A solid sphere of mass 1   kg rolls without slipping on a plane surface. Its kinetic energy is 7 ×…

Let velocity of centre of mass of sphere be $v$.

Total kinetic energy of solid sphere is $KE=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$, where, $I$ is moment of inertia of the solid sphere about the axis passing through it's centre of mass and $\omega =\frac{v}{R} \left(\text{pure rolling condition}\right)$.

Therefore,

 $7\times {10}^{-3}=\frac{1}{2}\times 1\times (v{)}^2+\frac{1}{2}\left[\frac{2}{5}\left(1\right)R^2\right]{\left(\frac{v}{R}\right)}^2$

$\Rightarrow \frac{1}{2}{v}^2\left(\frac{7}{5}\right)=7\times {10}^{-3}$

$\Rightarrow v^2={10}^{-2}$

$\Rightarrow v={10}^{-1}=0.1 \mathrm{m} {\mathrm{s}}^{-1}=10 \mathrm{cm} {\mathrm{s}}^{-1}$