Physics - W.E.P. Question with Solution | TestHub

Let $v_H$ and $v_L$ be the speeds of the stone at the highest and lowest points of the vertical circle, respectively.

The tension at the highest point (minimum tension) is given by:
$T_{min}=\frac{m{v}_H^2}{L}-mg$

The tension at the lowest point (maximum tension) is given by:
$T_{max}=\frac{m{v}_L^2}{L}+mg$

By the principle of conservation of mechanical energy between the highest and lowest points:
$\frac{1}{2}m{v}_L^2=\frac{1}{2}m{v}_H^2+mg(2L)$
$v_L^2=v_H^2+4gL$

Substituting $v_L^2$ into the expression for $T_{max}$:
$T_{max}=\frac{m(v_H^2+4gL)}{L}+mg=\frac{m{v}_H^2}{L}+5mg$

We are given that the ratio of maximum tension to minimum tension is 3:
$\frac{T_{max}}{T_{min}}=3$

$\frac{\frac{m{v}_H^2}{L}+5mg}{\frac{m{v}_H^2}{L}-mg}=3$

$\frac{m{v}_H^2}{L}+5mg=3\left(\frac{m{v}_H^2}{L}-mg\right)$

$\frac{m{v}_H^2}{L}+5mg=3\frac{m{v}_H^2}{L}-3mg$

$2\frac{m{v}_H^2}{L}=8mg$

$\frac{v_H^2}{L}=4g$

$v_H^2=4gL$

$v_H=2\sqrt{gL}$

Answer: 2 $\sqrt{\mathrm{gL}}$