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(D) Let the initial angular momentum be $L_1 = L$. Since the angular momentum is increased by $200\%$,the new angular momentum is $L_2 = L + 2.00L = 3

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(D) Let the initial angular momentum be $L_1 = L$. Since the angular momentum is increased by $200\%$,the new angular momentum is $L_2 = L + 2.00L = 3L$. The rotational kinetic energy $E$ is related to angular momentum $L$ by the formula $E = \frac{L^2}{2I}$,where $I$ is the moment of inertia. Assuming $I$ remains constant,the ratio of kinetic energies is $\frac{E_2}{E_1} = \frac{L_2^2}{L_1^2} = \frac{(3L)^2}{L^2} = 9$. The percentage increase in rotational kinetic energy is given by $\left( \frac{E_2 - E_1}{E_1} ight) 	imes 100\% = \left( \frac{E_2}{E_1} - 1 ight) 	imes 100\%$. Substituting the values,we get $(9 - 1) 	imes 100\% = 8 	imes 100\% = 800\%$. Thus,the rotational kinetic energy increases by $800\%$.

If the angular momentum of a body is increased by $200\%$,then the percentage increase in its rotational kinetic energy will be ........ $\%$.

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