JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Work, Power & Energy

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A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed

u

. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is

x (u^{2} - gL)

. The value of x is

(A)3
(B)2✓ Correct
(C)1
(D)5

Answer: (B)

Solution

Using energy conservation from lowest to horizontal position:

\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g L

, so

v^{2} = u^{2} - 2 g L

. At the lowest point velocity is horizontal (magnitude

u

), at horizontal position velocity is also horizontal (magnitude

v

). The magnitude of change in velocity is

∣Δ v ∣ = u^{2} + v^{2} = u^{2} + u^{2} - 2 g L = 2 (u^{2} - g L)

. Therefore

x = 2

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Work, Power & Energy

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