Let the speed of bob at lowest position be $v _{1}$ and at the highest position be $v _{2}$.

Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,

$\frac{1}{2} mv _{1}^{2}=\frac{1}{2} mv _{2}^{2}+ mg 2 l$

$\Rightarrow v _{1}^{2}= v _{2}^{2}+4 g l$ $……(1)$

Now $T _{\max }- mg =\frac{ mv _{1}^{2}}{l}$

$\Rightarrow T _{\max }= mg +\frac{ mv _{1}^{2}}{l}$

and $T _{\min }+ mg =\frac{ mv _{2}^{2}}{l}$

$\Rightarrow T _{\min }=\frac{ mv _{2}^{2}}{l}- mg$

$\frac{ T _{\max }}{ T _{\min }}=\frac{5}{1}$

$\Rightarrow \frac{ mg +\frac{ mv _{1}^{2}}{l}}{\frac{ mv _{2}^{2}}{l}- mg }=\frac{5}{1}$

$\Rightarrow mg +\frac{ mv _{1}^{2}}{l}=\left[\frac{ mv _{2}^{2}}{l}- mg \right] 5$

$\Rightarrow mg +\frac{ m }{l}\left[ v _{2}^{2}+4 g l\right]=\frac{5 mv _{2}^{2}}{l}-5 mg$

$\Rightarrow mg +\frac{ mv _{2}^{2}}{l}+4 mg =\frac{5 mv _{2}^{2}}{l}-5 mg$

$\Rightarrow 10 mg =\frac{4 mv _{2}^{2}}{l}$

$v _{2}{ }^{2}=\frac{10 \times 10 \times 1}{4}$

$\Rightarrow v _{2}^{2}=25 \Rightarrow v _{2}=5 m / s$

Thus, velocity of bob at highest position is $5 m / s$