Question 420: Properties of Matter

Question

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius $0.1 \mathrm{~mm}$ is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume $V_0$ to $\frac{100}{101} V_0$ with the movable piston. Considering air as an ideal gas, the height $(h)$ of the liquid column in the capillary above the liquid level in $\mathrm{cm}$ is _______.
[Given: Surface tension of the liquid is $0.075 \mathrm{~N} \mathrm{~m}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(\mathrm{g})$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]

Accepted Answer

$$
\begin{aligned}
& \mathrm{h}_0=\frac{2 \mathrm{~T} \cos 	heta}{ho \mathrm{gr}}=\frac{2 	imes 0.075 	imes 1}{10^3 	imes 10 	imes 10^{-4}}=15 \mathrm{~cm} \\
& \mathrm{P}_0 \mathrm{~V}_0=\mathrm{P} \frac{100 \mathrm{~V}_0}{101} \Rightarrow \mathrm{P}=\frac{101}{100} \mathrm{P}_0
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{P}_0-\frac{2 \mathrm{~T} \cos 	heta}{\mathrm{r}}+ho \mathrm{gh}=\mathrm{P}=\frac{101}{100} \mathrm{P}_0 \\
& \Rightarrow-ho \mathrm{gh}_0+ho \mathrm{gh}=\frac{\mathrm{P}_0}{100}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \mathrm{h}=\mathrm{h}_0+\frac{\mathrm{P}_0}{100 ho \mathrm{g}} \\
& =15 \mathrm{~cm}+\frac{10^5}{100 	imes 10^3 	imes 10}=25 \mathrm{~cm}
\end{aligned}
$$

Properties of Matter

Integer / Numeric Answer Type