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 :-

\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 A $\frac{ℓ_{1} - ℓ_{2}}{ℓ_{2} t_{1} - ℓ_{1} t_{2}}$
Suppose, height of liquid in each arm before rising the temperature is l.
ref. image
With the temperature rise height of liquid in each arm increases.
$l 1 > l$
$l 2 > l$
$l = (l 1 / 1 + £ t 1) = (l 2 + 1 + £ t 2)$
Simplyfying
We have
$£ = (l 1 - l 2) / (l 2 t 1 - l 1 t 2)$

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