Class 11 Physics: System of Particles and Rotational Motion MCQs (NCERT)

Since $\tau = \Delta L/\Delta t$, the torque is $(30 - 6)/8 = 3\ \text{N m}$. The value $24\ \text{N m}$ quotes the change in angular momentum itself without dividing by the time, $4.5\ \text{N m}$ adds the two angular momenta instead of subtracting them, and $0.75\ \text{N m}$ uses only the final value.

To balance the seesaw, the moments must be equal: 300 N × 2 m = W × 3 m. Solving gives W = 200 N. Other values result from miscalculating the balance of moments.

The two ends lie on opposite sides of the pivot, so two downward forces turn the rod in opposite senses, exactly as two children balance a seesaw. Equal arms of 5 m mean equal forces, so 50 N is needed. An upward force at the far end would turn the rod the same way as the first force and make matters worse.

Using the equation ω = ω₀ + αt, with final ω = 0, initial ω₀ = 10 rad/s, and α = -2 rad/s² (since it's deceleration), we solve for t: 0 = 10 - 2t, giving t = 5 s. Alternatives arise from using wrong signs or incorrect formula application.

For a thin spherical shell about a diameter, $I = \tfrac23MR^2 = \tfrac23(3)(0.04) = 0.08\ \text{kg m}^2$, so $K = \tfrac12I\omega^2 = \tfrac12(0.08)(100) = 4.0\ \text{J}$. The value $8.0\ \text{J}$ omits the factor $\tfrac12$, $2.4\ \text{J}$ uses $\tfrac25MR^2$, which belongs to a solid sphere, and $0.4\ \text{J}$ uses $\omega$ rather than $\omega^2$.

Only internal forces act, so the centre of mass stays exactly where it was, and the particles can meet nowhere else. Its distance from the $2\ \text{kg}$ particle is $(3 \times 10)/(2+3) = 6\ \text{m}$, so that particle covers $6\ \text{m}$. $4\ \text{m}$ is the distance travelled by the $3\ \text{kg}$ particle, $5\ \text{m}$ assumes they meet at the midpoint, and $10\ \text{m}$ is the whole separation.