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(C) The volume of a spherical body is given by $V = \frac{4}{3} \pi R^{3}$.Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 3 \frac{\Del

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Will decrease approximately by $0.67\%$

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(C) The volume of a spherical body is given by $V = \frac{4}{3} \pi R^{3}$.Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 3 \frac{\Delta R}{R}$.Given $\frac{\Delta V}{V} = 1\% = 0.01$,we have $3 \frac{\Delta R}{R} = 0.01$,which implies $\frac{\Delta R}{R} = \frac{0.01}{3} \approx 0.0033$ or $0.33\%$.Since there is no external torque,the angular momentum $L$ is conserved,where $L = I \omega$.For a solid sphere,$I = \frac{2}{5} M R^{2}$,so $L = \frac{2}{5} M R^{2} \omega$.Since $L$ and $M$ are constant,we have $R^{2} \omega = 	ext{constant}$.Taking the derivative,$2 \frac{\Delta R}{R} + \frac{\Delta \omega}{\omega} = 0$.Therefore,$\frac{\Delta \omega}{\omega} = -2 \frac{\Delta R}{R} = -2(0.33\%) = -0.66\% \approx -0.67\%$.Thus,the angular speed decreases by approximately $0.67\%$.

$A$ rigid spherical body is spinning around an axis without any external torque. Due to a change in temperature,the volume increases by $1\%$. Its angular speed:

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