A unit scale is to be prepared whose length d | vedclass

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(C) Let $\ell_{B}$ be the length of brass and $\ell_{i}$ be the length of iron.The condition is that the difference between their lengths remains cons

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$60$

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(C) Let $\ell_{B}$ be the length of brass and $\ell_{i}$ be the length of iron.The condition is that the difference between their lengths remains constant with temperature change $\Delta T$:$\ell_{B}(1 + \alpha_{B} \Delta T) - \ell_{i}(1 + \alpha_{i} \Delta T) = \ell_{B} - \ell_{i}$Expanding the terms:$\ell_{B} + \ell_{B} \alpha_{B} \Delta T - \ell_{i} - \ell_{i} \alpha_{i} \Delta T = \ell_{B} - \ell_{i}$This simplifies to:$\ell_{B} \alpha_{B} \Delta T = \ell_{i} \alpha_{i} \Delta T$Dividing by $\Delta T$:$\ell_{B} \alpha_{B} = \ell_{i} \alpha_{i}$Given $\ell_{B} = 40\,cm$,$\alpha_{B} = 1.8 	imes 10^{-5} K^{-1}$,and $\alpha_{i} = 1.2 	imes 10^{-5} K^{-1}$:$40 	imes 1.8 	imes 10^{-5} = \ell_{i} 	imes 1.2 	imes 10^{-5}$Solving for $\ell_{i}$:$\ell_{i} = \frac{40 	imes 1.8}{1.2} = \frac{40 	imes 3}{2} = 60\,cm$.

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Each side of a cubic metal box is $a$ at room temperature $T$. The coefficient of linear expansion of the metal sheet is $\alpha$. The metal box is heated uniformly by a small temperature $\Delta T$,so that its new temperature is $T + \Delta T$. Calculate the increase in the volume of the metal box.

Assertion :- $A$ brass disc is fast fitted in a hole in a steel plate. The system must be cooled to loosen the disc from the hole.
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