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(A) Let the initial height of the liquid in each arm at a reference temperature be $l_0$. When the temperatures of the two arms are raised to $t_1$ an

Vedclass · Accepted Answer

$\frac{{{l_1} - {l_2}}}{{{l_2}{t_1} - {l_1}{t_2}}}$

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(A) Let the initial height of the liquid in each arm at a reference temperature be $l_0$. When the temperatures of the two arms are raised to $t_1$ and $t_2$ respectively,the new heights $l_1$ and $l_2$ are given by the thermal expansion formula $l = l_0(1 + \gamma \Delta t)$. Assuming the reference temperature is $0^{\circ}C$,we have: $l_1 = l_0(1 + \gamma t_1)$ and $l_2 = l_0(1 + \gamma t_2)$. From these,we can write $l_0 = \frac{l_1}{1 + \gamma t_1} = \frac{l_2}{1 + \gamma t_2}$. Equating the two expressions for $l_0$: $l_1(1 + \gamma t_2) = l_2(1 + \gamma t_1)$ $l_1 + l_1 \gamma t_2 = l_2 + l_2 \gamma t_1$ $l_1 - l_2 = \gamma (l_2 t_1 - l_1 t_2)$ Therefore,the coefficient of volume expansion $\gamma$ is: $\gamma = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}$.

In a vertical $U-$tube containing a liquid,the two arms are maintained at different temperatures ${t_1}$ and ${t_2}$. The liquid columns in the two arms have heights ${l_1}$ and ${l_2}$ respectively. The coefficient of volume expansion of the liquid is equal to

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