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

Correct answer: C. 1 rad / s About fixed axis arrow Torque =I α aligned & α=(100)/(300) & α=(1)/(3) rad / s^2 aligned aligned & ω_i=0 & ω_f=ω_i+α t & at t=3 s & ω_f=(1)/(3) × 3=1 rad / s aligned

NEET 2023

Exam: NEET
Year: 2023
Subject: Physics
Chapter: Motion in Two Dimensions

NEET 2023 · Physics · Motion in Two Dimensions

A constant torque of 100 ~N ~m turns a wheel of moment of inertia...

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}

Options

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

Correct answer

C. $1 \mathrm{rad} / \mathrm{s}$

Step-by-step 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}

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