A liquid is completely filled in a cubical metal container of side length 1m. Find the new level of liquid if temperature is decreased by $ΔT$ .
( $γ_{ℓ}, γ_{c}$ – coefficient of volume expansion of liquid and container respectively)

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$(1 - γ_{ℓ} ΔT)$

$\frac{(1 - γ_{ℓ} Δ T)}{(1 - γ_{c} Δ T)}$

$\frac{3 - 3 γ_{ℓ} ΔT}{3 - 2 γ_{c} ΔT}$

$\frac{3 - 2 γ_{ℓ} ΔT}{3 - 3 γ_{c} ΔT}$

answer is C.

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Detailed Solution

$\begin{array}{l} \underset{}{V_{f i n a l}} (liquid) = 1 (1 - γ_{l} Δ T) \\ a n d \\ A_{base} = 1 (1 - \frac{2}{3} γ_{c} Δ T) \\ \Rightarrow H_{final} = \frac{V_{fina}}{A_{base}} = (\frac{3 - 3 γ_{l} Δ T}{3 - 2 γ_{c} Δ T}) \end{array}$

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