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Expansion of Liquids

In a vertical...

Question

In a vertical U-tube containing a liquid, the two arms are maintained at different temperature $t_{1} & t_{2}$ . The liquid columns in the two arms have length $l_{1} & l_{2}$ respectively. The coefficient of volume expansion of the liquid is equal to

A

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

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B

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

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C

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

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D

\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 $l .$
With rise in temperature,
$l_{1} = l (1 + γ t_{1}) \Rightarrow l = \frac{l_{1}}{1 + γ t_{1}}$
$l_{2} = l (1 + γ t_{2}) \Rightarrow l = \frac{l_{2}}{1 + γ t_{2}}$
Equating $l$ from above two equation, we get
$\frac{l_{1}}{1 + γ t_{1}} = \frac{l_{2}}{1 + γ t_{2}}$
$\Rightarrow l_{1} + γ l_{1} t_{2} = l_{2} + l_{2} t_{1} \Rightarrow γ = \frac{l_{1} - l_{2}}{l_{2} t_{1} - l_{1} t_{2}}$

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