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(B) The rotational kinetic energy $E$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $E = \frac{L^2}{2I}$.Since $I$ remai

Vedclass · Accepted Answer

$800$

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(B) The rotational kinetic energy $E$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $E = \frac{L^2}{2I}$.Since $I$ remains constant,we have $E \propto L^2$,which implies $\frac{E_2}{E_1} = \left( \frac{L_2}{L_1} ight)^2$.Given that the angular momentum is increased by $200\ \%$,the new angular momentum $L_2 = L_1 + 200\% 	ext{ of } L_1 = L_1 + 2L_1 = 3L_1$.Substituting this into the ratio: $\frac{E_2}{E_1} = \left( \frac{3L_1}{L_1} ight)^2 = 3^2 = 9$.Therefore,$E_2 = 9E_1$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100\% = \frac{9E_1 - E_1}{E_1} 	imes 100\% = 8 	imes 100\% = 800\%$.

If the angular momentum of a rotating body is increased by $200\ \%$,then its kinetic energy of rotation will be increased by .......... $\%$

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