A 1 litre glass flask contains some mercury. | vedclass

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(C) The volume of air in the flask remains constant at different temperatures. This implies that the increase in the volume of the glass flask must be

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$150$

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(C) The volume of air in the flask remains constant at different temperatures. This implies that the increase in the volume of the glass flask must be exactly equal to the increase in the volume of the mercury contained within it.Let $V_g$ be the total volume of the glass flask $(1000 \, cc)$ and $V_L$ be the volume of mercury.The condition for constant air volume is $\Delta V_g = \Delta V_L$.Using the formula for volume expansion,$\Delta V = V \gamma \Delta 	heta$,we have:$V_g \gamma_g \Delta 	heta = V_L \gamma_L \Delta 	heta$Given that the coefficient of linear expansion of glass is $\alpha_g = 9 	imes 10^{-6} /^{\circ}C$,the coefficient of volume expansion of glass is $\gamma_g = 3 \alpha_g = 3 	imes 9 	imes 10^{-6} = 27 	imes 10^{-6} /^{\circ}C$.The coefficient of volume expansion of mercury is $\gamma_L = 1.8 	imes 10^{-4} /^{\circ}C$.Substituting the values:$V_L = \frac{V_g \gamma_g}{\gamma_L} = \frac{1000 	imes 27 	imes 10^{-6}}{1.8 	imes 10^{-4}}$$V_L = \frac{27 	imes 10^{-3}}{1.8 	imes 10^{-4}} = \frac{270}{1.8} = 150 \, cc$.

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