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JEE Main 2019 (8 Apr, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2019 (8 Apr, Shift 2)

Q60JEE Main 2019 · 8 APR · SHIFT 2Difficult

Rotational Kinematics & Dynamics

A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5m. When released, it slips off the table in a very short time t = 0.01s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to :-

Figure

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Answer: (d)

  1. While slipping off, gravity gives a torque about the edge: α=mg(l/2)ml2/3=3g2l\alpha = \dfrac{mg(l/2)}{ml^2/3} = \dfrac{3g}{2l}.
  2. Angular speed gained: ω=αt=3×10×0.012×0.3=0.5\omega = \alpha t = \dfrac{3 \times 10 \times 0.01}{2 \times 0.3} = 0.5 rad/s.
  3. Fall time: t′=2h/g=2×5/10=1t' = \sqrt{2h/g} = \sqrt{2 \times 5/10} = 1 s.
  4. θ=ωt′=0.5×1=0.5\theta = \omega t' = 0.5 \times 1 = 0.5 rad, so the answer is (d).
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