JEE Advanced 2021 Paper 2 · Q10 · Angular Momentum

Correct answer:0.16

Solution

When the string becomes taut, the radial component of the bob's velocity is destroyed by the impulsive tension along the string; only the tangential component survives. Angular momentum about the suspension point is conserved across the jerk because the impulsive tension passes through that point: $J_f = J_i = 0.18$ kg m$^2$/s. After lift-off the bob moves in a circle of radius $L = 1$ m, so $J_f = m v_f L \Rightarrow 0.18 = 0.1 \times v_f \times 1 \Rightarrow v_f = 1.8$ m/s. Kinetic energy $K = (1/2) m v_f^2 = 0.5 \times 0.1 \times 1.8^2 = 0.162$ J $\approx 0.16$ J.

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