MHT CET · Physics · Rotational Motion
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 \(8 \mathrm{~s}\)
- B \(2 \mathrm{~s}\)
- C \(4 \mathrm{~s}\)
- D \(1 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \mathrm{I}=1.2 \mathrm{~kg}-\mathrm{m}^2, \mathrm{~K} . E .=1500 \mathrm{~J}, \alpha=25 \mathrm{rad} / \mathrm{s}^2 \\ & \text { K.E. }=\frac{1}{2} \mathrm{I} \omega^2 \\ & \therefore 1500=\frac{1}{2} \times 1.2 \times \omega^2 \\ & \omega=50 \mathrm{rad} / \mathrm{s} \\ & \omega=\omega_0+\alpha \mathrm{t} \\ & 50=0+25 \mathrm{t}=25 \mathrm{t} \\ & \mathrm{t}=2 \mathrm{~s} \end{aligned} \)
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