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JEE Main 2026 (22 Jan, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2026 (22 Jan, Shift 2)

Q8JEE Main 2026 · 22 JAN · SHIFT 2Medium

Rotational Kinematics & Dynamics

Two masses mm and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius r=0.1 mr = 0.1\text{ m}. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is ‾ m/s\underline{\qquad}\text{ m/s}. (Assume string does not slip and g=10 m/s2g = 10\text{ m/s}^{2})

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Pulley disc: I=12(30m)r2=15mr2I = \frac{1}{2}(30m)r^2 = 15mr^2, rotational KE =12(15m)v2= \frac{1}{2}(15m)v^2.
  2. Energy: (2m−m)gh=12(3m)v2+12(15m)v2=9mv2(2m - m)gh = \frac{1}{2}(3m)v^2 + \frac{1}{2}(15m)v^2 = 9mv^2
  3. v2=gh9=10×3.69=4v^2 = \frac{gh}{9} = \frac{10\times3.6}{9} = 4
  4. v=2 m/sv = 2\text{ m/s}
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