A glass flask of volume 1 , L at 0^{circ}C is | vedclass

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(B) The volume of mercury that spills out is equal to the difference in the increase in volume of the mercury and the increase in the internal volume

Vedclass · Accepted Answer

$15.2$

Vedclass · Answer

(B) The volume of mercury that spills out is equal to the difference in the increase in volume of the mercury and the increase in the internal volume of the glass flask.Given:Initial volume $V_0 = 1000 \, cc$ (since $1 \, L = 1000 \, cc$).Change in temperature $\Delta 	heta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.Coefficient of volume expansion of mercury $\gamma_L = 1.82 	imes 10^{-4} \, ^{\circ}C^{-1}$.Coefficient of linear expansion of glass $\alpha_g = 0.1 	imes 10^{-4} \, ^{\circ}C^{-1}$.The coefficient of volume expansion of glass is $\gamma_g = 3\alpha_g = 3 	imes 0.1 	imes 10^{-4} = 0.3 	imes 10^{-4} \, ^{\circ}C^{-1}$.The volume of mercury that spills out is given by the formula:$\Delta V = V_0 (\gamma_L - \gamma_g) \Delta 	heta$Substituting the values:$\Delta V = 1000 	imes (1.82 	imes 10^{-4} - 0.3 	imes 10^{-4}) 	imes 100$$\Delta V = 1000 	imes (1.52 	imes 10^{-4}) 	imes 100$$\Delta V = 15.2 \, cc$.Thus,$15.2 \, cc$ of mercury will spill out.

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