A liquid of coefficient of real expansion $γ$ is partly filled in a vessel of coefficient of linear expansion $γ / 3 .$ When the system is heated, then
a) the volume of space above liquid remains same
b) the level of liquid relative to vessel remains same
c) the fraction of volume of liquid in vessel remains same

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only (b) is correct

only (b) & (c) are correct

only (c) is true

all are true

answer is C.

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Detailed Solution

Let $A_{s}$ be the initial area of horizontal cross-section of vessel, $l_{s}$ its initial height. Let $l_{l}$ be the initial height of liquid. Let $V_{l}$ , $V_{s}$ be initial volumes of liquid and vessel respectively.

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Now $α_{v e s} = \frac{γ}{3} \Rightarrow γ_{v e s} = 3 α_{v e s} = γ$

Initial volume of vessel, $V_{s} = A_{s} l_{s}$

Initial volume of liquid, $V_{l} = A_{s} l_{l}$

Due to heating,

Change in area of vessel, $∆ A_{s} = A_{s} β ∆ θ$

Change in height of vessel, $∆ l_{s} = l_{s} α ∆ θ$

Change in volume of vessel, $∆ V_{s} = V_{s} γ ∆ θ$

Change in volume of liquid, $∆ V_{l} = V_{l} γ ∆ θ$

Since, $V_{s} \neq V_{l}$ , therefore, $∆ V_{s} \neq ∆ V_{l}$ . It means the volume of space above liquid (volume of solid minus volume of liquid) does not remain same.

New length of vessel, $l_{s}^{'} = l_{s} + ∆ l_{s} = l_{s} (1 + α ∆ θ)$

New area of vessel, $A_{s}^{'} = A_{s} + ∆ A_{s} = A_{s} (1 + β ∆ θ)$ ,

New volume of vessel, $V_{s}^{'} = V_{s} + ∆ V_{s} = V_{s} (1 + γ ∆ θ)$

New volume of liquid, $V_{l}^{'} = V_{l} + ∆ V_{l} = V_{l} (1 + γ ∆ θ)$

New height of liquid, $l_{l}^{'} = \frac{V_{l}^{'}}{A_{s}^{'}} = \frac{V_{l} (1 + γ ∆ θ)}{A_{s} (1 + β ∆ θ)} = l_{l} \frac{(1 + γ ∆ θ)}{(1 + β ∆ θ)}$

Now,

Initial level of liquid relative to vessel $= l_{s} - l_{l}$

Final level of liquid relative to vessel $= l_{s}^{'} - l_{l}^{'} = l_{s} (1 + α ∆ θ) - l_{l} (\frac{1 + γ ∆ θ}{1 + β ∆ θ})$

$= l_{s} (1 + α ∆ θ) - l_{l} (\frac{1 + 3 α ∆ θ}{1 + 2 α ∆ θ}) = l_{s} (1 + α ∆ θ) - l_{l} (1 + α ∆ θ)$

$= (l_{s} - l_{l}) (1 + α ∆ θ)$

So final level of liquid relative to vessel is not same as before.

Initial fraction of volume of liquid in vessel, $f = \frac{V_{l}}{V_{s}}$

Final fraction of volume of liquid in vessel, $f^{'} = \frac{V_{l}^{'}}{V_{s}^{'}} = \frac{V_{l} (1 + γ ∆ θ)}{V_{s} (1 + γ ∆ θ)} = \frac{V_{l}}{V_{s}} = f$