The angular momentum of rotating body is increased by 20%. What will be the increase in its rotational kinetic energy?

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40%

22%

11%

44%

answer is D.

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Detailed Solution

Kinetic energy $K . E . = \frac{L^{2}}{2 I} E α L^{2}$ \

$\frac{E + Δ E}{E} = {(\frac{120}{100})}^{2} (o r) \frac{Δ E}{E} = 0.44$
$\frac{Δ E}{E} \times 100 = 44 %$

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