A solid Copper sphere is spinning about a diameter with angular velocity 10 rad/s. Due to rise in temperature, the radius of sphere increase by 1%. What will be the change in its angular velocity?

Since no external torque is acting on the sphere, its angular momentum remains conserved.

$\begin{array}{l}\therefore \mathrm{I\omega }=\mathrm{L}=\mathrm{constant}\Rightarrow \frac{2}{5}{\mathrm{mR}}^2.\mathrm{\omega }=\mathrm{constant}\Rightarrow \mathrm{\omega }=\mathrm{c}.{\mathrm{R}}^{-2}\\ \therefore \frac{\mathrm{\Delta \omega }}{\mathrm{\omega }}=-2\left(\frac{\mathrm{\Delta R}}{\mathrm{R}}\right)\Rightarrow \frac{\mathrm{\Delta \omega }}{\mathrm{\omega }}\times 100\mathrm{\%}=-2\times 1\mathrm{\%}=-2\mathrm{\%}\end{array}$

So angular velocity decreases by 2%.