JEE Physics Mechanics and Thermal Physics MCQs Set 02

Answer: (c) 1 : 1

Explanation:
1. Heat lost by A equals heat gained by B, so \(m c_A(32-28) = m c_B(28-24)\).
2. This gives \(4c_A = 4c_B\), so \(c_A : c_B = 1 : 1\).

Teacher's Note:
1. With equal masses, the ratio of specific heats is the inverse of the ratio of temperature changes.
2. Here both changes are \(4^\circ\)C, so the liquids have the same specific heat.

Answer: (a) \(R_1 : R_2 = 3 : 2\)

Explanation:
1. By Newton's law of cooling, the rate of cooling is proportional to \(T - T_0\) when the temperature difference is small.
2. \(R_1 : R_2 = (70-40) : (60-40) = 30 : 20 = 3 : 2\).

Teacher's Note:
1. Compare the excess of each body over the surroundings, not the temperatures themselves.
2. Newton's law gives a rate proportional to this excess.

Answer: (b) 277 K

Explanation:
1. No heat is exchanged with the surroundings, so \(n_1 C_{v1}(T - T_1) + n_2 C_{v2}(T - T_2) = 0\). N\(_2\) has \(C_v = \tfrac{5}{2}R\) and He has \(C_v = \tfrac{3}{2}R\).
2. \(T_{mix} = \dfrac{2(\tfrac52 R)(300) + 1(\tfrac32 R)(200)}{2(\tfrac52 R) + \tfrac32 R} = \dfrac{1500 + 300}{6.5} \approx 277\) K.

Teacher's Note:
1. Use the molar heat capacity at constant volume of each gas, weighted by the number of moles.
2. Convert the temperatures to kelvin before substituting.

Answer: (b) 1.33

Explanation:
1. The statement gives \(P^3 \propto V^{-4}\), that is \(P^3 V^4 = \) constant.
2. Taking the cube root, \(P V^{4/3} = \) constant. Comparing with \(PV^{\gamma} = \) constant gives \(\gamma = 4/3 = 1.33\).

Teacher's Note:
1. In an adiabatic process \(PV^{\gamma}\) is constant, so reduce the given relation to this form.
2. \(\gamma = 4/3\) belongs to a polyatomic (non-linear) gas.

Answer: (d) 1

Explanation:
1. \(T_1 = 600\) K and \(T_2 = 500\) K, so the efficiency is \(\eta = 1 - \dfrac{500}{600} = \dfrac16\).
2. Work done \(W = \eta Q_1 = \dfrac16 \times 6 = 1\) kcal.

Teacher's Note:
1. Carnot efficiency depends only on the two reservoir temperatures in kelvin.
2. Work equals efficiency times heat absorbed from the source.

Answer: (a) 6

Explanation:
1. For two particles about the axis through the centre of mass, \(I = \mu d^2\) where the reduced mass is \(\mu = \dfrac{m_1 m_2}{m_1 + m_2} = \dfrac{2}{3}\) kg.
2. \(I = \dfrac23 \times 3^2 = 6\) kg m\(^2\).

Teacher's Note:
1. Check by locating the centre of mass: it is 2 m from the 1 kg mass and 1 m from the 2 kg mass.
2. Then \(I = 1(2)^2 + 2(1)^2 = 6\), which agrees.

Answer: (b) Sphere

Explanation:
1. For pure rolling down an incline, \(a = \dfrac{g\sin\theta}{1 + I/(mR^2)}\).
2. The disc has \(I/mR^2 = \tfrac12\) and the solid sphere has \(\tfrac25\). The sphere has the larger acceleration, so it reaches the bottom first, whatever the masses.

Teacher's Note:
1. The result depends only on the shape (the ratio \(I/mR^2\)), not on mass or radius.
2. The smaller this ratio, the faster the body rolls down.

Answer: (c) 32 N

Explanation:
1. At height \(h\), \(g' = g\left(\dfrac{R}{R+h}\right)^2\). With \(h = R/2\), \(g' = g\left(\dfrac{2}{3}\right)^2 = \dfrac{4g}{9}\).
2. Weight \(= \dfrac49 \times 72 = 32\) N.

Teacher's Note:
1. The distance from the centre of the earth is \(R + h = 1.5R\), and gravity falls as the inverse square of this distance.
2. Do not use the \(1 - 2h/R\) approximation for large heights.

Answer: (b) negative velocity and negative acceleration

Explanation:
1. The dots go from 1 to 5 towards smaller x, so the velocity is in the negative x-direction.
2. The gap between successive dots grows, so the speed increases. Speed increasing along the negative direction means the acceleration is also negative.

Teacher's Note:
1. Velocity and acceleration have the same sign when the speed increases, and opposite signs when it decreases.
2. Read the direction of motion from the order of the dots, not from their position.

Answer: (d) 390 Hz

Explanation:
1. Initially \(f_A = 384 \pm 6\), that is 390 Hz or 378 Hz.
2. Loading with wax lowers \(f_A\). If \(f_A\) were 378 Hz it would move further from 384 Hz and the beats would increase.
3. If \(f_A = 390\) Hz it drops towards 384 Hz and the beats can decrease to 4 (it becomes 388 Hz). So \(f_A = 390\) Hz.

Teacher's Note:
1. Loading a prong increases its mass and lowers its frequency.
2. Use the change in beat frequency to decide which of the two possible values is right.

Answer: (c) 9 : 4

Explanation:
1. Strain \(= \dfrac{F}{AY}\propto \dfrac{1}{r^2}\) for the same force and the same material.
2. \(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{r_2^2}{r_1^2} = \dfrac{9}{4}\).

Teacher's Note:
1. Strain depends on the cross-sectional area but not on the length of the wire.
2. The length ratio is extra information here.

Answer: (c) 24 m s\(^{-1}\)

Explanation:
1. Volume is conserved: \(\tfrac43\pi R^3 = 64 \cdot \tfrac43\pi r^3\), so \(R = 4r\).
2. Terminal velocity is proportional to \(r^2\), so \(v_T' = 1.5 \times 4^2 = 24\) m s\(^{-1}\).

Teacher's Note:
1. Terminal velocity \(v_T = \dfrac{2r^2(\rho - \sigma)g}{9\eta}\) varies as the square of the radius.
2. First find the radius of the big drop from conservation of volume.

Answer: (d) 14 m

Explanation:
1. Energy just before impact is \(mg(20)\). 70% of it remains after the collision.
2. \(mgh' = 0.7 \times mg(20)\), so \(h' = 14\) m.

Teacher's Note:
1. Energy lost as heat reduces the energy but the ball still rises until all its kinetic energy becomes potential energy.
2. Height is proportional to the energy retained.

Answer: (c) 480

Explanation:
1. Force on the gun equals the momentum given to bullets per second, \(F = n\,m v\) where \(n\) is the number fired per second.
2. \(n = \dfrac{400}{0.05 \times 1000} = 8\) per second, so \(8 \times 60 = 480\) per minute.

Teacher's Note:
1. Convert 50 g to 0.05 kg before using the momentum of one bullet.
2. Do not forget to change bullets per second into bullets per minute.

Answer: (b) a parabola

Explanation:
1. \(T = 2\pi\sqrt{L/g}\), so \(L = \dfrac{g}{4\pi^2}T^2\).
2. \(L\) is proportional to \(T^2\), so the graph between \(L\) and \(T\) is a parabola.

Teacher's Note:
1. The graph between \(L\) and \(T^2\) would be a straight line through the origin.
2. A graph of \(L\) against \(T\) itself is a parabola.

Answer: (b) \((\alpha p + \beta q + \gamma r)\%\)

Explanation:
1. For a product of powers, the maximum relative error is \(\dfrac{\Delta x}{x} = \alpha\dfrac{\Delta M}{M} + \beta\dfrac{\Delta L}{L} + \gamma\dfrac{\Delta T}{T}\).
2. The exponents enter with their magnitudes and the errors always add, so the error is \((\alpha p + \beta q + \gamma r)\%\) (taking the exponents as positive).

Teacher's Note:
1. Errors never cancel in the worst case, so they are added whatever the sign of the exponent.
2. Use the magnitude of each exponent.

Answer: (a) \(\dfrac12\)

Explanation:
1. Centripetal acceleration is \(a = \omega^2 r\). The time periods are equal, so \(\omega\) is the same.
2. \(\dfrac{a_1}{a_2} = \dfrac{r_1}{r_2} = \dfrac{4}{8} = \dfrac12\).

Teacher's Note:
1. Mass does not appear in the acceleration.
2. For equal \(\omega\), acceleration is directly proportional to the radius.

Answer: (d) Power delivered by gravitational force at point 4 is negative.

Explanation:
1. Tension acts along the string towards the centre and the velocity is tangential, so the power of tension \(\vec T\cdot\vec v = 0\) at every point. This rules out options (a) and (b).
2. At point 3 the velocity is vertically upward while gravity is downward, so the power of gravity is negative, not zero.
3. At point 4 the bob is still moving upward (the upward component of the velocity is non-zero) against gravity, so the power of gravity is negative. This matches option (d).

Teacher's Note:
1. Power is \(\vec F\cdot\vec v\). A force perpendicular to the velocity does no work.
2. Gravity does negative work when the bob rises and positive work when it falls.

Answer: (a) 100%

Explanation:
1. An increase of 300% makes the new kinetic energy 4 times the old one.
2. Since \(p = \sqrt{2mK}\), the momentum becomes \(\sqrt4 = 2\) times, an increase of 100%.

Teacher's Note:
1. A 300% increase means the final value is 400% of the original, not 300%.
2. Momentum varies as the square root of kinetic energy.