A solid whose volume does not change with temperature floats in liquid at two different temperatures $t_{1}^{0} C$ and $t_{2}^{0} C$ of the liquid. Fractions of the volume of the solid remain submerged are $f_{1}$ and $f_{2}$ respectively. The coefficient of absolute expansion of the liquid is equal to:

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

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

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

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\frac{f_{1} \times f_{2}}{f_{2} t_{1} + f_{1} t_{2}}

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Solution

The correct option is B $\frac{f_{1} - f_{2}}{f_{1} t_{1} - f_{2} t_{2}}$
Let the density of liquid at $0^{\circ} C$ be $ρ_{0}$ .
$⟹ ρ_{t_{1}^{\circ} C} = \frac{ρ_{0}}{1 + γ t_{1}}$
Also $ρ_{t_{2}^{\circ} C} = \frac{ρ_{0}}{1 + γ t_{2}}$
The buoyant force at the both temperatures would be same and equal to the weight of the solid.
$⟹ f_{1} V ρ_{t_{1}^{\circ} C} g = f_{2} V ρ_{t_{2}^{\circ} C} g$
$⟹ \frac{f_{1}}{1 + γ t_{1}} = \frac{f_{2}}{1 + γ t_{2}}$
$⟹ γ = \frac{f_{1} - f_{2}}{f_{1} t_{1} - f_{2} t_{2}}$

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