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A stationary horizontal disc is free to rotate about its axis. When a torque is applied, its kinetic energy as a function of the angle rotated is kθ². If its moment of inertia is I, the angular acceleration of the disc is

Asked in JEE Main 9th April 1st Shift 2019 · Torque and angular acceleration

Answer: (4) (2k)/Iθ

Step-by-step solution

1/2 Iω²=kθ²⇒ ω²=(2k)/Iθ², so ω=θ√(2k)/I.

α=(dω)/(dt)=(dω)/(dθ)ω=√(2k)/I·θ√(2k)/I=(2k)/Iθ.

Why the other options are wrong

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