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If the K.E of a rotating body about an axis is decreased by 36%. Its angular momentum about that axis

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a

Decrease by 20%

b

increase by 20%

c

Decrease by 72%

d

increase by 72%

answer is D.

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Detailed Solution

$\begin{array}{l} \begin{array}{ll} K . E = \frac{L^{2}}{2 I} \\ K \cdot {EαL}^{2} \\ Lα \sqrt{K \cdot E} \\ \frac{L_{1}}{L_{2}} = \sqrt{\frac{K \cdot E_{1}}{K \cdot E_{2}}} \\ \frac{L_{1}}{L_{2}} = \sqrt{\frac{100}{64}} = \frac{10}{8} \\ \frac{L_{1}}{L_{2}} = \frac{5}{4} \\ \frac{ΔL}{L} = \frac{L_{2} - L_{1}}{L_{1}} = \frac{4 - 5}{5} = \frac{- 1}{5} \\ = - 0.2 = - 20 % \end{array} \\ Decrease by 20 % \end{array}$

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