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

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Accepted Answer

Correct answer: B. (b) : From the work-energy principle we get, 12l ^2= work done ⇒ 12 I 4 ^2( (20 )^2- (10 )^2)=12000 ⇒ 12 I 4 ^2 300=12000 ⇒I= 12000 24 ^2 300= 12000 24 10 300=2 kg m^2

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

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