Given,

$\frac{T_{1}}{T_{2}} = 4$ $T_{1} = 4T_{2}$

Here

$T_{1}$ is maximium tension

$T_{2}$ is minimum tension

The difference between the maximum tension and minimum tension for a body in the vertical circle is

$\Rightarrow T_{1} = 4T_{2} = 6 mg$ $\Rightarrow 4T_{2} – T_{1} = 6 mg$ $\Rightarrow 3T_{2} = 6 mg$ $\Rightarrow T_{2} = 2 mg$

Now,

The highest point minimum tension can be

$T_{2} = \frac{mv^{2}}{r} – mg$ $2 mg = \frac{mv^{2}}{r} – mg$ $3mg = \frac{mv^{2}}{r}$

Therefore, $v = \sqrt{3mgr}$

Here r = $L = \frac{10}{3r}$ $v = \sqrt{3 * 10 * \frac{10}{3}}$ $v = \sqrt{100}m/s$

Therefore, $10 ms^{-1}$

Explore more such questions and answers at BYJU’S.