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Question

If the angular momentum of any rotating ball increases by $$200\%$$, then the increase in kinetic energy is?


A
400%
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B
800%
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C
200%
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D
100%
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Solution

The correct option is B $$800\%$$
$$\displaystyle E=\frac{L^2}{2l}$$
$$\therefore \displaystyle E\propto L^2$$
$$\displaystyle\frac{E_2}{E_1}=\left(\displaystyle\frac{L_2}{L_1}\right)^2$$
$$\displaystyle\frac{E_2}{E_1}=\left[\displaystyle\frac{L_1+200\% of L_1}{L_1}\right]$$
$$=\left[\displaystyle\frac{L_1+2L_1}{L_1}\right]^2=(3)^2$$
$$E_2=9E_1$$
Increment in kinetic energy
$$\Delta E=E_2-E_1$$
$$=9E_1 - E_1$$
$$\Delta E=8E_1$$
$$\displaystyle\frac{\Delta E}{E_1}=8$$
or percentage increase$$=800\%$$.

Physics

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