JEE Main 2025 Jan 23 Shift 2, Physics Q7: Law of Rotation And Work-Energy Theorem

JEE Main2025Jan 23, Shift 2Physics

A circular disk of radius $R$ meter and mass $M$ kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $θ (t) = 5 t^{2} - 8 t$ , where $θ (t)$ is the angular position of the rotating disc as a function of time $t$ .

How much power is delivered by the applied torque, when $t = 2$ s?

A
$60 M R^{2}$
B
$72 M R^{2}$
C
$108 M R^{2}$
D
$8 M R^{2}$

Solution

Differentiate twice to get $ω$ and $α$ :

$ω = \frac{d θ}{d t} = 10 t - 8, α = \frac{d ω}{d t} = 10 rad/s^{2} .$

At $t = 2$ s: $ω = 20 - 8 = 12 rad/s$ .

Moment of inertia of a disc about the perpendicular axis through its centre: $I = \frac{M R ^{2}}{2}$ .

Torque: $τ = I α = \frac{M R ^{2}}{2} \cdot 10 = 5 M R^{2}$ .

Power: $P = τ ω = 5 M R^{2} \cdot 12 = 60 M R^{2}$ W.

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JEE Main 2025 Jan 23 Shift 2, Physics Q7: Law of Rotation And Work-Energy Theorem

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