The ratio of linear expansivity to the coeffi | vedclass

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(B) Let the length of the sides of a square sheet be $L$. The area of the sheet is $A = L^2$.When the temperature increases by $\Delta T$, the new len

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$0.5$

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(B) Let the length of the sides of a square sheet be $L$. The area of the sheet is $A = L^2$.When the temperature increases by $\Delta T$, the new length becomes $L' = L(1 + \alpha \Delta T)$, where $\alpha$ is the coefficient of linear expansion.The new area becomes $A' = (L')^2 = L^2(1 + \alpha \Delta T)^2$.Expanding this, we get $A' = A(1 + 2\alpha \Delta T + \alpha^2 \Delta T^2)$.Since $\alpha$ is very small, $\alpha^2$ is negligible, so $A' \approx A(1 + 2\alpha \Delta T)$.The coefficient of superficial expansion $\beta$ is defined by the relation $A' = A(1 + \beta \Delta T)$.Comparing the two expressions, we find $\beta = 2\alpha$.Therefore, the ratio of linear expansivity $(\alpha)$ to the coefficient of superficial expansion $(\beta)$ is $\frac{\alpha}{\beta} = \frac{\alpha}{2\alpha} = 0.5$.

The ratio of linear expansivity to the coefficient of superficial expansion of a rectangular sheet of a solid is

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