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

$\begin{array}{l}\begin{array}{ll} & \mathrm{K}.\mathrm{E}=\frac{{\mathrm{L}}^2}{2\mathrm{I}}\\ & \mathrm{K}\cdot {\mathrm{E\alpha L}}^2\\ & \mathrm{L\alpha }\sqrt{\mathrm{K}\cdot \mathrm{E}}\\ & \frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}=\sqrt{\frac{\mathrm{K}\cdot {\mathrm{E}}_1}{\mathrm{K}\cdot {\mathrm{E}}_2}}\\ & \frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}=\sqrt{\frac{100}{64}}=\frac{10}{8}\\ & \frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}=\frac{5}{4}\\ & \frac{\mathrm{\Delta L}}{\mathrm{L}}=\frac{{\mathrm{L}}_2-{\mathrm{L}}_1}{{\mathrm{L}}_1}=\frac{4-5}{5}=\frac{-1}{5}\\ & =-0.2=-20\mathrm{\%}\end{array}\\ \text{ Decrease by }20\mathrm{\%}\end{array}$