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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 2 April · Morning Shift · Physics · Rotational Motion

A cord of negligible mass is wound around the rim of a wheel...

A cord of negligible mass is wound 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 :
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of

Options

  1. A20rad/s20 \mathrm{rad} / \mathrm{s}
  2. B30rad/s30 \mathrm{rad} / \mathrm{s}
  3. C10rad/s10 \mathrm{rad} / \mathrm{s}
  4. D0rad/s0 \mathrm{rad} / \mathrm{s}

Correct answer

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

Step-by-step solution

WF=20×1=20 J∴ΔKE=20 J=12Iω2I=MR2=10×0.12=0.1 kg m2∴20=12×0.1×ω2⇒ω=20rad/sec\begin{aligned} & \mathrm{W}_{\mathrm{F}}=20 \times 1=20 \mathrm{~J} \\ & \therefore \quad \Delta \mathrm{KE}=20 \mathrm{~J}=\frac{1}{2} \mathrm{I} \omega^2 \\ & \mathrm{I}=\mathrm{MR}^2=10 \times 0.1^2=0.1 \mathrm{~kg} \mathrm{~m}^2 \\ & \therefore \quad 20=\frac{1}{2} \times 0.1 \times \omega^2 \\ & \Rightarrow \omega=20 \mathrm{rad} / \mathrm{sec}\end{aligned}
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