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(A) The decrease in length due to cooling is given by $\Delta l = l \alpha \Delta T$.Given $l = 1\;m$,$\alpha = 2 	imes 10^{-5}\;K^{-1}$,and $\Delta

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$60$

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(A) The decrease in length due to cooling is given by $\Delta l = l \alpha \Delta T$.Given $l = 1\;m$,$\alpha = 2 	imes 10^{-5}\;K^{-1}$,and $\Delta T = (210 - 160) = 50\;K$.Thus,$\Delta l = 1 	imes 2 	imes 10^{-5} 	imes 50 = 10^{-3}\;m$.To restore the original length,the mass $M$ must produce an extension equal to $\Delta l$.Using Young's modulus formula: $Y = \frac{F/A}{\Delta l/l}$,where $F = Mg$.Substituting the values: $2 	imes 10^{11} = \frac{Mg / (3 	imes 10^{-6})}{10^{-3} / 1}$.$Mg = 2 	imes 10^{11} 	imes 3 	imes 10^{-6} 	imes 10^{-3} = 6 	imes 10^{2} 	imes 10^{-3} = 0.6\;N$.Since $g = 10\;ms^{-2}$,$M = \frac{0.6}{10} = 0.06\;kg$. Wait,re-evaluating the calculation: $Mg = 2 	imes 10^{11} 	imes 3 	imes 10^{-9} = 600\;N$. $M = \frac{600}{10} = 60\;kg$.

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