) and \(AE = v_2 t\).
5. In \(\Delta ABC\), \(\sin i = BC/AC = v_1 t / AC\).
6. In \(\Delta AEC\), \(\sin r = AE/AC = v_2 t / AC\).
7. Ratio: \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \mu\).
ii. On passing light of wavelength 7500 Å through two pinholes 0.5mm apart... Find the distance between the fifth bright band on one side... and the sixth dark band on the other side.
Solution:
Given: \(\lambda = 7500 \AA = 7.5 \times 10^{-7} \text{ m}\), \(d = 0.5 \text{ mm} = 5 \times 10^{-4} \text{ m}\), \(D = 200 \text{ cm} = 2 \text{ m}\).
Fringe width \(X = \frac{\lambda D}{d} = \frac{7.5 \times 10^{-7} \times 2}{5 \times 10^{-4}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}\).
Position of 5th Bright ($n=5$): \(y_{5B} = 5X = 15 \text{ mm}\).
Position of 6th Dark ($m=6$): \(y_{6D} = (2m-1)\frac{X}{2} = (11)\frac{3}{2} = 16.5 \text{ mm}\).
Since they are on opposite sides, Total Distance = \(y_{5B} + y_{6D}\).
Distance = \(15 + 16.5 = 31.5 \text{ mm}\).
iii. Derive an expression for magnetic induction at a point near infinitely long straight conductor carrying current.
Solution (using Ampere's Law):
Consider a long straight wire carrying current I. Imagine an Amperian loop (circle) of radius r around the wire.
\(\oint B \cdot dl = \mu_0 I\).
Due to symmetry, B is constant along the loop.
\(B (2\pi r) = \mu_0 I\).
\(B = \frac{\mu_0 I}{2\pi r}\).
iv. A hydrogen atom undergoes a transition from a state with n = 4 to a state with n = 1. Calculate the change in the angular momentum of the electron and the wavelength of the emitted photon.
Solution:
1. Change in Angular Momentum (\(\Delta L\)):
\(L_n = \frac{nh}{2\pi}\).
\(\Delta L = L_4 - L_1 = \frac{h}{2\pi}(4 - 1) = \frac{3h}{2\pi}\).
\(\Delta L = \frac{3 \times 6.63 \times 10^{-34}}{2 \times 3.142} \approx 3.16 \times 10^{-34} \text{ Js}\).
2. Wavelength (\(\lambda\)):
\(\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = \frac{15R}{16}\).
\(\frac{1}{\lambda} = \frac{15 \times 1.097 \times 10^7}{16}\).
\(\lambda = \frac{16}{15 \times 1.097} \times 10^{-7} \approx 0.972 \times 10^{-7} \text{ m} = 972 \AA\).
Q.8. (A) and (B) [Attempt Set 1 OR Set 2]
Set 1
A. Explain principle, construction and working of transformer.
Principle: Mutual Induction. A changing current in the primary coil creates a changing magnetic flux that induces an EMF in the secondary coil.
Construction: Consists of a laminated soft iron core with two separate coils (Primary and Secondary) wound around it.
Working: AC voltage applied to primary creates alternating flux. This flux links with secondary. \(e_p = -N_p \frac{d\phi}{dt}\), \(e_s = -N_s \frac{d\phi}{dt}\). Ratio \(\frac{e_s}{e_p} = \frac{N_s}{N_p}\).
B. ...Wavelength 4950 Å, stopping potential 0.6 V... Second source stopping potential 1.1 V. Find work function and second wavelength.
Solution:
Equation: \(K_{max} = eV_s = \frac{hc}{\lambda} - W_0\).
Case 1: \(\lambda_1 = 4950 \AA\), \(V_{s1} = 0.6 \text{ V}\).
Energy \(E_1 = \frac{12400}{4950} \approx 2.505 \text{ eV}\).
\(0.6 = 2.505 - W_0 \Rightarrow W_0 = 1.905 \text{ eV}\).
Case 2: \(V_{s2} = 1.1 \text{ V}\).
\(1.1 = E_2 - 1.905 \Rightarrow E_2 = 3.005 \text{ eV}\).
\(\lambda_2 = \frac{12400}{3.005} \approx 4126 \AA\).
Set 2
A. Describe Davisson Germer experiment to explain wave nature.
Description: An electron gun fires electrons at a Nickel crystal. The scattered electrons are detected by a movable detector. A graph of intensity vs angle is plotted. A peak is observed at 54V and \(50^\circ\). The wavelength calculated using de Broglie hypothesis matches the wavelength calculated using Bragg's law for X-ray diffraction, confirming the wave nature of electrons.
B. Three capacitors of capacities 2 µF, 4 µF and 6 µF are connected in series and in parallel respectively. Compare the effective capacities in the two combinations.
Solution:
\(C_1=2, C_2=4, C_3=6\) µF.
Series (\(C_s\)):
\(\frac{1}{C_s} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6+3+2}{12} = \frac{11}{12}\).
\(C_s = \frac{12}{11} \approx 1.09 \mu\text{F}\).
Parallel (\(C_p\)):
\(C_p = 2 + 4 + 6 = 12 \mu\text{F}\).
Comparison (Ratio):
\(\frac{C_p}{C_s} = \frac{12}{12/11} = 11:1\).