JEE Physics System Of Particles and Rotational Motion MCQs Set 09

Answer: (a) 50 rad

Explanation:
1. \( \theta = \tfrac12\alpha t^2 = \tfrac12(4)(25) = 50 \) rad.

Teacher's Note:
1. Rotational kinematics mirrors linear kinematics with θ, ω, α.

Answer: (c) \( \pi/2 \) rad/s²

Explanation:
1. \( \omega_0 = 300\times\dfrac{2\pi}{60} = 10\pi \) rad/s; \( \theta = 50\times2\pi = 100\pi \) rad.
2. \( \omega_0^2 = 2\alpha\theta \Rightarrow 100\pi^2 = 200\pi\alpha \Rightarrow \alpha = \pi/2 \) rad/s².

Teacher's Note:
1. Convert rpm to rad/s first.

Answer: (c) 20

Explanation:
1. Average speed = 60 rpm = 1 rev/s, so in 20 s: 20 revolutions. Check: \( \theta = \tfrac{\omega_0}{2}t = \tfrac{4\pi}{2}(20) = 40\pi \) rad = 20 rev.

Teacher's Note:
1. Average of initial and final angular speed applies for uniform retardation.

Answer: (c) 20 rad/s²

Explanation:
1. \( \alpha = \tau/I = 10/0.5 = 20 \) rad/s².

Teacher's Note:
1. Use τ = Iα.

Answer: (b) 2 rad/s²

Explanation:
1. \( \tau = 10\times0.5 = 5 \) N m, \( I = MR^2 = 2.5 \) kg m². \( \alpha = 2 \) rad/s².

Teacher's Note:
1. Torque = force x perpendicular distance.

Answer: (c) \( \dfrac{3g}{2l} \)

Explanation:
1. \( \tau = mg\tfrac{l}{2} = \tfrac{ml^2}{3}\alpha \Rightarrow \alpha = \dfrac{3g}{2l} \).

Teacher's Note:
1. Weight acts at the centre, at l/2 from the hinge.

Answer: (c) \( 5\hat k \) N m

Explanation:
1. \( \vec\tau = \vec r\times\vec F = (1)(3) - (-1)(2) = 5 \) along \( \hat k \).

Teacher's Note:
1. Use the determinant (cross product) formula.

Answer: (b) 4 N m

Explanation:
1. \( \tau = rF\sin30^\circ = 0.4\times20\times0.5 = 4 \) N m.

Teacher's Note:
1. Only the component of force perpendicular to r produces torque.

Answer: (c) doubles

Explanation:
1. \( L = rp \), so L is proportional to p.

Teacher's Note:
1. Angular momentum is r x p.

Answer: (c) 24 kg m²/s

Explanation:
1. \( L = mvd = 2\times3\times4 = 24 \) kg m²/s.

Teacher's Note:
1. The perpendicular distance from the origin to the line of motion is 4 m.

Answer: (b) 20 kg m²/s

Explanation:
1. \( L = I\omega = 20 \).

Teacher's Note:
1. Direct formula.

Answer: (d) increases

Explanation:
1. No external torque, so \( I\omega \) is constant; I decreases, so ω increases.

Teacher's Note:
1. Conservation of angular momentum.

Answer: (a) \( \omega/2 \)

Explanation:
1. \( I\omega = 2I\omega' \Rightarrow \omega' = \omega/2 \).

Teacher's Note:
1. Angular momentum is conserved in the coupling.

Answer: (c) 1/2

Explanation:
1. \( K_i = \tfrac12I\omega^2 \), \( K_f = \tfrac12(2I)(\omega/2)^2 = \tfrac14I\omega^2 \). Loss = half.

Teacher's Note:
1. Energy is lost to friction while the discs reach a common speed.

Answer: (c) torque

Explanation:
1. \( \vec\tau = d\vec L/dt \).

Teacher's Note:
1. Rotational analogue of F = dp/dt.

Answer: (d) zero net torque

Explanation:
1. Rotational equilibrium requires the net torque about any point to be zero.

Teacher's Note:
1. Translational equilibrium is the separate condition of zero net force.

Answer: (d) 2 N m

Explanation:
1. Couple = F x separation = 5 x 0.4 = 2 N m.

Teacher's Note:
1. A couple has no net force but a net torque.

Answer: (d) \( \dfrac{W}{2}\cot\theta \)

Explanation:
1. Torque about the foot: \( N_w L\sin\theta = W\tfrac L2\cos\theta \Rightarrow N_w = \tfrac W2\cot\theta \). Friction balances the wall reaction.

Teacher's Note:
1. Take torques about the foot of the ladder.

Answer: (b) 2.5 N

Explanation:
1. Rod weight acts at 0.5 m, i.e. 0.1 m on the other side of the pivot: \( W(0.4) = 10(0.1) \Rightarrow W = 2.5 \) N.

Teacher's Note:
1. Torques about the pivot must balance.

Answer: (d) \( L/4 \)

Explanation:
1. \( K = \tfrac12I\omega^2 \). With \( K' = K/2 \) and \( \omega' = 2\omega \): \( I' = I/8 \).
2. \( L' = I'\omega' = \tfrac I8\cdot2\omega = L/4 \).

Teacher's Note:
1. Use K and ω to get the new I.

Answer: (d) Planck's constant

Explanation:
1. Both are kg m²/s (J s).

Teacher's Note:
1. Energy is J, torque is N m.