Q108 / 200 · NEET 2023 (Manipur) · Physics

A constant torque of $100 \mathrm{~N} \mathrm{~m}$ turns a wheel of moment of inertia $300 \mathrm{~kg} \mathrm{~m}^2$ about an axis passing through its centre. Starting from rest, its angular velocity after $3 \mathrm{~s}$ is

  • A
    $10 \mathrm{rad} / \mathrm{s}$
  • B
    $15 \mathrm{rad} / \mathrm{s}$
  • Correct answer
    C
    $1 \mathrm{rad} / \mathrm{s}$
  • D
    $5 \mathrm{rad} / \mathrm{s}$

Solution

About fixed axis $\rightarrow$ Torque $=I \alpha$ $\begin{aligned} & \alpha=\frac{100}{300} \\ & \alpha=\frac{1}{3} \mathrm{rad} / \mathrm{s}^2\end{aligned}$ $\begin{aligned} & \Rightarrow \omega_i=0 \\ & \omega_f=\omega_i+\alpha t \\ & \text { at } t=3 \mathrm{~s} \\ & \omega_f=\frac{1}{3} \times 3=1 \mathrm{rad} / \mathrm{s}\end{aligned}$

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

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