JEE Main 2019 — Physics solution

JEE Main2019Physics

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 acceleration of 20 rad/s²must be applied about the axis for a duration of

A
2.5 s
B
2 s
C
5 s
D
3 s

Solution

$l = 150; α = 20 r a d / s^{2} ∴ ω = a t E = \frac{1}{2} l ω^{2} = 1200 \frac{1}{2} \times 1.5 \times (20 t)^{2} = 1200 J \Rightarrow t = 2 s$

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