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JEE Main 2025 (2 Apr, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2025 (2 Apr, Shift 1)

Q20JEE Main 2025 · 2 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

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:

Figure

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Answer: (a)

  1. Spokes massless, so the wheel is a ring: I=MR2=10(0.1)2=0.1 kg m2I = MR^2 = 10(0.1)^2 = 0.1\text{ kg m}^2.
  2. Work by pull =F s=20×1=20= F\,s = 20\times1 = 20 J =12Iω2= \frac{1}{2}I\omega^2
  3. ω2=400.1=400⇒ω=20 rad/s\omega^2 = \frac{40}{0.1} = 400 \Rightarrow \omega = 20\text{ rad/s}
  4. So the answer is (a) 20 rad/s.
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