A solid sphere of mass m and radius r is released from rest from the given position. If it rolls without sliding on the circular track of radius R, its speed when it reaches its lowest position will be

When the sphere reaches its lowest position it loses gravitational potential energy by $∆$U and gains a kinetic energy by $∆$K, where $∆$U = -mg(R - r)

$\begin{array}{l}\text{ and }\mathrm{\Delta }K=\frac{1}{2}m{v}_c^2\left(1+\frac{K^2}{R^2}\right)=\frac{1}{2}m{v}^2\left(1+\frac{2}{5}\right)=\frac{7}{10}m{v}^2\\ \text{ According to conservation of energy, }\mathrm{\Delta }U+\mathrm{\Delta }K=0\\ \text{ or }-mg(R-r)+\frac{7}{10}m{v}^2=0\\ \text{ or }v=\sqrt{\frac{10}{7}g(R-r)}\end{array}$