A small bob tied at one end of a thin string of length 1m is...

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,

$\dfrac{1}{2}m{v}_{1}^{2} = \dfrac{1}{2}m{v}_{2}^{2} + mg2l$

$⇒ {v}_{1}^{2} = {v}_{2}^{2} + 4gl . . . . \left(1\right)$

Now, {T}_\max - mg = \dfrac{m{v}_{1}^{2}}{l}

⇒ {T}_\max = mg + \dfrac{m{v}_{1}^{2}}{l}

\text{ and } {T}_\min + mg = \dfrac{m{v}_{2}^{2}}{l}

⇒ {T}_\min = \dfrac{m{v}_{2}^{2}}{l} - mg

\dfrac{{T}_\max}{{T}_\min} = \dfrac{5}{1}

$⇒ \dfrac{mg + \dfrac{m{v}_{1}^{2}}{l}}{\dfrac{m{v}_{2}^{2}}{l} - mg} = \dfrac{5}{1}$

$⇒ mg + \dfrac{m{v}_{1}^{2}}{l} = \left[\dfrac{m{v}_{2}^{2}}{l} - mg\right]5$

$⇒ mg + \dfrac{m}{l}\left[{v}_{2}^{2} + 4gl\right] = \dfrac{5m{v}_{2}^{2}}{l} - 5mg$

$⇒ mg + \dfrac{m{v}_{2}^{2}}{l} + 4mg = \dfrac{5m{v}_{2}^{2}}{l} - 5mg$

$⇒ 10mg = \dfrac{4m{v}_{2}^{2}}{l}$

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

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