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Dimensional Analysis

In a vertical...

Question

In a vertical U-tube containing a liquid, the two arms are maintained at different temperature $t_{1}$ 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

\frac{l_{1} - l_{2}}{l_{2} t_{1} + l_{1} t_{2}}

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\frac{l_{1} - l_{2}}{l_{1} t_{1} - l_{2} t_{2}}

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\frac{l_{1} + l_{2}}{l_{2} t_{1} + l_{1} t_{2}}

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\frac{l_{1} + l_{2}}{l_{1} t_{1} + l_{2} t_{2}}

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Solution

The correct option is B $\frac{l_{1} - l_{2}}{l_{2} t_{1} + l_{1} t_{2}}$
R.E.F image
Volume Expansion
Suppose, height of liquid in each
arm before rising the temperature
with temperature rise height of
liquid in each arm increase
i,e. $l_{1} > 1$ and $l_{2} > 1$
also
$l = \frac{l_{1}}{1 + γ t_{1}} = \frac{l_{2}}{1 + γ t_{2}} | p$
$l_{1} + γ l_{1} t_{2} = l_{2} + γ l_{2} t_{1} P$
$γ = \frac{l_{1} - l_{2}}{l_{2} t_{1} - l_{1} t_{2}} \to$ Required answer

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