Skip to content
dronacharya.ai
← JEE Advanced 2023 · Paper 2
PhysicsInteger+4Question 28 of 51

JEE Advanced 2023 · Paper 2, Physics question 11

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 V0V_0 to 100101V0\dfrac{100}{101}V_0 with the movable piston. Considering air as an ideal gas, the height (h)(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−10.075\ \mathrm{N\,m^{-1}}, atmospheric pressure is 105 N m−210^5\ \mathrm{N\,m^{-2}}, acceleration due to gravity (g)(g) is 10 m s−210\ \mathrm{m\,s^{-2}}, density of the liquid is 103 kg m−310^3\ \mathrm{kg\,m^{-3}} and contact angle of capillary surface with the liquid is zero]

A container holds liquid with trapped air above it, sealed by a weightless movable piston (pushed down, arrows). A capillary tube passes through an airtight hole in the piston into the liquid; its top is open to the outside air above the piston. Liquid in the capillary stands a height h above the liquid level in the container. g acts downward.
Answer and worked solution

Official answer: 25

Pressure of the trapped air. Boyle's law gives
P=101100×105=1.01×105 Pa.P=\frac{101}{100}\times10^5=1.01\times10^5\ \text{Pa}.
The capillary is open to the atmosphere at the top, so the liquid below sees an excess pressure of 10001000 Pa.

Capillary rise. Both this excess pressure and surface tension hold the column up:
ρgh=(P−P0)+2Tr=1000+2×0.07510−4=1000+1500=2500 Pa.\rho gh=(P-P_0)+\frac{2T}{r}=1000+\frac{2\times0.075}{10^{-4}}=1000+1500=2500\ \text{Pa}.

h=2500103×10=0.25 m=25 cm.h=\frac{2500}{10^3\times10}=0.25\ \text{m}=25\ \text{cm}.

Topics:CapillaritySurface tensionIdeal gas equation
← Previous questionNext question →