JEE Physics Mechanics and Thermal Physics MCQs Set 03

Answer: (c) \(\mathrm{rad\,s^{-2}}\)

Explanation:
1. Angular acceleration is the rate of change of angular velocity, so its unit is rad s\(^{-1}\) per second.
2. Hence the unit is \(\mathrm{rad\,s^{-2}}\).

Teacher's Note:
1. Angular acceleration \(\alpha = \Delta\omega/\Delta t\).
2. Radian is a dimensionless unit, so \(\alpha\) has dimension \(\mathrm{T^{-2}}\).

Answer: (c) The acceleration of the particle is 2a

Explanation:
1. Differentiate: \(v = u + 2a(t-2)\).
2. Differentiate again: acceleration \(= 2a\).
3. At \(t = 2\), \(x = 0\) so the particle is at the origin, and the velocity at \(t=0\) is \(u - 4a\), not u.

Teacher's Note:
1. Compare with \(x = ut + \tfrac12 at^2\): the coefficient of the squared term is half the acceleration.
2. Always differentiate twice to get acceleration from a position equation.

Answer: (b) less than 6 kg

Explanation:
1. For the Atwood machine \(a = \dfrac{(5-1)g}{5+1} = \dfrac{2g}{3}\).
2. Tension \(T = \dfrac{2m_1m_2 g}{m_1+m_2} = \dfrac{2\times 5}{6}g\).
3. The balance supports the pulley with a force \(2T = \dfrac{20}{6}\) kgf \(\approx 3.3\) kg, which is less than 6 kg.

Teacher's Note:
1. When the masses accelerate, the total weight is not fully carried by the support, because the heavier mass accelerates downwards.
2. The reading equals 6 kg only when the masses are at rest or move uniformly.

Answer: (b) increases linearly with time

Explanation:
1. \(v = \dfrac{dS}{dt} = \dfrac{2t}{3}\) and \(a = \dfrac23\) m s\(^{-2}\).
2. \(F = ma = 2\) N, a constant.
3. \(P = Fv = 2\times\dfrac{2t}{3} = \dfrac{4t}{3}\), which increases linearly with t.

Teacher's Note:
1. Power of a constant force is proportional to the velocity.
2. Starting from rest with uniform acceleration, velocity grows linearly, hence so does power.

Answer: (b) may pass through the centre of mass

Explanation:
1. In pure rotation the body turns about a fixed axis. This axis need not pass through the centre of mass (for example a door turning about its hinges).
2. It can pass through the centre of mass (a wheel on its axle), so the statement that it "may" do so is the correct one.
3. It need not pass through any particle either (a ring about its central axis).

Teacher's Note:
1. Do not confuse the axis of pure rotation with the axis of a freely moving body, which passes through the centre of mass.
2. "Must" statements are usually false because of counter examples.

Answer: (a) \(\sqrt{Rg\left(\dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}\right)}\)

Explanation:
1. At the minimum speed the car tends to slide down, so friction acts up the incline.
2. Resolve: \(N\cos\theta + f\sin\theta = mg\) and \(N\sin\theta - f\cos\theta = mv^2/R\) with \(f = \mu N\).
3. Dividing, \(\dfrac{v^2}{Rg} = \dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}\).

Teacher's Note:
1. For maximum speed friction acts down the slope and the signs of \(\mu\) flip.
2. Check by putting \(\mu = 0\): the result reduces to \(v = \sqrt{Rg\tan\theta}\).

Answer: (c) 108 km/hr

Explanation:
1. Friction supplies the centripetal force, so \(v_{max} = \sqrt{\mu R g} = \sqrt{0.3\times300\times10} = 30\) m s\(^{-1}\).
2. Convert: \(30\times 3.6 = 108\) km/hr.

Teacher's Note:
1. On a level road \(\mu mg = mv^2/R\).
2. Multiply m/s by 3.6 to get km/hr.

Answer: (d) \(\dfrac53 MR^2\)

Explanation:
1. About a diameter, \(I_{cm} = \dfrac23 MR^2\) for a hollow sphere.
2. Parallel axis theorem for a tangent at distance R: \(I = \dfrac23 MR^2 + MR^2 = \dfrac53 MR^2\).

Teacher's Note:
1. The parallel axis theorem adds \(Md^2\) to the moment of inertia about the centre of mass.
2. The solid sphere gives \(\tfrac75MR^2\) by the same method.

Answer: (b) \(\dfrac{M}{M+6m}\omega_0\)

Explanation:
1. No external torque, so angular momentum is conserved.
2. Initially \(I_1 = \dfrac{ML^2}{12}\). Finally \(I_2 = \dfrac{ML^2}{12} + 2m\left(\dfrac L2\right)^2 = \dfrac{L^2}{12}(M + 6m)\).
3. \(\omega = \dfrac{I_1}{I_2}\omega_0 = \dfrac{M}{M+6m}\omega_0\).

Teacher's Note:
1. When mass moves outward the moment of inertia rises, so angular speed falls.
2. Apply \(I_1\omega_1 = I_2\omega_2\).

Answer: (c) \(\sqrt{\dfrac{GM}{\sqrt3 R}}\)

Explanation:
1. Side of the triangle \(= \sqrt3 R\), so each pair attracts with \(F = \dfrac{GM^2}{3R^2}\).
2. The two forces on a body make \(30^\circ\) with the line to the centre, so the net force is \(2F\cos30^\circ = \dfrac{GM^2}{\sqrt3R^2}\).
3. This supplies \(\dfrac{Mv^2}{R}\), giving \(v = \sqrt{\dfrac{GM}{\sqrt3 R}}\).

Teacher's Note:
1. The side of an equilateral triangle inscribed in a circle of radius R is \(\sqrt3 R\).
2. Only the component of each force towards the centre adds up.

Answer: (b) \(\left(\dfrac{GM}{R}\right)^{1/2}\)

Explanation:
1. Initially the body is at distance 2R from the centre, finally at R.
2. Energy conservation: \(\tfrac12 v^2 = GM\left(\dfrac1R - \dfrac1{2R}\right) = \dfrac{GM}{2R}\).
3. So \(v = \sqrt{GM/R}\).

Teacher's Note:
1. Use \(U = -GMm/r\) with distances measured from the centre of the earth.
2. Height R above the surface means distance 2R from the centre.

Answer: (b) 0.004

Explanation:
1. Shear strain at the surface is \(\phi = \dfrac{r\theta}{L}\).
2. \(\phi = \dfrac{0.01\times0.8}{2} = 0.004\).

Teacher's Note:
1. Strain is zero on the axis and maximum on the surface in torsion.
2. Use \(r\theta = L\phi\), the arc traced by the surface line.

Answer: (d) \(\dfrac{2}{\sqrt5}\) s

Explanation:
1. Speed of efflux \(v = \sqrt{2gh}\). The level falls at \(-\dfrac{dh}{dt} = \dfrac{a}{A}\sqrt{2gh}\).
2. Integrating, \(t = \dfrac{A}{a}\sqrt{\dfrac2g}\left(\sqrt{h_1} - \sqrt{h_2}\right)\).
3. \(t = 10\times\sqrt{0.2}\times(0.9 - 0.7) = 2\sqrt{0.2} = \dfrac{2}{\sqrt5}\) s.

Teacher's Note:
1. Convert heights to metres before substituting.
2. The time depends on the difference of the square roots of the heights, not on the heights themselves.

Answer: (a) 10,000 N

Explanation:
1. Thermal strain prevented \(= \alpha\Delta T = 10^{-6}\times100 = 10^{-4}\).
2. Force \(F = YA\alpha\Delta T = 10^{11}\times10^{-3}\times10^{-4} = 10^4\) N.

Teacher's Note:
1. The force does not depend on the length of the bar.
2. Stress = Y \(\times\) strain with the strain equal to \(\alpha\Delta T\).

Answer: (c) 200 K

Explanation:
1. Wien's law: \(\lambda_m T = \) constant.
2. \(T = \dfrac{1.75\times1640}{14.35} = 200\) K.

Teacher's Note:
1. Longer peak wavelength means a colder body.
2. Wien's constant is about \(2.9\times10^{-3}\) m K.

Answer: (c) 2R

Explanation:
1. \(\Delta U = Q - W = 450\) J. For a monoatomic gas \(\Delta U = n\cdot\tfrac32R\,\Delta T\), so \(n\Delta T = \dfrac{300}{R}\).
2. \(C = \dfrac{Q}{n\Delta T} = \dfrac{600R}{300} = 2R\).

Teacher's Note:
1. First find \(\Delta U\) from the first law, then \(n\Delta T\) from \(C_v\).
2. Molar heat capacity of the process is heat per mole per kelvin.

Answer: (b) \(\eta\)

Explanation:
1. \(\eta = 1 - \dfrac{T_2}{T_1}\).
2. With both temperatures doubled the ratio \(\dfrac{2T_2}{2T_1}\) is unchanged, so the efficiency stays \(\eta\).

Teacher's Note:
1. Carnot efficiency depends only on the ratio of the temperatures.
2. Scaling both temperatures by the same factor changes nothing.

Answer: (a) \(\dfrac{21}{8}\) s

Explanation:
1. Starting from opposite extremes, the initial phase difference is \(\pi\).
2. They are first in phase when the faster pendulum has gained a half cycle: \(\left(\dfrac{2\pi}{3} - \dfrac{2\pi}{7}\right)t = \pi\).
3. \(\dfrac{8\pi}{21}t = \pi\) gives \(t = \dfrac{21}{8}\) s.

Teacher's Note:
1. Equal phase needs a phase difference of an even multiple of \(\pi\); opposite extremes already give \(\pi\).
2. The first meeting occurs when the relative phase changes by \(\pi\).

Answer: (a) 3 beats per second with intensity ratio between maxima and minima equal to 9

Explanation:
1. Frequencies are \(f_1 = 250\) Hz and \(f_2 = 253\) Hz, so the beat frequency is 3 per second.
2. \(\dfrac{I_{max}}{I_{min}} = \left(\dfrac{A_1 + A_2}{A_1 - A_2}\right)^2 = \left(\dfrac{6}{2}\right)^2 = 9\).

Teacher's Note:
1. From \(y = A\sin 2\pi f t\), \(f = \omega/2\pi\).
2. Intensity is proportional to the square of the amplitude.

Answer: (d) \(6.4\times10^{7}\)

Explanation:
1. Intensity \(I = \dfrac{P}{4\pi R^2}\).
2. \(I = \dfrac{3.9\times10^{26}}{4\pi(6.96\times10^8)^2} = \dfrac{3.9\times10^{26}}{6.09\times10^{18}} \approx 6.4\times10^{7}\ \mathrm{W\,m^{-2}}\).

Teacher's Note:
1. At the solar surface use the radius of the sun, not the earth-sun distance.
2. Intensity falls as the inverse square of the distance from the source.