Q117 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics

The moment of inertia of a body about a given axis is $1.2 \mathrm{~kg} \cdot \mathrm{m}^2$. Initially the body is at rest. In order to produce rotational kinetic energy of $1500 \mathrm{~J}$, an angular acceleration of $25 \mathrm{rad} / \mathrm{s}^2 \mathrm{must}$ be applied about an axis for a time duration of

  • A
    $8 \mathrm{~s}$
  • Correct answer
    B
    $2 \mathrm{~s}$
  • C
    $4 \mathrm{~s}$
  • D
    $1 \mathrm{~s}$

Solution

$$ \begin{aligned} & \mathrm{I}=1.2 \mathrm{~kg}-\mathrm{m}^2, \mathrm{~K} . E .=1500 \mathrm{~J}, \alpha=25 \mathrm{rad} / \mathrm{s}^2 \\ & \text { K.E. }=\frac{1}{2} \mathrm{I} \omega^2 \\ & \therefore 1500=\frac{1}{2} \times 1.2 \times \omega^2 \\ & \omega=50 \mathrm{rad} / \mathrm{s} \\ & \omega=\omega_0+\alpha \mathrm{t} \\ & 50=0+25 \mathrm{t}=25 \mathrm{t} \\ & \mathrm{t}=2 \mathrm{~s} \end{aligned} $$

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

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