A stone of mass 1 kg tied to a light inextensible string of length $\mathrm{L}=\frac{5}{3} \mathrm{~m}$ is rotating in a circular path of radius L in a vertical plane. If the ratio of maximum tension in the string to the minimum tension in the string is 3 , the speed of the stone at the highest point of the circle is ( $\mathrm{g}=$ acceleration due to gravity)

The forces acting on the stone at the highest point yield the minimum tension, satisfying $T_{\min} + mg = \frac{mv_H^2}{L}$, where $v_H$ is the speed at the highest point. Rearranging gives $T_{\min} = \frac{mv_H^2}{L} - mg$.

At the lowest point, the centripetal force requires $T_{\max} - mg = \frac{mv_L^2}{L}$, giving $T_{\max} = \frac{mv_L^2}{L} + mg$.

By conservation of mechanical energy, taking the lowest point as zero potential energy, $\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + 2mgL$. Multiplying by $2/m$ yields $v_L^2 = v_H^2 + 4gL$.

Substituting into the expression for $T_{\max}$:

$T_{\max} = \frac{m(v_H^2 + 4gL)}{L} + mg = \frac{mv_H^2}{L} + 4mg + mg = \frac{mv_H^2}{L} + 5mg$

Given $T_{\max}/T_{\min} = 3$, we substitute the expressions:

$\frac{\frac{mv_H^2}{L} + 5mg}{\frac{mv_H^2}{L} - mg} = 3$

Letting $X = \frac{mv_H^2}{L}$, the equation becomes $\frac{X + 5mg}{X - mg} = 3$, leading to $X + 5mg = 3(X - mg)$. Solving gives $X = 4mg$.

Thus $\frac{mv_H^2}{L} = 4mg$, so $v_H^2 = 4gL$ and $v_H = 2\sqrt{gL}$.

Final speed at highest point: $2\sqrt{gL}$