HSC Maharashtra Board Question Paper Solution: March 2013 (Physics - I)
Below are the comprehensive solutions for the Physics Section I of the HSC Maharashtra Board question paper from March 2013.
PHYSICS: SECTION – I
Q. 1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
The moment of inertia of a thin uniform rod of mass M and length L, about an axis passing through a point, midway between the centre and one end, perpendicular to its length is ______.
Correct Answer: (B) \(\frac{7}{48} ML^2\)
The moment of inertia of a thin rod about an axis passing through its centre of mass (IC) perpendicular to its length is \(I_C = \frac{ML^2}{12}\).
The point midway between the centre and one end is at a distance \(h = \frac{L}{2} \times \frac{1}{2} = \frac{L}{4}\) from the centre.
Using the Theorem of Parallel Axes: \(I_O = I_C + Mh^2\)
\(I_O = \frac{ML^2}{12} + M(\frac{L}{4})^2 = \frac{ML^2}{12} + \frac{ML^2}{16}\)
LCM of 12 and 16 is 48.
\(I_O = \frac{4ML^2 + 3ML^2}{48} = \frac{7}{48}ML^2\)
‘n’ droplets of equal size of radius r coalesce to form a bigger drop of radius R. The energy liberated is equal to _______. (T = Surface tension of water)
Correct Answer: (A) \(4\pi R^2 T [n^{1/3} - 1]\)
Volume remains constant: \(n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = n^{1/3}r \implies r = R n^{-1/3}\).
Initial Surface Area \(A_1 = n \times 4\pi r^2\).
Final Surface Area \(A_2 = 4\pi R^2\).
Change in area \(\Delta A = A_1 - A_2 = 4\pi (n r^2 - R^2)\).
Substitute \(r = R n^{-1/3}\):
\(A_1 = n 4\pi (R n^{-1/3})^2 = 4\pi R^2 n^{1/3}\).
Energy Liberated \(E = T \times \Delta A = T(A_1 - A_2) = T(4\pi R^2 n^{1/3} - 4\pi R^2)\).
\(E = 4\pi R^2 T [n^{1/3} - 1]\).
The buckling of a beam is found to be more if _______.
Correct Answer: (D) the depth of the beam is small.
Buckling stability depends on the Geometrical Moment of Inertia \(I\). For a rectangular beam, \(I = \frac{bd^3}{12}\), where \(b\) is breadth and \(d\) is depth. Since \(d\) is cubed, a smaller depth significantly reduces the moment of inertia, making the beam less stable and more prone to buckling.
When a transverse wave on a string is reflected from the free end, the phase change produced is _______.
Correct Answer: (A) zero rad
When a wave is reflected from a free end (an end that is free to move), there is no phase reversal. The crest reflects as a crest.
The number of degrees of freedom for a rigid diatomic molecule is _______.
Correct Answer: (B) 5
A rigid diatomic molecule has 3 translational degrees of freedom and 2 rotational degrees of freedom (about the two axes perpendicular to the bond axis). Total = 3 + 2 = 5.
Two particles perform linear simple harmonic motion along the same path of length 2A and period T as shown in the graph below. The phase difference between them is _______.
Correct Answer: (C) \(\frac{\pi}{2}\) rad
Particle 1 starts from the mean position (0) and moves towards positive extreme. Equation: \(x_1 = A \sin(\omega t)\).
Particle 2 starts from the positive extreme position (A) and moves towards mean. Equation: \(x_2 = A \cos(\omega t) = A \sin(\omega t + \frac{\pi}{2})\).
The phase difference is \(\phi = (\omega t + \frac{\pi}{2}) - \omega t = \frac{\pi}{2}\) radians.
The light from the Sun is found to have a maximum intensity near the wavelength of 470 nm. Assuming the surface of the Sun as a black body, the temperature of the Sun is _______. [Wien’s constant \(b = 2.898 \times 10^{-3}\) mK]
Correct Answer: (C) 6166 K
According to Wien's Displacement Law: \(\lambda_{max} T = b\)
Given \(\lambda_{max} = 470 \text{ nm} = 470 \times 10^{-9} \text{ m}\).
\(T = \frac{b}{\lambda_{max}} = \frac{2.898 \times 10^{-3}}{470 \times 10^{-9}}\)
\(T = \frac{2.898}{4.7} \times 10^4 = 0.61659 \times 10^4 \approx 6166 \text{ K}\).
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Q. 2. Attempt any SIX: [12]
1. Kepler’s Law of Orbit (First Law):
All planets move in elliptical orbits around the Sun, with the Sun situated at one of the foci of the ellipse.
2. Kepler’s Law of Equal Areas (Second Law):
The line joining a planet and the Sun sweeps out equal areas in equal intervals of time. (This implies that the areal velocity of the planet is constant).
A car of mass 1500 Kg rounds a curve of radius 250m at 90 Km/hour. Calculate the centripetal force acting on it.
Given:
Mass \(m = 1500 \text{ kg}\)
Radius \(r = 250 \text{ m}\)
Velocity \(v = 90 \text{ km/hr} = 90 \times \frac{5}{18} \text{ m/s} = 25 \text{ m/s}\)
Formula:
Centripetal Force \(F_{cp} = \frac{mv^2}{r}\)
Calculation:
\(F_{cp} = \frac{1500 \times (25)^2}{250}\)
\(F_{cp} = \frac{1500 \times 625}{250}\)
\(F_{cp} = 1500 \times 2.5\)
\(F_{cp} = 3750 \text{ N}\)
Answer: The centripetal force acting on the car is 3750 N.
(Simplified Representation: Double walled hollow sphere, aperture, conical projection opposite to aperture, evacuated space between walls, inner surface coated with lamp black.)
A mass M attached to a spring oscillates with a period of 2 seconds. If the mass is increased by 2 Kg, the period increases by 1 second. Find the initial mass, assuming that Hooke’s law is obeyed.
Given:
Initial period \(T_1 = 2 \text{ s}\), Mass \(m_1 = M\)
Final period \(T_2 = 2 + 1 = 3 \text{ s}\), Mass \(m_2 = M + 2 \text{ kg}\)
Formula: \(T = 2\pi \sqrt{\frac{m}{k}}\)
Therefore, \(T \propto \sqrt{m}\)
Calculation:
\(\frac{T_1}{T_2} = \sqrt{\frac{m_1}{m_2}}\)
\(\frac{2}{3} = \sqrt{\frac{M}{M+2}}\)
Squaring both sides:
\(\frac{4}{9} = \frac{M}{M+2}\)
\(4(M + 2) = 9M\)
\(4M + 8 = 9M\)
\(5M = 8\)
\(M = \frac{8}{5} = 1.6 \text{ kg}\)
Answer: The initial mass M is 1.6 kg.
| Free Vibrations | Forced Vibrations |
|---|---|
| 1. Vibrations of a body under the influence of its own restoring force after being disturbed from equilibrium. | 1. Vibrations of a body under the influence of an external periodic force. |
| 2. The frequency depends on the natural properties (dimensions, elastic constants) of the body (Natural Frequency). | 2. The frequency is equal to the frequency of the external driving force. |
| 3. Amplitude decreases with time due to damping (eventually stops) unless in vacuum. | 3. Amplitude remains constant as long as the external force is applied. |
The surface tension of water at 0°C is 75.5 dyne/cm. Find surface tension of water at 25°C. [ \(\alpha\) for water = 0.0021/°C ]
Given:
\(T_0 = 75.5 \text{ dyne/cm}\)
\(\Delta t = 25 - 0 = 25^\circ\text{C}\)
\(\alpha = 0.0021 /^\circ\text{C}\)
Formula:
\(T = T_0(1 - \alpha \Delta t)\)
Calculation:
\(T_{25} = 75.5 [1 - (0.0021 \times 25)]\)
\(T_{25} = 75.5 [1 - 0.0525]\)
\(T_{25} = 75.5 [0.9475]\)
\(T_{25} \approx 71.54 \text{ dyne/cm}\)
Answer: Surface tension at 25°C is approx 71.54 dyne/cm.
Consider a rectangular frame of wire ABCD with a movable wire PQ. A soap film is formed on the frame.
1. Surface tension \(T\) acts perpendicular to PQ, pulling it inwards. Force \(F = T \times 2L\) (factor of 2 because the film has two surfaces).
2. To move the wire PQ outwards by a small distance \(dx\) isothermally, an external force \(F' = F\) must be applied.
Work done \(dW = F' \times dx = T \times 2L \times dx\).
3. The increase in surface area of the film is \(dA = 2 \times (L \times dx) = 2Ldx\).
4. Therefore, \(dW = T \times dA\).
5. This work done is stored as potential energy called Surface Energy (E).
\(E = T \times dA\)
6. Surface Energy per unit area = \(\frac{E}{dA} = T\).
Conclusion: Surface tension is numerically equal to the surface energy per unit area.
A wheel of moment of inertia 1 Kgm² is rotating at a speed of 40 rad/s. Due to friction on the axis, the wheel comes to rest in 10 minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.
Given:
\(I = 1 \text{ kg m}^2\), \(\omega_0 = 40 \text{ rad/s}\), \(\omega_f = 0\)
Total time to stop \(t = 10 \text{ min} = 600 \text{ s}\).
Step 1: Find angular retardation (\(\alpha\))
\(\omega_f = \omega_0 + \alpha t\)
\(0 = 40 + \alpha(600) \implies \alpha = -\frac{40}{600} = -\frac{1}{15} \text{ rad/s}^2\).
Step 2: Find velocity 2 minutes before stopping
"Two minutes before it comes to rest" means time remaining is 2 minutes (120 s).
Let us consider the motion in reverse from rest, or calculate the velocity at \(t' = 10 - 2 = 8 \text{ min} = 480 \text{ s}\).
\(\omega_{new} = \omega_0 + \alpha t'\)
\(\omega_{new} = 40 - \frac{1}{15}(480)\)
\(\omega_{new} = 40 - 32 = 8 \text{ rad/s}\).
Step 3: Calculate Angular Momentum (\(L\))
\(L = I \omega_{new}\)
\(L = 1 \times 8 = 8 \text{ kg m}^2/\text{s}\).
Answer: The angular momentum is 8 kg m²/s.
Q. 3. Attempt any THREE: [9]
A particle of mass m, just completes the vertical circular motion. Derive the expression for the difference in tensions at the highest and the lowest points.
Consider a particle of mass \(m\) tied to a string of length \(r\) performing vertical circular motion.
At Highest Point (H):
Forces downwards: Tension \(T_H\) and Weight \(mg\).
Centripetal force: \(T_H + mg = \frac{mv_H^2}{r}\) ...(1)
At Lowest Point (L):
Forces: Tension \(T_L\) (up) and Weight \(mg\) (down).
Centripetal force: \(T_L - mg = \frac{mv_L^2}{r}\) ...(2)
Energy Conservation:
Total Energy at H = Total Energy at L
\(\frac{1}{2}mv_H^2 + mg(2r) = \frac{1}{2}mv_L^2 + 0\)
Multiply by 2/r: \(\frac{mv_H^2}{r} + 4mg = \frac{mv_L^2}{r}\)
Substitute \(\frac{mv^2}{r}\) terms from (1) and (2):
\((T_H + mg) + 4mg = (T_L - mg)\)
\(T_H + 5mg = T_L - mg\)
\(T_L - T_H = 6mg\)
Result: The difference in tensions at the lowest and highest points is 6mg.
The Earth is rotating with angular velocity \(\omega\) about its own axis. R is the radius of the Earth. If \(R\omega^2 = 0.03386 \text{ m/s}^2\), calculate the weight of a body of mass 100 gram at latitude 25°. (\(g = 9.8 \text{ m/s}^2\))
Given:
\(R\omega^2 = 0.03386 \text{ m/s}^2\)
Mass \(m = 100 \text{ g} = 0.1 \text{ kg}\)
Latitude \(\phi = 25^\circ\)
\(g = 9.8 \text{ m/s}^2\)
Formula:
Effective gravity \(g' = g - R\omega^2 \cos^2 \phi\)
Weight \(W' = m g'\)
Calculation:
\(\cos(25^\circ) \approx 0.9063\)
\(\cos^2(25^\circ) \approx (0.9063)^2 \approx 0.8214\)
\(g' = 9.8 - 0.03386 \times 0.8214\)
\(g' = 9.8 - 0.0278\)
\(g' = 9.7722 \text{ m/s}^2\)
Weight \(W' = 0.1 \times 9.7722 = 0.97722 \text{ N}\)
Answer: The weight of the body is approximately 0.977 N.
Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find kinetic energy for a solid sphere.
Derivation:
Total Kinetic Energy = Translational KE + Rotational KE
\(E = E_{trans} + E_{rot}\)
\(E = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\)
Since \(v = R\omega\) (no slipping) and \(I = MK^2\) (where K is radius of gyration):
\(E = \frac{1}{2} M v^2 + \frac{1}{2} (MK^2) (\frac{v}{R})^2\)
\(E = \frac{1}{2} M v^2 (1 + \frac{K^2}{R^2})\)
For Solid Sphere:
Moment of inertia \(I = \frac{2}{5} MR^2\), so \(K^2 = \frac{2}{5} R^2\) or \(\frac{K^2}{R^2} = \frac{2}{5} = 0.4\).
Substituting this into the expression:
\(E_{sphere} = \frac{1}{2} M v^2 (1 + \frac{2}{5})\)
\(E_{sphere} = \frac{1}{2} M v^2 (\frac{7}{5})\)
\(E_{sphere} = \frac{7}{10} M v^2\)
A steel wire of diameter \(1 \times 10^{-3}\) m is stretched by a force of 20 N. Calculate the strain energy per unit volume. (\(Y_{steel} = 2 \times 10^{11} \text{ N/m}^2\))
Given:
Diameter \(d = 10^{-3} \text{ m} \implies \text{Radius } r = 0.5 \times 10^{-3} \text{ m}\)
Force \(F = 20 \text{ N}\)
Young's Modulus \(Y = 2 \times 10^{11} \text{ N/m}^2\)
Formula:
Strain Energy per unit volume \(u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} \frac{(\text{Stress})^2}{Y}\)
Stress \(\sigma = \frac{F}{A} = \frac{F}{\pi r^2}\)
Calculation:
Area \(A = \pi (5 \times 10^{-4})^2 = 3.142 \times 25 \times 10^{-8} \approx 7.855 \times 10^{-7} \text{ m}^2\)
Stress \(\sigma = \frac{20}{7.855 \times 10^{-7}} \approx 2.546 \times 10^7 \text{ N/m}^2\)
\(u = \frac{1}{2} \frac{(2.546 \times 10^7)^2}{2 \times 10^{11}}\)
\(u = \frac{1}{4} \frac{6.48 \times 10^{14}}{10^{11}}\)
\(u = \frac{1}{4} (6480) = 1620 \text{ J/m}^3\)
Answer: Strain energy per unit volume is 1620 J/m³ (approx).
Q. 4. Answer the following: [7]
Define an ideal simple pendulum. Show that, under certain conditions, simple pendulum performs linear simple harmonic motion.
Definition: An ideal simple pendulum is a heavy point mass suspended from a rigid support by a massless, inextensible, and flexible string.
Derivation of SHM:
Consider a bob of mass \(m\) displaced by a small angle \(\theta\).
Forces acting on the bob:
1. Tension \(T'\) along the string.
2. Weight \(mg\) downwards.
Resolve \(mg\) into components: \(mg \cos\theta\) (balances tension) and \(mg \sin\theta\) (restoring force).
Restoring force \(F = -mg \sin\theta\).
Condition: If \(\theta\) is very small, \(\sin\theta \approx \theta \approx \frac{x}{L}\) (where x is displacement arc length, L is length).
\(F = -mg (\frac{x}{L})\)
\(F = -(\frac{mg}{L}) x\)
Since \(m, g, L\) are constants, \(F \propto -x\).
This is the condition for Linear SHM. Thus, for small amplitudes, a simple pendulum performs SHM.
A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of 72 Km/hour. [Speed of sound in air = 340 m/s.]
Given:
Actual frequency \(n = 640 \text{ Hz}\)
Velocity of sound \(v = 340 \text{ m/s}\)
Velocity of source \(v_s = 72 \text{ km/hr} = 20 \text{ m/s}\)
Velocity of observer \(v_o = 0\)
Case 1: Train moves towards observer
Apparent freq \(n_1 = n (\frac{v}{v - v_s})\)
\(n_1 = 640 (\frac{340}{340 - 20}) = 640 (\frac{340}{320}) = 640 (1.0625) = 680 \text{ Hz}\).
Case 2: Train moves away from observer
Apparent freq \(n_2 = n (\frac{v}{v + v_s})\)
\(n_2 = 640 (\frac{340}{340 + 20}) = 640 (\frac{340}{360}) = 640 (0.9444) \approx 604.44 \text{ Hz}\).
Difference:
\(\Delta n = n_1 - n_2 = 680 - 604.44 = 75.56 \text{ Hz}\).
Answer: The difference in apparent frequencies is 75.56 Hz.
OR
With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction.
Harmonics in Open Pipe:
Let length of pipe be \(L\), speed of sound \(v\).
1. Fundamental Mode (1st Harmonic):
Antinodes at both ends, node in middle. Length \(L = \frac{\lambda_1}{2}\).
Frequency \(n_1 = \frac{v}{\lambda_1} = \frac{v}{2L}\).
2. Second Harmonic (1st Overtone):
\(L = \lambda_2\).
Frequency \(n_2 = \frac{v}{\lambda_2} = \frac{v}{L} = 2(\frac{v}{2L}) = 2n_1\).
3. Third Harmonic (2nd Overtone):
\(L = \frac{3\lambda_3}{2}\).
Frequency \(n_3 = \frac{v}{\lambda_3} = \frac{3v}{2L} = 3n_1\).
Since frequencies \(n_1, 2n_1, 3n_1, ...\) are obtained, all harmonics (odd and even) are present.
Definition of End Correction:
The antinode is not formed exactly at the open end of the pipe but at a small distance outside it. This small distance is called end correction (\(e\)). For an open pipe, total correction is \(2e\), so corrected length \(L_{corr} = L + 2e\).
Calculate the kinetic energy of 10 gram of Argon molecules at 127°C. [Universal gas constant R = 8320 J/k mole K, Atomic weight of Argon = 40]
Given:
Mass \(m = 10 \text{ g}\)
Atomic weight \(M_0 = 40 \text{ g/mole}\) (or 40 kg/kmole)
Temperature \(T = 127^\circ\text{C} = 127 + 273 = 400 \text{ K}\)
\(R = 8320 \text{ J/k mole K} = 8.32 \text{ J/mole K}\).
Calculation of Moles:
Number of moles \(n = \frac{m}{M_0} = \frac{10}{40} = 0.25 \text{ mole}\).
Formula:
Argon is a monoatomic gas. Degrees of freedom \(f = 3\).
Kinetic Energy \(E = \frac{f}{2} nRT = \frac{3}{2} nRT\).
Calculation:
\(E = 1.5 \times 0.25 \times 8.32 \times 400\)
\(E = 1.5 \times 0.25 \times 3328\)
\(E = 0.375 \times 3328\)
\(E = 1248 \text{ J}\).
Answer: The kinetic energy is 1248 Joules.
