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JEE Main 2019 (9 Apr, Shift 2) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from JEE Main 2019 (9 Apr, Shift 2)

Q45JEE Main 2019 · 9 APR · SHIFT 2Easy

Rotational Kinetic Energy, Work & Power

Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular accleration of 20 rad/s2 must be applied about the axis for a duration of :-

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Answer: (d)

$\frac{1}{2}I\omega^2 = 1200 \Rightarrow \omega^2 = \frac{2400}{1.5} = 1600$ , $\omega = 40$ rad/s
From rest: $t = \frac{\omega}{\alpha} = \frac{40}{20} = 2$ s
So the answer is (d) 2 s.

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