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Physics · Rotational Dynamics

JEE Main 2025 — 2 April, Morning Shift — Question 63

A cord of negligible mass is around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest.

If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m , would be :

Question figure
  1. Option A:

    10rad/s10 \mathrm{rad} / \mathrm{s}

  2. Option B:

    20rad/s20 \mathrm{rad} / \mathrm{s}

    Correct
  3. Option C:

    30rad/s30 \mathrm{rad} / \mathrm{s}

  4. Option D:

    0rad/s0 \mathrm{rad} / \mathrm{s}

Answer: B

Step-by-step solution

W=F.S=20×1=20 JW=F . S=20 \times 1=20 \mathrm{~J}

20 J=12(MR2)ω220=12×10×1100×ω2ω=20rad/s\begin{aligned} & 20 \mathrm{~J}=\frac{1}{2}\left(M R^{2}\right) \omega^{2} & \\ 20=\frac{1}{2} \times 10 \times \frac{1}{100} \times \omega^{2} &\\ \omega=20 \mathrm{rad} / \mathrm{s} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling

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