A steel wire of length 20 text{ cm} and area | vedclass

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(B) The thermal stress developed in a wire fixed at both ends when its temperature changes is given by $\sigma = Y \alpha \Delta t$. Since stress $\si

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$44$

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(B) The thermal stress developed in a wire fixed at both ends when its temperature changes is given by $\sigma = Y \alpha \Delta t$. Since stress $\sigma = \frac{T}{A}$, the change in tension $T$ is given by $T = Y A \alpha \Delta t$. Given: Young's modulus $Y = 2.0 	imes 10^{11} 	ext{ N/m}^2$. Area $A = 1 	ext{ mm}^2 = 10^{-6} 	ext{ m}^2$. Coefficient of linear expansion $\alpha = 1.1 	imes 10^{-5} {}^{\circ} 	ext{C}^{-1}$. Change in temperature $\Delta t = 40^{\circ} 	ext{C} - 20^{\circ} 	ext{C} = 20^{\circ} 	ext{C}$. Substituting these values: $T = (2.0 	imes 10^{11}) 	imes (10^{-6}) 	imes (1.1 	imes 10^{-5}) 	imes (20)$. $T = 2.0 	imes 10^{11} 	imes 10^{-6} 	imes 1.1 	imes 10^{-5} 	imes 20$. $T = 2.0 	imes 1.1 	imes 20 	imes 10^{11-6-5}$. $T = 44 	imes 10^0 = 44 	ext{ N}$. Thus, the change in tension is $44 	ext{ N}$.

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