An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 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 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 cm is \ \ \ \ . [Given: Surface tension of the liquid is 0.075 N m -1 , atmospheric pressure is 10 5 N m -2 , acceleration due to gravity ( g ) is 10 m s -2 , density of the liquid is 10 3 kg m -3 and contact angle of capillary surface with the liquid is zero]

Physics · Mechanical Properties of Matter

JEE Advanced 2023 — Paper 2 — Question 24

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 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 $\mathrm{V}_{0}$ to $\frac{100}{101} \mathrm{~V}_{0}$ 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 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]

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{P}_{0} \mathrm{~V}_{0}=\mathrm{P}_{\mathrm{A}}\left(\frac{100}{101} \mathrm{~V}_{0}\right)$ $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{0}\left(\frac{101}{100}\right)$ $\because \mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{D}}$ $\Rightarrow \mathrm{P}_{\mathrm{D}}=\left(\frac{101}{100}\right) \mathrm{P}_{0}$ Also, $\mathrm{P}_{\mathrm{B}}-\mathrm{P}_{\mathrm{C}}=\frac{2 \mathrm{~T}}{\mathrm{r}}$ $\mathrm{P}_{\mathrm{C}}=\mathrm{P}_{\mathrm{B}}-\frac{2 \mathrm{~T}}{\mathrm{r}}=\mathrm{P}_{0}-\frac{2 \mathrm{~T}}{\mathrm{r}}$ (since, $\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{0}$ ) $\mathrm{P}_{\mathrm{D}}=\left(\frac{101}{100}\right) \mathrm{P}_{0}=\mathrm{P}_{\mathrm{C}}+\rho g h$ $\Rightarrow\left(\frac{101}{100}\right) \mathrm{P}_{0}=\left(\mathrm{P}_{0}-\frac{2 \mathrm{~T}}{\mathrm{r}}\right)+\mathrm{pgh}$ Solving we get, $\mathrm{h}=0.25 \mathrm{~m}=25 \mathrm{~cm}$ .

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy