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 $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ 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]
IIT 2023, Medium
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$h_0=\frac{2 T \cos \theta}{\rho g r}=\frac{2 \times 0.075 \times 1}{10^3 \times 10 \times 10^{-4}}=15 \mathrm{~cm}$
$P_0 V_0=P \frac{100 V_0}{101} \Rightarrow P=\frac{101}{100} \mathrm{P}_0$
$P_0-\frac{2 T \cos \theta}{\mathrm{r}}+\rho g \mathrm{gh}=\mathrm{P}=\frac{101}{100} \mathrm{P}_0 $
$\Rightarrow-\rho g \mathrm{~h}_0+\rho g \mathrm{gh}=\frac{\mathrm{P}_0}{100}$
$\Rightarrow \mathrm{h}=\mathrm{h}_0+\frac{\mathrm{P}_0}{100 \mathrm{\rho g}}$
$=15 \mathrm{~cm}+\frac{10^5}{100 \times 10^3 \times 10}=25 \mathrm{~cm}$
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