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A glass flask of volume one litre is filled completely with mercury at $0^{\circ} \mathrm{C}$. The flask is now heated to $100^{\circ} \mathrm{C}$. Coefficient of volume expansion of mercury is $1.82 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ and coefficient of linear expansion of glass is $0.1 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. During this process, amount of mercury which overflows is
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Verified Answer
The correct answer is:
$15.2 \mathrm{cc}$
Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by
$$
\begin{aligned}
\Delta V & =V_0\left[\gamma_L-\gamma_g\right] \Delta \theta \\
& =V\left[\gamma_m-3 \alpha_g\right] \Delta \theta
\end{aligned}
$$
Given, $\gamma_m=182 \times 10^{-6} /{ }^{\circ} \mathrm{C}, \alpha_g=10 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
$$
\begin{array}{ll}
\Delta \theta=100^{\circ} \mathrm{C}, \quad V=1 \mathrm{~L} \\
\therefore \quad & \Delta V=1\left[\left(182 \times 10^{-6}-3 \times 10 \times 10^{-6}\right)\right] \times 100 \\
& \Delta V=15.2 \mathrm{CC}
\end{array}
$$
$$
\begin{aligned}
\Delta V & =V_0\left[\gamma_L-\gamma_g\right] \Delta \theta \\
& =V\left[\gamma_m-3 \alpha_g\right] \Delta \theta
\end{aligned}
$$
Given, $\gamma_m=182 \times 10^{-6} /{ }^{\circ} \mathrm{C}, \alpha_g=10 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
$$
\begin{array}{ll}
\Delta \theta=100^{\circ} \mathrm{C}, \quad V=1 \mathrm{~L} \\
\therefore \quad & \Delta V=1\left[\left(182 \times 10^{-6}-3 \times 10 \times 10^{-6}\right)\right] \times 100 \\
& \Delta V=15.2 \mathrm{CC}
\end{array}
$$
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