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NEET 2022 (Phase 2) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from NEET 2022 (Phase 2)

Q32NEET 2022 · PHASE 2Easy

Rotational Kinetic Energy, Work & Power

An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is:

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Answer: (b)

  1. ω1=2π\omega_1 = 2\pi rad/s (60 rpm), ω2=12π\omega_2 = 12\pi rad/s (360 rpm)
  2. W=12I(ω22−ω12)=12I(4π2)(36−1)=70π2IW = \frac{1}{2}I(\omega_2^2 - \omega_1^2) = \frac{1}{2}I(4\pi^2)(36 - 1) = 70\pi^2 I
  3. I=48470π2=484690.8≈0.7I = \frac{484}{70\pi^2} = \frac{484}{690.8} \approx 0.7 kg m2^2
  4. So the answer is (b) 0.7 kg m2^2.
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