A glass flask contains some mercury at room t | vedclass

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(C) Let $V_m$ be the volume of mercury and $V_f$ be the volume of the glass flask.Since the volume of air inside the flask remains constant at differe

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$2000$

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(C) Let $V_m$ be the volume of mercury and $V_f$ be the volume of the glass flask.Since the volume of air inside the flask remains constant at different temperatures,the change in the volume of mercury must be equal to the change in the volume of the flask.$\Delta V_m = \Delta V_f$$\gamma_m V_m \Delta T = \gamma_f V_f \Delta T$$\gamma_m V_m = \gamma_f V_f$We know that the coefficient of volume expansion of the flask $\gamma_f = 3 \alpha_{glass}$.Given $\gamma_m = 1.8 	imes 10^{-4} (^{\circ}C)^{-1}$,$\alpha_{glass} = 9 	imes 10^{-6} (^{\circ}C)^{-1}$,and $V_m = 300 \, cm^3$.Substituting the values:$1.8 	imes 10^{-4} 	imes 300 = 3 	imes (9 	imes 10^{-6}) 	imes V_f$$5.4 	imes 10^{-2} = 2.7 	imes 10^{-5} 	imes V_f$$V_f = \frac{5.4 	imes 10^{-2}}{2.7 	imes 10^{-5}} = 2 	imes 10^3 = 2000 \, cm^3$.

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