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(B) Given: Mass $M = 2 \ kg$, length $l = 2 \ m$, area $A = 1 \ mm^2 = 1 	imes 10^{-6} \ m^2$, maximum elongation $\Delta l = 2 \ mm = 2 	imes 10^{-

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$1.2 	imes 10^{11} \ Nm^{-2}$

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(B) Given: Mass $M = 2 \ kg$, length $l = 2 \ m$, area $A = 1 \ mm^2 = 1 	imes 10^{-6} \ m^2$, maximum elongation $\Delta l = 2 \ mm = 2 	imes 10^{-3} \ m$, $g = 10 \ ms^{-2}$.For the tension to be zero at the highest point, the velocity at the bottom $v$ must be $\sqrt{5gl}$.$v = \sqrt{5 	imes 10 	imes 2} = \sqrt{100} = 10 \ ms^{-1}$.The maximum tension $T_{\max}$ occurs at the lowest point of the vertical circle:$T_{\max} = mg + \frac{Mv^2}{l} = (2 	imes 10) + \frac{2 	imes (10)^2}{2} = 20 + 100 = 120 \ N$.Using the formula for Young's modulus $Y = \frac{T_{\max} \cdot l}{A \cdot \Delta l}$:$Y = \frac{120 	imes 2}{1 	imes 10^{-6} 	imes 2 	imes 10^{-3}} = \frac{240}{2 	imes 10^{-9}} = 120 	imes 10^9 = 1.2 	imes 10^{11} \ Nm^{-2}$.

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