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Physics · Rotational Dynamics

JEE Main 2024 — 9 April, Shift 1 — Question 54

A string is wrapped around the rim of a wheel of moment of inertia 0.40kgm20.40 \mathrm{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 . The angular velocity of the wheel after 10 s is xrad/s\mathrm{x} \mathrm{rad} / \mathrm{s}, where x is \qquad .

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

τ=FR=Iα⇒40×0.1=0.4α\tau=\mathrm{FR}=\mathrm{I} \alpha \Rightarrow 40 \times 0.1=0.4 \alpha

α=10rad/s2\alpha=10 \mathrm{rad} / \mathrm{s}^{2}

Wf=10×10=100rad/s\mathrm{W}_{\mathrm{f}}=10 \times 10=100 \mathrm{rad} / \mathrm{s}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling

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