A Bakelite beaker has volume capacity of 500 cc at 30°C . When it is partially filled with V_m volume (at 30^oC ) of mercury, it is found that the unfilled volume of the beaker remains constant as the temperature is varied. If $γ_{(beaker)} = 6 \times {10^{- 6}}^{\circ} C^{- 1}$ and $γ_{(mercury)} = 1 .5 \times {10^{- 4}}^{\circ} C^{- 1}$ where $γ$ is the coefficient of volume expansion, then V_m (in cc) is close to

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Detailed Solution

Given: V = 500 cc, T = 30°C
$\begin{array}{l} γ_{beaker} = 6 \times 10^{- 6} /^{\circ} C \\ γ_{mercuy} = 1 .5 \times 10^{- 4} /^{\circ} C \\ Now, V^{'} = V (1 + γ_{b} ΔT) \Rightarrow ∆ V = {Vγ}_{b} ΔT \\ and V_{m}^{'} = V_{m} (1 + γ_{m} ΔT) \Rightarrow ∆ V_{m} = {Vγ}_{m} ΔT \end{array}$

Since the volume of air not changing the change in volume of mercury and change in volume of Bakelite beaker are same.

${∆V=∆V}_{m} ⇒ V_{m} = V \frac{γ_{b}}{γ_{m}} = \frac{500 \times 6 \times 10^{- 6}}{1 .5 \times 10^{- 4}} = \frac{5 \times 6}{1 .5} = 20 cc$

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