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(D) The maximum tension $T_{max}$ occurs at the bottom of the vertical circle,and the minimum tension $T_{min}$ occurs at the top.The expressions for

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$10$

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(D) The maximum tension $T_{max}$ occurs at the bottom of the vertical circle,and the minimum tension $T_{min}$ occurs at the top.The expressions for tension are:$T_{max} = \frac{mv_B^2}{L} + mg$$T_{min} = \frac{mv_T^2}{L} - mg$Given the ratio $\frac{T_{max}}{T_{min}} = 4$,we have:$\frac{\frac{mv_B^2}{L} + mg}{\frac{mv_T^2}{L} - mg} = 4$Using the conservation of energy between the top and bottom points:$\frac{1}{2}mv_B^2 = \frac{1}{2}mv_T^2 + mg(2L)$$v_B^2 = v_T^2 + 4gL$Substituting $v_B^2$ into the tension ratio equation:$\frac{v_T^2 + 4gL + gL}{v_T^2 - gL} = 4$$\frac{v_T^2 + 5gL}{v_T^2 - gL} = 4$$v_T^2 + 5gL = 4v_T^2 - 4gL$$3v_T^2 = 9gL$$v_T^2 = 3gL$Substituting the values $g = 10\, m/s^2$ and $L = \frac{10}{3}\, m$:$v_T^2 = 3 	imes 10 	imes \frac{10}{3} = 100$$v_T = 10\, m/s$.

Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

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