A steel wire of length L at 40^{circ} C is su | vedclass

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(A) The extension produced by the mass $m$ is $\Delta L = \frac{FL}{AY} = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta L = L \alpha \Delt

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(A) The extension produced by the mass $m$ is $\Delta L = \frac{FL}{AY} = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta L = L \alpha \Delta T$.Since the wire regains its original length,the extension due to the mass must equal the contraction due to cooling:$\frac{mgL}{AY} = L \alpha \Delta T \Rightarrow mg = AY \alpha \Delta T$.Given $r = 1 \ mm = 10^{-3} \ m$,so $A = \pi r^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6} \ m^2$.Given $\Delta T = 40^{\circ} C - 30^{\circ} C = 10^{\circ} C$,$\alpha = 10^{-5} /^{\circ} C$,$Y = 10^{11} \ N/m^2$,and $g \approx 10 \ m/s^2$.Substituting the values:$m = \frac{A Y \alpha \Delta T}{g} = \frac{(\pi 	imes 10^{-6}) 	imes 10^{11} 	imes 10^{-5} 	imes 10}{10} = \pi \approx 3.14 \ kg$.Rounding to the nearest integer,$m \approx 3 \ kg$.

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