JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Fluid Mechanics

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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}

\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 Nm^{- 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]

Answer: 25

Solution

Initial capillary rise:

h_{0} = \frac{2 T c o s θ}{ρ g r} = \frac{2 \times 0.075 \times 1}{1 0 ^{3} \times 10 \times 1 0 ^{- 4}} = 15

cm.

For isothermal compression:

P_{0} V_{0} = P \cdot \frac{100 V _{0}}{101}

, so

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

.

Pressure balance at the liquid surface in capillary:

P_{0} - \frac{2 T c o s θ}{r} + ρ g h = P

P_{0} - ρ g h_{0} + ρ g h = \frac{101}{100} P_{0}

ρ g (h - h_{0}) = \frac{P _{0}}{100}

h = h_{0} + \frac{P _{0}}{100 ρ g} = 15 + \frac{1 0 ^{5}}{100 \times 1 0 ^{3} \times 10} = 15 + 10 = 25

cm.

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Fluid Mechanics

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