JEE Main 2025 Jan 22 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

JEE Main2025Jan 22, Shift 1Physics

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 ______.

Solution

The string becomes slack at the topmost point D, so the bob just barely completes the vertical circle. Hence $v_{A}^{2} = 5 g ℓ$ at the bottom.

B is at height $ℓ /2$ above A (since OB makes $6 0^{\circ}$ with OA, so B's height above A is $ℓ - ℓ cos 6 0^{\circ} = ℓ /2$ ).

Energy conservation from A to B: $\frac{1}{2} m (5 g ℓ) = K E_{B} + m g \cdot \frac{ℓ}{2}$ , giving $K E_{B} = \frac{5 m g ℓ}{2} - \frac{m g ℓ}{2} = 2 m g ℓ$ .

C lies $ℓ /2$ below D (OC makes $6 0^{\circ}$ with OD). At the top D, the bob has $v_{D}^{2} = g ℓ$ , so $K E_{D} = \frac{1}{2} m g ℓ$ . From D to C the bob drops by $ℓ /2$ :

$K E_{C} = K E_{D} + m g \cdot \frac{ℓ}{2} = \frac{m g ℓ}{2} + \frac{m g ℓ}{2} = m g ℓ$ .

$∴ \frac{K E _{B}}{K E _{C}} = \frac{2 m g ℓ}{m g ℓ} = 2$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

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