JEE Physics Mechanics and Thermal Physics MCQs Set 01

Answer: (b) 3 s

Explanation:
1. End to end (series): thermal resistance \( R_s = 2R \), time \( t_s = 12 \) s.
2. Side by side (parallel): \( R_p = \frac{R}{2} \).
3. Heat \( Q = \frac{\Delta T}{R_{th}}\, t \) is the same, so \( t \propto R_{th} \): \( t_p = 12 \times \frac{R/2}{2R} = 3 \) s.

Teacher's Note:
1. Thermal resistances combine just like electrical resistances.
2. Parallel combination is one fourth of the series value, so the time is one fourth.

Answer: (a) \( 6P_0V_0 \)

Explanation:
1. Work in a cycle equals the area of triangle ABC.
2. Base \( BC = 5V_0 - 2V_0 = 3V_0 \)\, height \( AB = 5P_0 - P_0 = 4P_0 \).
3. Area \( = \frac12 \times 3V_0 \times 4P_0 = 6P_0V_0 \). The cycle C to A to B to C is clockwise, so the work is positive.

Teacher's Note:
1. A clockwise cycle on a P-V diagram gives positive work done by the gas.
2. Always subtract the correct volume and pressure differences from the axis labels.

Answer: (b) \( 819^\circ \)C

Explanation:
1. \( v_{rms} \propto \sqrt{T} \). Doubling the speed needs \( T_2 = 4T_1 \).
2. \( T_1 = 273 \) K, so \( T_2 = 1092 \) K.
3. \( T_2 = 1092 - 273 = 819^\circ \)C.

Teacher's Note:
1. Always convert to kelvin before using ratios of temperature.
2. Speed doubles when temperature becomes four times.

Answer: (c) 1

Explanation:
1. \( f_1 = \frac{1000\pi}{2\pi} = 500 \) Hz and \( f_2 = \frac{998\pi}{2\pi} = 499 \) Hz.
2. Beat frequency \( = f_1 - f_2 = 1 \) per second.

Teacher's Note:
1. Divide the coefficient of t by \( 2\pi \) to get the frequency.
2. Beats per second equal the difference of frequencies.

Answer: (d) solid ball

Explanation:
1. \( \frac{K_{rot}}{K_{trans}} = \frac{\frac12 I\omega^2}{\frac12 mv^2} = \frac{I}{mR^2} \) (since \( v = R\omega \)).
2. This equals 0.4 \( = \frac25 \), so \( I = \frac25 mR^2 \), the moment of inertia of a solid sphere.

Teacher's Note:
1. Ring: ratio 1; solid cylinder: 1/2; hollow sphere: 2/3; solid sphere: 2/5.
2. Rolling without slipping links v and \( \omega \).

Answer: (a) 14

Explanation:
1. Rotational power \( P = \vec{\tau} \cdot \vec{\omega} \).
2. \( P = (2)(2) + (2)(1) + (2)(4) = 4 + 2 + 8 = 14 \) (in SI units, W).

Teacher's Note:
1. Power is the dot product of torque and angular velocity, as \( P = \vec{F}\cdot\vec{v} \) is for linear motion.
2. Multiply matching components and add.

Answer: (b) \( 2.4 \times 10^5 \) N, upwards

Explanation:
1. By Bernoulli's principle the pressure above the roof is lower: \( \Delta P = \frac12 \rho v^2 = \frac12 \times 1.2 \times 40^2 = 960 \) Pa.
2. Force \( = \Delta P \times A = 960 \times 250 = 2.4 \times 10^5 \) N.
3. Pressure inside is higher than outside, so the force is upwards.

Teacher's Note:
1. Fast-moving air has lower pressure, which is why roofs are lifted in storms.
2. Direction is from the high-pressure side to the low-pressure side.

Answer: (b) its kinetic energy is doubled

Explanation:
1. In a circular orbit \( KE = \frac{GMm}{2r} \) and \( PE = -\frac{GMm}{r} \), so total energy \( = -KE \).
2. Doubling the KE makes the total energy \( -KE + KE = 0 \) (new KE = 2KE, PE unchanged), which is the condition for escape.
3. Equivalently the speed must become \( \sqrt{2} \) times the orbital speed, not \( \sqrt{1.5} \) times.

Teacher's Note:
1. Escape requires total energy of zero or more.
2. Escape velocity \( = \sqrt{2} \times \) orbital velocity.

Answer: (d) 30 m/s

Explanation:
1. On a frictionless banked road \( v^2 = rg\tan\theta \).
2. \( v^2 = 90 \times 10 \times \tan 45^\circ = 900 \), so \( v = 30 \) m/s.

Teacher's Note:
1. The mass of the car does not matter.
2. With \( \theta = 45^\circ \), \( \tan\theta = 1 \).

Answer: (c) both A and B strike the ground with the same velocity

Explanation:
1. When A returns to the level of the building it has speed v directed downward, the same as the initial speed of B.
2. From there both fall through the same height, so by \( v_f^2 = v^2 + 2gh \) both reach the ground with the same speed \( \sqrt{v^2 + 2gh} \).

Teacher's Note:
1. Energy conservation gives the same final speed whatever the direction of projection.
2. Only the time of flight differs.

Answer: (c) downward with increasing speed

Explanation:
1. Reading 220 N is less than the true weight 240 N, so \( mg - N = ma \) with a net downward acceleration.
2. Downward acceleration with downward motion means the speed is increasing.
3. (Upward motion with decreasing speed would also give a downward acceleration, but it is not offered.)

Teacher's Note:
1. Apparent weight less than true weight means the acceleration is downward.
2. Speed increases when velocity and acceleration are in the same direction.

Answer: (b) 7%

Explanation:
1. \( \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \).
2. \( = \frac{5}{100} + \frac{0.2}{10} = 0.05 + 0.02 = 0.07 \), i.e. 7%.

Teacher's Note:
1. For a quotient, the maximum fractional errors add.
2. Convert each fractional error into a percentage before adding.