NEET · Physics · STD 11 - 6. system of particles and rotational motion
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 \(.............\,kg - m ^2\)
- A \(0.7\)
- B \(3.22\)
- C \(30.8\)
- D \(0.07\)
Answer & Solution
Correct Answer
(A) \(0.7\)
Step-by-step Solution
Detailed explanation
\(\omega_{ i }=60 rpm =60 \times \frac{2 \pi}{60}=2 \pi rad / s\) \(\omega_{ f }=360 rpm =360 \times \frac{2 \pi}{60}=12 \pi rad / s\)
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