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When a trapeze artist, spinning in air, folds his body, his moment of inertia decreases from I1 to I2.

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When a trapeze artist, spinning in air, folds his body, his moment of inertia decreases from I1 to I2. The work done by him in this process is proportional to

(a) I1 - I2

(b) 1/I2 - 1/I1

(c) 1/(I1 - I2)

(d) I1I2 /(I1 - I2)

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Correct Answer is: (b) 1/I2 - 1/I1

The angular momentum L will be conserved.

Initial rotational KE = Ei = L2/2I1.

Final rotational KE =Ef = L2/2I.

Work done = Ef - Ei = L2/2 [1/I2 - 1/I1].

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