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A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which is has rotated, is given as kθ2 . If its moment of inertia is I then the angular acceleration of the disc is

 (1) k/I  θ (2)  k/2I θ (3)  k/4I θ (4)  2k/I θ

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Correct option  (4)  2k/I θ


Given that energy ⇒ (1/2)I ω2 = kθ2

(1/2)Iω2 = kθ2

By differentiate

(1/2) I × 2ω dω/dt = 2kθ dθ/dt

I ω dω/dθ = 2kθ

α= 2kθ/I

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