OMTEX AD 2

HSC Physics October 2017 Board Question Paper Solution | Maharashtra Board 12th

Maharashtra Board Class 12 Physics Question Paper Solution July 2017

Physics Board Question Paper

July 2017 - Maharashtra State Board
Total Marks: 70 Time: 3 Hours
Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Maharashtra Board Physics Paper
SECTION – I
Q.1. Attempt any SIX: [12]
i. If the angular speed of the earth is \(7.26 \times 10^{-5}\) rad/s and radius of the earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.

Given:
Angular speed \(\omega = 7.26 \times 10^{-5}\) rad/s
Radius \(R = 6400 \text{ km} = 6.4 \times 10^6\) m
Mass \(m = 1\) kg

Formula:
The weight at latitude \(\lambda\) is \(W_\lambda = mg - mR\omega^2 \cos^2 \lambda\).
At equator, \(\lambda = 0^\circ\), so \(W_e = mg - mR\omega^2\).
At pole, \(\lambda = 90^\circ\), so \(W_p = mg\).

Calculation:
Change in weight \(\Delta W = W_p - W_e = mR\omega^2\).

\(\Delta W = 1 \times (6.4 \times 10^6) \times (7.26 \times 10^{-5})^2\)
\(\Delta W = 6.4 \times 10^6 \times 52.7076 \times 10^{-10}\)
\(\Delta W = 6.4 \times 52.7076 \times 10^{-4}\)
\(\Delta W \approx 337.3 \times 10^{-4}\)
\(\Delta W \approx 0.03373 \text{ N}\)

Answer: The change in weight is 0.0337 N.

ii. A small body of mass 0.3 kg oscillates in vertical plane with the help of a string 0.5 m long with a constant speed of 2 m/s. It makes an angle of 60° with the vertical. Calculate tension in the string (g = 9.8 m/s²).

Given:
Mass \(m = 0.3\) kg, Length \(r = 0.5\) m, Speed \(v = 2\) m/s, Angle \(\theta = 60^\circ\).

Concept:
Assuming the angle is measured from the downward vertical (standard for such problems unless specified otherwise), the forces acting along the string are Tension \(T\) (inward) and the component of weight \(mg \cos \theta\) (outward). The net radial force provides centripetal acceleration.

\(T - mg \cos \theta = \frac{mv^2}{r}\)
\(T = \frac{mv^2}{r} + mg \cos 60^\circ\)
\(T = \frac{0.3 \times (2)^2}{0.5} + 0.3 \times 9.8 \times 0.5\)
\(T = \frac{1.2}{0.5} + 1.47\)
\(T = 2.4 + 1.47\)
\(T = 3.87 \text{ N}\)

Answer: The tension in the string is 3.87 N.

iii. Two soap bubbles have radii in the ratio 4:3. What is the ratio of work done to blow these bubbles?

Solution:
Work done to blow a soap bubble is \(W = T \times 2 \times 4\pi r^2\) (since a bubble has two surfaces).

Therefore, \(W \propto r^2\).

\(\frac{W_1}{W_2} = \left( \frac{r_1}{r_2} \right)^2\)
\(\frac{W_1}{W_2} = \left( \frac{4}{3} \right)^2 = \frac{16}{9}\)

Answer: The ratio of work done is 16:9.

iv. At what temperature will average kinetic energy of gas be exactly half of its value at N.T.P.?

Given:
At N.T.P., \(T_1 = 273.15\) K.
Kinetic Energy \(E \propto T\).
Condition: \(E_2 = \frac{1}{2} E_1\).

\(\frac{E_2}{E_1} = \frac{T_2}{T_1}\)
\(\frac{1}{2} = \frac{T_2}{273.15}\)
\(T_2 = \frac{273.15}{2} = 136.575 \text{ K}\)

Answer: At 136.575 K.

v. Define surface tension and surface energy.

Surface Tension: It is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of the liquid.
S.I. Unit: N/m.

Surface Energy: The extra potential energy possessed by the molecules of the surface layer of a liquid due to unbalanced cohesive forces is called surface energy. It is equal to the work done in increasing the surface area by unity.
S.I. Unit: Joule (J).

vi. Prove that \(g_h = g(1 - \frac{2h}{R})\) where \(g_h\) is the acceleration due to gravity at altitude h and h << R.

Proof:
1. Acceleration due to gravity on the earth's surface: \(g = \frac{GM}{R^2}\).
2. Acceleration due to gravity at height \(h\): \(g_h = \frac{GM}{(R+h)^2}\).
3. Dividing \(g_h\) by \(g\):
\(\frac{g_h}{g} = \frac{R^2}{(R+h)^2} = \frac{R^2}{R^2(1 + \frac{h}{R})^2} = (1 + \frac{h}{R})^{-2}\).
4. Using Binomial expansion for \(h \ll R\), neglect higher order terms:
\((1 + x)^n \approx 1 + nx\). Here \(x = h/R\) and \(n = -2\).
\(\frac{g_h}{g} = 1 - \frac{2h}{R}\).
5. Therefore, \(g_h = g \left( 1 - \frac{2h}{R} \right)\).

vii. Explain the physical significance of radius of gyration.

The radius of gyration (\(K\)) is the distance from the axis of rotation to a point where the entire mass of the body can be assumed to be concentrated so that its moment of inertia remains the same.

Significance:
1. It measures the distribution of mass about the axis of rotation.
2. A smaller \(K\) implies mass is distributed close to the axis (easier to rotate), while a larger \(K\) implies mass is distributed far from the axis.

viii. Draw neat, labelled diagram showing different forces acting on a vehicle moving along a banked road.

(Note: In the exam, a drawing is required. Here is the description of the necessary elements.)

  • Diagram Description:
    • A cross-section of a road inclined at angle \(\theta\) to the horizontal.
    • A vehicle (represented as a block or car) on the slope.
    • Weight (mg): Acting vertically downwards through the center of gravity.
    • Normal Reaction (N): Acting perpendicular to the road surface upwards.
    • Friction (f): Acting along the road surface, downwards/inwards (if speed is high) or upwards/outwards (if speed is low) to prevent skidding. Typically for optimum speed, friction is zero. For maximum safety speed derivation, friction acts downwards along the slope.
    • Components of N: \(N \cos\theta\) (vertical) and \(N \sin\theta\) (horizontal).

HSC Physics Board Papers with Solution

Q.2. Attempt any THREE: [9]
i. Prove Kirchhoff’s law of radiation theoretically.

Statement: The coefficient of absorption of a body is equal to its coefficient of emission at any given temperature. (\(a = e\)).

Proof Setup: Consider an enclosure at constant temperature \(T\). Inside, place two bodies: one ordinary body \(A\) and one perfectly black body \(B\).

Let:
\(R\) = Emissive power of body \(A\).
\(R_b\) = Emissive power of black body \(B\).
\(a\) = Coefficient of absorption of \(A\).
\(Q\) = Radiant heat incident per unit area per unit time on both.

Equilibrium for A:
Heat absorbed = \(aQ\).
Heat emitted = \(R\).
At equilibrium, \(R = aQ \Rightarrow Q = R/a\).

Equilibrium for Black Body B:
\(a = 1\) for black body.
Heat absorbed = \(1 \cdot Q = Q\).
Heat emitted = \(R_b\).
At equilibrium, \(R_b = Q\).

Conclusion:
From both equations, \(R/a = R_b \Rightarrow a = R/R_b\).
By definition, coefficient of emission \(e = R/R_b\).
Therefore, \(a = e\).

ii. Within the elastic limit, find the work done by a stretching force on a wire.

Consider a wire of length \(L\) and area \(A\). Let \(F\) be the force applied, stretching it by \(l\).

Young's modulus \(Y = \frac{F/A}{l/L} \Rightarrow F = \frac{YA l}{L}\).

Work done \(dW\) to stretch by small amount \(dx\) is \(dW = F dx\).
Total work \(W = \int_0^l F dx = \int_0^l \frac{YA x}{L} dx\).
\(W = \frac{YA}{L} \left[ \frac{x^2}{2} \right]_0^l = \frac{1}{2} \frac{YA l^2}{L}\).

Rearranging: \(W = \frac{1}{2} \left( \frac{YAl}{L} \right) \cdot l = \frac{1}{2} F \cdot l\).

Result: Work Done = \(\frac{1}{2} \times \text{Load} \times \text{Extension}\).

iii. A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces ‘Y’ beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of 90 Hz. Find ‘Y’ and frequency of the first and the last tuning forks.

Given:
Number of forks \(n = 12\).
Beat frequency = \(Y\).
Relation: \(n_{last} = 2 \times n_{first}\) (Octave).
Frequency of 5th fork \(n_5 = 90\) Hz.

Formula: Frequency of kth fork in AP: \(n_k = n_1 + (k-1)Y\).

1. For the last fork (12th):
\(n_{12} = n_1 + (12-1)Y = n_1 + 11Y\).
Since \(n_{12} = 2n_1\), we have:
\(2n_1 = n_1 + 11Y \Rightarrow n_1 = 11Y\).

2. For the 5th fork:
\(n_5 = n_1 + (5-1)Y = n_1 + 4Y\).
Substitute \(n_1 = 11Y\):
\(90 = 11Y + 4Y = 15Y\).
\(Y = 90 / 15 = 6\) beats/sec.

3. Find frequencies:
\(n_1 = 11Y = 11 \times 6 = 66\) Hz.
\(n_{12} = 2n_1 = 2 \times 66 = 132\) Hz.

Answer: \(Y = 6\), First frequency = 66 Hz, Last frequency = 132 Hz.

iv. A uniform solid sphere has radius 0.2 m and density \(8 \times 10^3\) kg/m³. Find the moment of inertia about the tangent to its surface. (\(\pi = 3.142\))

Given:
\(R = 0.2\) m, \(\rho = 8 \times 10^3\) kg/m³.

Mass Calculation:
\(M = \text{Density} \times \text{Volume} = \rho \times \frac{4}{3}\pi R^3\).
\(M = 8000 \times \frac{4}{3} \times 3.142 \times (0.2)^3\).
\(M = 8000 \times 1.333 \times 3.142 \times 0.008\).
\(M \approx 268.12\) kg.

Moment of Inertia (Tangent):
By Parallel Axis Theorem: \(I_{tan} = I_{CM} + MR^2\).
For solid sphere, \(I_{CM} = \frac{2}{5}MR^2\).
\(I_{tan} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2\).

\(I_{tan} = \frac{7}{5} \times 268.12 \times (0.2)^2\)
\(I_{tan} = 1.4 \times 268.12 \times 0.04\)
\(I_{tan} = 15.01 \text{ kg m}^2\)

Answer: Moment of inertia is approx 15.01 kg m².

Q.3. A. Define linear simple harmonic motion. Assuming the expression for displacement of a particle starting from extreme position, explain graphically the variation of velocity and acceleration w.r.t. time.

Definition: Linear SHM is a periodic motion in which the restoring force (or acceleration) is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.

Expressions (Starting from Extreme Position):
Displacement: \(x = A \cos(\omega t)\)
Velocity: \(v = \frac{dx}{dt} = -A\omega \sin(\omega t)\)
Acceleration: \(a = \frac{dv}{dt} = -A\omega^2 \cos(\omega t)\)

Graphical Variation:
1. Displacement curve: A cosine wave starting at max amplitude \(A\).
2. Velocity curve: A negative sine wave starting at 0. Phase difference of \(\pi/2\) with displacement.
3. Acceleration curve: A negative cosine wave starting at max negative value \(-\omega^2 A\). Phase difference of \(\pi\) with displacement.

Q.3. B. A clock regulated by seconds pendulum, keeps correct time. During summer, length of pendulum increases to 1.005 m. How much will the clock gain or loose in one day? (g = 9.8 m/s² and \(\pi = 3.142\))

Given:
Seconds pendulum period \(T = 2\) s.
New length \(L' = 1.005\) m.
Standard length \(L\) (calculated from \(T=2\)):
\(L = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 4}{4 \times (3.142)^2} \approx 0.993\) m.

Calculation:
New Period \(T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.005}{9.8}} \approx 2.0121\) s.
Since \(T' > T\), the clock runs slow (loses time).

Number of seconds in 1 day = 86400.
Number of oscillations correct clock makes = \(86400 / 2 = 43200\).
Time taken by slow clock for 43200 oscillations = \(43200 \times 2.0121 = 86922.7\) s.

Actually, the question asks gain or loss in one day (86400s).
Number of oscillations actually performed by slow clock in 1 day = \(86400 / 2.0121 \approx 42940\).
Time registered by clock = \(42940 \times 2 = 85880\) s.
Loss = True Time - Registered Time = \(86400 - 85880 = 520\) s.

Answer: The clock will lose approximately 520 seconds in one day.

OR
A. Discuss different modes of vibrations in an air column of a pipe open at both the ends. State the cause of end correction. Find the end correction for the pipe open at both the ends in fundamental mode.

Modes of Vibration (Open Pipe):
1. Fundamental Mode: Antinodes at both ends, one node in middle. \(L = \lambda/2 \Rightarrow n = v/2L\).
2. Second Harmonic (1st Overtone): \(L = \lambda \Rightarrow n_1 = v/L = 2n\).
3. Third Harmonic (2nd Overtone): \(L = 3\lambda/2 \Rightarrow n_2 = 3v/2L = 3n\).
frequencies are in ratio 1:2:3 (All harmonics are present).

Cause of End Correction:
The antinode is not formed exactly at the open end but slightly outside because air particles at the open end are not free to move in all directions due to the resistance of the outside medium. This distance is called end correction (\(e \approx 0.3d\)).

Formula:
For open pipe (two ends): Corrected length \(L_{corr} = l + 2e\).
\(v = n \lambda = n (2)(l+2e)\).
\(e = \frac{v}{2n} - l\) (if solving for e).

B. What should be tension applied to a wire of length 1 m and mass 10 gram, if it has to vibrate with fundamental frequency of 50 Hz?

Given:
Length \(L = 1\) m, Mass \(M = 10\) g = 0.01 kg, Frequency \(n = 50\) Hz.
Linear density \(\mu = M/L = 0.01 / 1 = 0.01\) kg/m.

Formula: \(n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\).

\(50 = \frac{1}{2(1)} \sqrt{\frac{T}{0.01}}\)
\(100 = \sqrt{\frac{T}{0.01}}\)
Squaring both sides:
\(10000 = \frac{T}{0.01}\)
\(T = 10000 \times 0.01 = 100 \text{ N}\).

Answer: The tension applied should be 100 N.

Q.4. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
  • i. A body of mass ‘m’ performs uniform circular motion along a circular path of radius ‘r’ with velocity ‘v’. If its angular momentum is L, then the centripetal force acting on it is _______.
    • (A) \(mL^2/r^3\)
    • (B) \(L^2/mr\)
    • (C) \(L^2/mr^2\)
    • (D) \(L^2/mr^3\)
    Answer: (D)
    Explanation: \(L = mvr \Rightarrow v = L/mr\). Centripetal Force \(F = mv^2/r = m(L/mr)^2/r = m(L^2/m^2r^2)/r = L^2/mr^3\).
  • ii. If the Earth completely loses its gravity, then for any body _______.
    • (A) both mass and weight become zero.
    • (B) neither mass nor weight become zero.
    • (C) weight becomes zero but not the mass.
    • (D) mass becomes zero but not the weight.
    Answer: (C)
    Explanation: Weight is the force of gravity (\(mg\)). If \(g=0\), Weight=0. Mass is intrinsic matter content and remains constant.
  • iii. If a rigid body of radius ‘R’ starts from rest and rolls down an inclined plane of inclination ‘\(\theta\)’ then linear acceleration of body rolling down the plane is _______.
    Answer: (A) or (C) [Formatting Dependent]
    Correct Formula: \(a = \frac{g \sin \theta}{1 + K^2/R^2}\). Looking at the options in the image, Option (C) matches this form: \(\frac{g \sin \theta}{1 + K^2/R^2}\).
  • iv. 1000 tiny mercury droplets coalesce to form a bigger drop. In this process, temperature of the drop _______ .
    • (A) increases
    • (B) may increase or decrease
    • (C) decreases
    • (D) does not change
    Answer: (A)
    Explanation: Surface area decreases, so surface energy is released as heat.
  • v. Doppler effect is not applicable when _______.
    • (A) source and observer are at rest.
    Answer: (A)
    Explanation: Doppler effect requires relative motion. If both are at rest, frequency remains unchanged.
  • vi. If the total kinetic energy per unit volume of gas enclosed in a container is E, the pressure exerted by the gas is _______.
    • (D) \(\frac{2}{3}E\)
    Answer: (D)
    Explanation: \(P = \frac{2}{3}E\).
  • vii. Two wires of the same material have radii \(r_A\) and \(r_B\) respectively. The radius of wire A is twice the radius of wire B. If they are stretched by same load then stress on wire B is _______.
    • (D) four times that of A.
    Answer: (D)
    Explanation: Stress \(\propto 1/r^2\). \(r_A = 2r_B\). Stress\(_A \propto 1/4r_B^2\). Stress\(_B \propto 1/r_B^2\). So Stress\(_B = 4 \times\) Stress\(_A\).
SECTION – II
Q.5. Attempt any SIX: [12]
i. Thorium \({}_{90}\text{Th}^{232}\) is disintegrated into lead \({}_{82}\text{Pb}^{200}\). Find the number of \(\alpha\) and \(\beta\) particles emitted in disintegration.

Solution:
Reaction: \({}_{90}\text{Th}^{232} \rightarrow {}_{82}\text{Pb}^{200} + n_1({}_{2}\text{He}^{4}) + n_2({}_{-1}\text{e}^{0})\).
Mass Balance: \(232 = 200 + 4n_1 \Rightarrow 4n_1 = 32 \Rightarrow n_1 = 8\) (\(\alpha\) particles).
Charge Balance: \(90 = 82 + 2n_1 - n_2 \Rightarrow 90 = 82 + 16 - n_2\).
\(90 = 98 - n_2 \Rightarrow n_2 = 8\) (\(\beta\) particles).

Answer: 8 \(\alpha\) particles and 8 \(\beta\) particles.

ii. If the work function of a metal is 3 eV, calculate the threshold wavelength of that metal.

Given: \(\Phi = 3 \text{ eV} = 3 \times 1.6 \times 10^{-19} \text{ J}\).

\(\lambda_0 = \frac{hc}{\Phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3 \times 1.6 \times 10^{-19}}\)
\(\lambda_0 = \frac{6.63 \times 10^{-26}}{1.6 \times 10^{-19}}\)
\(\lambda_0 = 4.14375 \times 10^{-7} \text{ m} = 4143.75 \text{ \AA}\)
iii. Three capacitors of capacities 8 µF, 8 µF and 4 µF are connected in a series and potential difference of 120 volt is maintained across the combination. Calculate the charge on capacitor of capacity 4 µF.

Solution:
In series, charge \(Q\) is same on all capacitors.
Equivalent capacitance \(1/C_s = 1/8 + 1/8 + 1/4 = 2/8 + 2/8 = 4/8 \Rightarrow C_s = 2 \mu\text{F}\).
Total Charge \(Q = C_s V = 2 \mu\text{F} \times 120 \text{ V} = 240 \mu\text{C}\).
Charge on 4 µF capacitor is also 240 µC.

iv. If the total energy of radiation of frequency \(10^{14}\) Hz is 6.63 J, calculate the number of photons in the radiation.

Solution:
Energy of one photon \(E = h\nu = 6.63 \times 10^{-34} \times 10^{14} = 6.63 \times 10^{-20}\) J.
Total Energy \(E_{total} = N \times E\).
\(N = \frac{6.63}{6.63 \times 10^{-20}} = 10^{20}\).

Answer: \(10^{20}\) photons.

v. Distinguish between diamagnetic and paramagnetic substances.
Diamagnetic Paramagnetic
1. Weakly repelled by magnets. 1. Weakly attracted by magnets.
2. Susceptibility is small and negative. 2. Susceptibility is small and positive.
3. Independent of temperature. 3. Inversely proportional to temperature.
Q.6. Attempt any THREE: [9]
iii. When a resistance of 12 ohm is connected across a cell, its terminal potential difference is balanced by 120 cm length of potentiometer wire. When the resistance of 18 ohm is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

Given:
Case 1: \(R_1 = 12 \Omega\), \(L_1 = 120\) cm.
Case 2: \(R_2 = 18 \Omega\), \(L_2 = 150\) cm.
Formula: \(V = kL = E \frac{R}{R+r}\).

\(\frac{L_1}{L_2} = \frac{R_1/(R_1+r)}{R_2/(R_2+r)}\)
\(\frac{120}{150} = \frac{12}{18} \times \frac{18+r}{12+r}\)
\(\frac{4}{5} = \frac{2}{3} \times \frac{18+r}{12+r}\)
\(\frac{12}{10} = \frac{18+r}{12+r} \Rightarrow 1.2(12+r) = 18+r\)
\(14.4 + 1.2r = 18 + r \Rightarrow 0.2r = 3.6 \Rightarrow r = 18 \Omega\).

For open circuit length (\(L_0\)):
\(E = kL_0\). From Case 1: \(E \frac{12}{12+18} = k(120)\).
\(E \frac{12}{30} = k(120) \Rightarrow E(0.4) = k(120) \Rightarrow E = k(300)\).
So \(L_0 = 300\) cm.

Answer: Balancing length \(L_0 = 300\) cm, Internal resistance \(r = 18 \Omega\).

iv. Find the ratio of longest wavelength in Paschen series to shortest wavelength in Balmer series.

Solution:
Longest Paschen (\(n=4 \rightarrow 3\)): \(\frac{1}{\lambda_P} = R(\frac{1}{9} - \frac{1}{16}) = R\frac{7}{144}\).
Shortest Balmer (\(n=\infty \rightarrow 2\)): \(\frac{1}{\lambda_B} = R(\frac{1}{4} - 0) = \frac{R}{4}\).

\(\frac{\lambda_P}{\lambda_B} = \frac{R/4}{7R/144} = \frac{1}{4} \times \frac{144}{7} = \frac{36}{7}\).

Answer: Ratio is 36:7.

Q.7. A. State the principle of working of transformer... Derive expression...

Principle: Mutual Induction.
Relation: \(\frac{E_s}{E_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k\).

Q.7. B. Find the magnetization of a bar magnet of length 10 cm and cross-sectional area 4 cm², if the magnetic moment is 2 Am².

Given: \(L = 0.1\) m, \(A = 4 \times 10^{-4}\) m², \(m = 2\) Am².

Magnetization \(M_z = \frac{\text{Net Moment}}{\text{Volume}}\)
\(V = A \times L = 4 \times 10^{-5}\) m³.
\(M_z = \frac{2}{4 \times 10^{-5}} = 0.5 \times 10^5 = 5 \times 10^4\) A/m.
OR
Q.7. B. The refractive indices of glass and water w.r.t. air are 3/2 and 4/3 respectively. Determine the refractive index of glass w.r.t. water.
\({}_w\mu_g = \frac{{}_a\mu_g}{{}_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8} = 1.125\).
Q.8. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
  • i. The logic gate which produces LOW output when one of the input is HIGH and produces HIGH output only when all of its inputs are LOW is called _______.
    Answer: (C) NOR gate
  • ii. For efficient radiation and reception of signal with wavelength \(\lambda\), the transmitting antennas would have length comparable to ______.
    Answer: (D) \(\lambda/4\) of frequency used (Note: Quarter wave monopole is standard).
  • iii. Cyclotron can not accelerate ______.
    Answer: (B) neutrons (Neutral particles cannot be accelerated by electric fields).
  • iv. In series LCR circuit at resonance, phase difference between current and e.m.f. of source is ______.
    Answer: (D) zero rad
  • v. When unknown resistance is determined by meter bridge, the error due to contact resistance is minimized ______.
    Answer: (B) by interchanging the positions of known and unknown resistance.
  • vi. The ratio of kinetic energy of an electron in Bohr’s orbit to its total energy in the same orbit is ______.
    Answer: (A) – 1 (KE = -Total Energy).
  • vii. Using monochromatic light of wavelength \(\lambda\) in Young’s double slit experiment, the eleventh dark fringe is obtained on the screen for a phase difference of ______.
    Answer: (D) 21\(\pi\) rad
    Explanation: Phase diff for nth dark = \((2n-1)\pi\). For \(n=11\): \((22-1)\pi = 21\pi\).
Labels: , ,