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$\text{Hint: } T_{\text{max}} = mg + \frac{mv_1^2}{l} \text{ and } T_{\text{min}} + mg = \frac{mv_2^2}{l}$ $\text{Step: Find the velocity of the bob at the highest position.}$ $\text{Let the speed of the bob at the lowest position be } v_1 \text{ and at the highest position be } v_2.$ $\text{The maximum tension is at the lowest position and minimum tension at the highest position.}$ $\text{Now, using the conservation of mechanical energy we get:}$ $\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg2l$ $\Rightarrow v_1^2 = v_2^2 + 4gl \quad \ldots (1)$ $\text{Now } T_{\text{max}} - mg = \frac{mv_1^2}{l}$ $\Rightarrow T_{\text{max}} = mg + \frac{mv_1^2}{l}$ $T_{\text{min}} + mg = \frac{mv_2^2}{l}$ $\Rightarrow T_{\text{min}} = \frac{mv_2^2}{l} - mg$ $\Rightarrow \frac{mv_2^2}{l} - mg = \frac{1}{5} \left[ mg + \frac{mv_1^2}{l} \right]$ $\Rightarrow mg + \frac{mv_1^2}{l} = \left[ \frac{mv_2^2}{l} - mg \right] 5$ $\Rightarrow mg + \frac{m}{l} [v_2^2 + 4gl] = \frac{5mv_2^2}{l} - 5mg$ $\Rightarrow mg + \frac{mv_2^2}{l} + 4mg = \frac{5mv_2^2}{l} - 5mg$ $\Rightarrow 10mg = \frac{4mv_2^2}{l}$ $\Rightarrow v_2^2 = \frac{10 \times 10 \times 1}{4}$ $\Rightarrow v_2^2 = 25$ $\Rightarrow v_2 = 5 \text{ m/s}$ $\text{Thus, the velocity of the bob at the highest position is } 5 \text{ m/s.}$ $\text{Hence, option (2) is the correct answer.}$