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MHT CET 2019 (2 May, Shift 2) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2019 (2 May, Shift 2)

Q44MHT CET 2019 · 2 MAY · SHIFT 2Easy

Rotational Kinetic Energy, Work & Power

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is (π2=10)(\pi^2 = 10)

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Answer: (b)

  1. Work = change in KE: W=12I(ω22−ω12)=2π2I(n22−n12)W = \frac{1}{2}I(\omega_2^2 - \omega_1^2) = 2\pi^2 I(n_2^2 - n_1^2)
  2. 12000=2(10)I(400−100)=6000 I12000 = 2(10)I(400 - 100) = 6000\,I
  3. I=2I = 2 kg m2^2
  4. So the answer is (b) 2 kg m22\text{ kg m}^2.
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