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Q.

The angular momentum of rotating body is increased by 20% . The increase in rotational K.E is _____ %

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answer is 44.

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Detailed Solution

$K . E = \frac{L^{2}}{2 I} \Rightarrow {EαL}^{2}$

$\frac{ΔE}{E} = {(\frac{120}{100})}^{2} - 1 \Rightarrow \frac{ΔE}{E} = 0.44 \Rightarrow \frac{ΔE}{E} \times 100 = 44 %$

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