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(B) Let $v_h$ be the speed at the highest point and $v_l$ be the speed at the lowest point.At the highest point,the tension $T_h$ is given by $T_h = \

Vedclass · Accepted Answer

$2 \sqrt{gL}$

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(B) Let $v_h$ be the speed at the highest point and $v_l$ be the speed at the lowest point.At the highest point,the tension $T_h$ is given by $T_h = \frac{mv_h^2}{L} - mg$.At the lowest point,the tension $T_l$ is given by $T_l = \frac{mv_l^2}{L} + mg$.Using the conservation of energy between the highest and lowest points: $\frac{1}{2}mv_l^2 = \frac{1}{2}mv_h^2 + mg(2L)$,which simplifies to $v_l^2 = v_h^2 + 4gL$.Substituting $v_l^2$ into the expression for $T_l$: $T_l = \frac{m(v_h^2 + 4gL)}{L} + mg = \frac{mv_h^2}{L} + 4mg + mg = \frac{mv_h^2}{L} + 5mg$.Given the ratio $\frac{T_l}{T_h} = 3$,we have $\frac{\frac{mv_h^2}{L} + 5mg}{\frac{mv_h^2}{L} - mg} = 3$.Let $x = \frac{v_h^2}{L}$. Then $\frac{x + 5g}{x - g} = 3 \implies x + 5g = 3x - 3g \implies 2x = 8g \implies x = 4g$.Since $x = \frac{v_h^2}{L}$,we have $\frac{v_h^2}{L} = 4g$,which gives $v_h^2 = 4gL$.Therefore,$v_h = 2\sqrt{gL}$.

List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

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