A circular copper ring at 30^{circ} C has a h | vedclass

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Question

(A) Let $A_c$ be the area of the copper ring and $A_s$ be the area of the steel rod at temperature $T_0 = 30^{\circ} C$.Given: $A_c = 9.98 \ cm^2$, $A

Vedclass · Accepted Answer

$167.6$

Vedclass · Answer

(A) Let $A_c$ be the area of the copper ring and $A_s$ be the area of the steel rod at temperature $T_0 = 30^{\circ} C$.Given: $A_c = 9.98 \ cm^2$, $A_s = 10 \ cm^2$.Coefficients of linear expansion: $\alpha_c = 17 	imes 10^{-6} /{ }^{\circ} C$, $\alpha_s = 11 	imes 10^{-6} /{ }^{\circ} C$.Coefficients of area expansion: $\beta_c = 2\alpha_c = 34 	imes 10^{-6} /{ }^{\circ} C$, $\beta_s = 2\alpha_s = 22 	imes 10^{-6} /{ }^{\circ} C$.For the ring to slip onto the rod, the area of the ring must be equal to or greater than the area of the rod at the new temperature $T = T_0 + \Delta T$.$A_c(1 + \beta_c \Delta T) = A_s(1 + \beta_s \Delta T)$$9.98(1 + 34 	imes 10^{-6} \Delta T) = 10(1 + 22 	imes 10^{-6} \Delta T)$$9.98 + 9.98 	imes 34 	imes 10^{-6} \Delta T = 10 + 10 	imes 22 	imes 10^{-6} \Delta T$$(339.32 - 220) 	imes 10^{-6} \Delta T = 10 - 9.98$$119.32 	imes 10^{-6} \Delta T = 0.02$$\Delta T = \frac{0.02}{119.32 	imes 10^{-6}} \approx 167.6^{\circ} C$.

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