A small bob tied at one end of a thin string of length 1   m is describing a vertical circle so that…

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,

$\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_2^2+mg2l$

$\Rightarrow v_1^2=v_2^2+4 gl ....\left(1\right)$

Now, $T_{\max }-mg=\frac{m{v}_1^2}{l}$

$\Rightarrow T_{\max }=mg+\frac{m{v}_1^2}{l}$

$\& T_{\min }+mg=\frac{m{v}_2^2}{l}$

$\Rightarrow T_{\min }=\frac{m{v}_2^2}{l}-mg$

$\frac{T_{\max }}{T_{\min }}=\frac{5}{1}$

$\Rightarrow \frac{mg+\frac{m{v}_1^2}{l}}{\frac{m{v}_2^2}{l}-mg}=\frac{5}{1}$

$\Rightarrow mg+\frac{m{v}_1^2}{l}=\left[\frac{m{v}_2^2}{l}-mg\right]5$

$\Rightarrow mg+\frac{m}{l}\left[v_2^2+4 gl\right]=\frac{5m{v}_2^2}{l}-5mg$

$\Rightarrow mg+\frac{m{v}_2^2}{l}+4mg=\frac{5m{v}_2^2}{l}-5mg$

$\Rightarrow 10 mg=\frac{4m{v}_2^2}{l}$

$v_2=\frac{10\times 10\times 1}{4}$

Thus, the velocity of the bob at the highest position is $5 \mathrm{m} {\mathrm{s}}^{-1}.$