A metal rod of cross-sectional area 10^{-4} , | vedclass

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(A) The length of the rod remains unchanged. This means the contraction due to cooling is equal to the elongation due to the hanging weight.Thermal st

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$15$

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(A) The length of the rod remains unchanged. This means the contraction due to cooling is equal to the elongation due to the hanging weight.Thermal strain = Strain caused by the weight$\alpha \Delta 	heta = \frac{\Delta l}{l}$Since Young's modulus $Y = \frac{F/A}{\Delta l/l}$,we have $\frac{\Delta l}{l} = \frac{F}{YA}$.Substituting this into the thermal strain equation:$\alpha \Delta 	heta = \frac{F}{YA}$$\Delta 	heta = \frac{F}{YA \alpha}$Given $F = 5000 \, N$,$Y = 4 	imes 10^{12} \, N/m^{2}$,$A = 10^{-4} \, m^{2}$,and $\alpha = 2.5 	imes 10^{-6} \, K^{-1}$:$\Delta 	heta = \frac{5000}{4 	imes 10^{12} 	imes 10^{-4} 	imes 2.5 	imes 10^{-6}}$$\Delta 	heta = \frac{5000}{1000} = 5^{\circ} C$Since $\Delta 	heta = 20^{\circ} C - T = 5^{\circ} C$,we get $T = 15^{\circ} C$.

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