c
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 \(5 m / s\)