JEE Advanced 2023 — Paper 2 — PHY-III solution

JEE Advanced2023Paper 2PHY-III

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 $\frac{100}{101} V_{0}$ 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 cm is_______.

[Given: Surface tension of the liquid is $0.075$ N m $^{- 1}$ , atmospheric pressure is $1 0^{5}$ N m $^{- 2}$ , acceleration due to gravity ( $g$ ) is $10$ m s $^{- 2}$ , density of the liquid is $1 0^{3}$ kg m $^{- 3}$ and contact angle of capillary surface with the liquid is zero]

Your answer

Solution

Let $P_{0} = 1 0^{5}$ Pa be the atmospheric pressure (which acts on the column of air inside the capillary above the meniscus, since the capillary opens to the atmosphere above the piston).

After isothermal compression, the trapped air below the piston obeys $P_{0} V_{0} = P \cdot \frac{100 V _{0}}{101}$ , so the new air pressure trapped below the piston is

$P = \frac{101}{100} P_{0}$ .

Pressure balance at the meniscus. With contact angle zero, the meniscus is concave with radius of curvature $r$ (capillary radius). Pressure just below the meniscus inside the capillary is

$P_{liq}^{(in)} = P_{0} - \frac{2 T}{r}$ .

Equating pressures at the level of the outer liquid surface (continuous static fluid) gives

$P_{0} - \frac{2 T}{r} + ρ g h = P = \frac{101}{100} P_{0}$ .

Rearrange:

$ρ g h = \frac{P _{0}}{100} + \frac{2 T}{r}$ .

Compute each term. The standard capillary rise piece is $h_{0} = \frac{2 T}{ρ g r} = \frac{2 \times 0.075}{1 0 ^{3} \times 10 \times 1 0 ^{- 4}} = 0.15$ m = $15$ cm.

The compression-driven piece is $\frac{P _{0}}{100 ρ g} = \frac{1 0 ^{5}}{100 \times 1 0 ^{3} \times 10} = 0.10$ m = $10$ cm.

$∴ h = 15 + 10 = 25 cm$ .

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