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A glass flask of volume $200 \,cm ^3$ is just filled with mercury at $20^{\circ} C$. The amount of mercury that will overflow when the temperature of the system is raised to $100^{\circ} C$ is ........ $cm ^3$ $\left(\gamma_{\text {glase }}=1.2 \times 10^{-5} / C ^{\circ}, \gamma_{\text {mercury }}=1.8 \times 10^{-4} / C^{\circ}\right)$
$2.15$
$2.69$
$2.52$
$2.52$
Solution
(b)
$\gamma_{\text {glass }}=\frac{\Delta V }{ V }$
$=1.2 \times 10^{-5}$
$\Delta V _{\text {glass }}=\beta_{\text {glass }} V _0 \Delta T$
$=\left(1.2 \times 10^{-5}\right)(200)(100-20)$
$=0.19 \,cm ^3$
$\Delta V _{\text {mercury }}=\beta_{\text {mercury }} V _0 \Delta T$
$=\left(1.8 \times 10^{-5}\right)(200)(100-20)$
$=2.9\,cm ^3$
$\Delta V _{\text {mercury }}-\Delta V _{\text {glass }}=2.9-0.19=2.7$
