A bob of mass m is suspended at a point O by a light string of length l and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v₀ at the point ' A ', the string becomes slack when, the bob reaches at the point ' D '. The ratio of the kinetic energy of the bob at the points B and C is _________. (a) 4 (b) 1 (c) 2 (d) 3

Answer: (c). String goes slack at the top D, so tension is zero there: mg = (mvD²)/(l) ⇒ KED = 1/2mgl. Heights above A: hB = l(1-cos 60°) = l/2, hC = l + lcos 60° = 3l/2, hD = 2l. KEB = KED + mg(2l - l/2) = 1/2mgl + 3/2mgl = 2mgl KEC = KED + mg(2l - 3l/2) = 1/2mgl + 1/2mgl = mgl So KEB: KEC = 2, the answer is (c) 2.

A bob of mass m is suspended at a point O by a light string of length… | JEE Main 2025 (22 Jan, Shift 1)

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Critical Speed & Looping

A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $v_0$ at the point ' A ', the string becomes slack when, the bob reaches at the point ' $D$ '. The ratio of the kinetic energy of the bob at the points B and C is _________.

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Answer: (c)

String goes slack at the top D, so tension is zero there: $mg = \frac{mv_D^2}{l} \Rightarrow KE_D = \frac{1}{2}mgl$ .
Heights above A: $h_B = l(1-\cos 60^\circ) = \frac{l}{2}$ , $h_C = l + l\cos 60^\circ = \frac{3l}{2}$ , $h_D = 2l$ .
$KE_B = KE_D + mg(2l - \frac{l}{2}) = \frac{1}{2}mgl + \frac{3}{2}mgl = 2mgl$
$KE_C = KE_D + mg(2l - \frac{3l}{2}) = \frac{1}{2}mgl + \frac{1}{2}mgl = mgl$
So $KE_B : KE_C = 2$ , the answer is (c) 2.

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