A string is wrapped around the rim of a wheel of moment of inertia 0.40 kgm^2 and radius 10 ~cm . The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 ~N

Correct answer: 0 aligned & =FR=I α 40 × 0.1=0.4 α & α=10 rad / s^2 & ~W_ f =10 × 10=100 rad / s aligned

JEE Main 20249 AprilMorning Shift

JEE Main 2024 · 9 April · Morning Shift · Physics · Rotational Motion

A string is wrapped around the rim of a wheel of moment of inertia...

A string is wrapped around the rim of a wheel of moment of inertia

0.40 \mathrm{kgm}^2

and radius

10 \mathrm{~cm}

. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of

40 \mathrm{~N}

. The angular velocity of the wheel after

10 \mathrm{~s}

x \mathrm{rad} / \mathrm{s}

, where

x

is _______

\begin{aligned} & \tau=\mathrm{FR}=\mathrm{I} \alpha \Rightarrow 40 \times 0.1=0.4 \alpha \\ & \alpha=10 \mathrm{rad} / \mathrm{s}^2 \\ & \mathrm{~W}_{\mathrm{f}}=10 \times 10=100 \mathrm{rad} / \mathrm{s}\end{aligned}