Total kinetic energy of solid sphere of mass $m$ rolling with velocity $v$ without slipping is

The correct option is A $\frac{7}{10}m{v}^2$
A rolling body has linear kinetic energy along with rotational kinetic energy.
$K.E_{Trans}=\frac{1}{2}m{v}^2,K.E_{Rotational}=\frac{1}{2}I{\omega }^2$
Thus, total K.E of rolling body is
$K.E_{total}=K.E_{Trans}+K.E_{Rotational}K.E_{total}=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2{\omega }^2\right)$
As we know that in rolling without slipping, $velocity\left(v\right)=radiusofsphere\left(r\right)\times angularvelocity\left(\omega \right)v=r\times \omega$ $\therefore K.E_{total}=\frac{7}{10}m{v}^2$