A bakelite beaker has a volume capacity of 50 | vedclass

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(A) Let $V_{0}$ be the total capacity of the beaker and $V_{m}$ be the volume of mercury at $30^{\circ} C$. The unfilled volume is $\Delta V = V_{0} -

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let $V_{0}$ be the total capacity of the beaker and $V_{m}$ be the volume of mercury at $30^{\circ} C$. The unfilled volume is $\Delta V = V_{0} - V_{m}$.When the temperature increases by $\Delta T$,the new capacity of the beaker is $V_{0}' = V_{0}(1 + \gamma_{b} \Delta T)$ and the new volume of mercury is $V_{m}' = V_{m}(1 + \gamma_{m} \Delta T)$.The unfilled volume remains constant,so $\Delta V' = \Delta V$,which implies $V_{0}' - V_{m}' = V_{0} - V_{m}$.Substituting the expressions for $V_{0}'$ and $V_{m}'$:$V_{0}(1 + \gamma_{b} \Delta T) - V_{m}(1 + \gamma_{m} \Delta T) = V_{0} - V_{m}$$V_{0} + V_{0} \gamma_{b} \Delta T - V_{m} - V_{m} \gamma_{m} \Delta T = V_{0} - V_{m}$$V_{0} \gamma_{b} \Delta T = V_{m} \gamma_{m} \Delta T$$V_{m} = \frac{V_{0} \gamma_{b}}{\gamma_{m}}$Given $V_{0} = 500\, cc$,$\gamma_{b} = 6 	imes 10^{-6}{ }^{\circ} C^{-1}$,and $\gamma_{m} = 1.5 	imes 10^{-4}{ }^{\circ} C^{-1}$:$V_{m} = \frac{500 	imes 6 	imes 10^{-6}}{1.5 	imes 10^{-4}} = \frac{3000 	imes 10^{-6}}{1.5 	imes 10^{-4}} = \frac{3 	imes 10^{-3}}{1.5 	imes 10^{-4}} = 2 	imes 10^{1} = 20\, cc$.

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