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(A) The thermal strain produced due to the temperature change is given by $	ext{Strain} = \frac{\Delta l}{l} = \alpha \Delta 	heta$.Using Hooke's La

Vedclass · Accepted Answer

$4000$

Vedclass · Answer

(A) The thermal strain produced due to the temperature change is given by $	ext{Strain} = \frac{\Delta l}{l} = \alpha \Delta 	heta$.Using Hooke's Law,the thermal stress is $	ext{Stress} = Y 	imes 	ext{Strain} = Y \alpha \Delta 	heta$.The tension $T$ in the rod is given by $T = 	ext{Stress} 	imes 	ext{Area} = Y \alpha \Delta 	heta 	imes A$.Given the diameter $d = 1\,cm = 10^{-2}\,m$,the cross-sectional area $A = \frac{\pi d^2}{4} = \frac{\pi (10^{-2})^2}{4} = \frac{\pi 	imes 10^{-4}}{4}\,m^2$.The temperature change is $\Delta 	heta = 25\,^{\circ}C - 0\,^{\circ}C = 25\,^{\circ}C$.Substituting the values: $T = (2 	imes 10^{11}) 	imes (10^{-5}) 	imes (25) 	imes \left(\frac{\pi 	imes 10^{-4}}{4}ight)$.$T = \frac{2 	imes 10^{11} 	imes 10^{-5} 	imes 25 	imes \pi 	imes 10^{-4}}{4} = \frac{50 	imes \pi 	imes 10^2}{4} = 12.5 	imes 3.14159 	imes 100 \approx 3927\,N$.Rounding to the nearest option,$T \approx 4000\,N$.

$A$ steel rod of diameter $1\,cm$ is clamped firmly at each end when its temperature is $25\,^{\circ}C$ so that it cannot contract on cooling. The tension in the rod at $0\,^{\circ}C$ is approximately ......... $N$ $(\alpha = 10^{-5}/\,^{\circ}C, Y = 2 \times 10^{11}\,N/m^2)$

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