JEE Main 2026 Jan 28 Shift 2, Physics Q25: Work Energy Theorem And Its Applications

JEE Main2026Jan 28, Shift 2Physics

A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is ______ J. ( $g = 10$ m/s $^{2}$ )

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Solution

Let $M = 3$ kg (wheel), $R = 5$ m, $m = 3$ kg (hanging block), $h = 3$ m. Treat the wheel as a uniform disc: $I = \frac{1}{2} M R^{2}$ . String constraint: $v = ω R$ .

Conservation of energy: $m g h = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) ω^{2} + \frac{1}{2} m v^{2}$ .

Using $ω R = v$ and $M = m$ : $m g h = \frac{1}{4} m v^{2} + \frac{1}{2} m v^{2} = \frac{3}{4} m v^{2}$ .

$g h = \frac{3}{4} v^{2} \Rightarrow v^{2} = \frac{4 \times 10 \times 3}{3} = 40$ .

KE of wheel $= \frac{1}{2} I ω^{2} = \frac{1}{4} M v^{2} = \frac{1}{4} \times 3 \times 40 = 30$ J.

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JEE Main 2026 Jan 28 Shift 2, Physics Q25: Work Energy Theorem And Its Applications

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