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(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.$\Delta V = V_0 (\gamma_m -

Vedclass · Accepted Answer

$15.2$

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(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.$\Delta V = V_0 (\gamma_m - \gamma_g) \Delta 	heta$Given that $\gamma_g = 3 \alpha_g$, where $\alpha_g$ is the coefficient of linear expansion of glass.$\gamma_g = 3 	imes (0.1 	imes 10^{-4} /{ }^{\circ} C) = 0.3 	imes 10^{-4} /{ }^{\circ} C = 30 	imes 10^{-6} /{ }^{\circ} C$Given $\gamma_m = 1.82 	imes 10^{-4} /{ }^{\circ} C = 182 	imes 10^{-6} /{ }^{\circ} C$, $V_0 = 1 	ext{ litre} = 1000 	ext{ cc}$, and $\Delta 	heta = 100^{\circ} C$.$\Delta V = 1000 	imes (182 	imes 10^{-6} - 30 	imes 10^{-6}) 	imes 100$$\Delta V = 1000 	imes (152 	imes 10^{-6}) 	imes 100 = 15.2 	ext{ cc}$Thus, the amount of mercury that overflows is $15.2 	ext{ cc}$.

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