Q83 / 90 · JEE Main 2019 (09 Apr Shift 2) · Physics

Moment of inertia of a body about a given axis is $1.5 k g m^{2} .$ Initially the body is at rest. In order to produce a rotational kinetic energy of $1200 J,$ the angular acceleration of $20 r a d / s^{2}$ must be applied about the axis for a duration of:

Solution

We know that,

Rotational kinetic energy, $K = \frac{1}{2} I ω^{2}$

$\Rightarrow \frac{1}{2} I ω^{2} = 1200$
$\Rightarrow \frac{1}{2} \times 1 .5 \times ω^{2} = 1200$
$\Rightarrow ω = 40 rad s^{- 1}$

We also know that, $ω = α t$
$\Rightarrow t = \frac{40}{20} = 2 s$

Hence, the correct answer is option: 'B'.