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(C) Let $u$ be the velocity at the bottom and $v$ be the velocity at the top. The tension at the bottom is $T_{max} = \frac{m u^2}{r} + mg$ and the te

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$\sqrt{50} \,ms^{-1}$

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(C) Let $u$ be the velocity at the bottom and $v$ be the velocity at the top. The tension at the bottom is $T_{max} = \frac{m u^2}{r} + mg$ and the tension at the top is $T_{min} = \frac{m v^2}{r} - mg$.Using conservation of energy between the bottom and top: $\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg(2r) \Rightarrow u^2 = v^2 + 4rg$.Substituting $u^2$ into the $T_{max}$ expression: $T_{max} = \frac{m(v^2 + 4rg)}{r} + mg = \frac{m v^2}{r} + 5mg$.Given $\frac{T_{max}}{T_{min}} = 4$, so $\frac{\frac{m v^2}{r} + 5mg}{\frac{m v^2}{r} - mg} = 4$.Let $x = \frac{m v^2}{r}$. Then $\frac{x + 5mg}{x - mg} = 4 \Rightarrow x + 5mg = 4x - 4mg \Rightarrow 3x = 9mg \Rightarrow x = 3mg$.Thus, $\frac{m v^2}{r} = 3mg \Rightarrow v^2 = 3rg$.Given $r = \frac{5}{3} \,m$ and $g = 10 \,ms^{-2}$, $v^2 = 3 	imes \frac{5}{3} 	imes 10 = 50$.Therefore, $v = \sqrt{50} \,ms^{-1}$.

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