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. [2023]

Question

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 $0.075 {Nm}^{- 1}$ , atmospheric pressure is $10^{5} {Nm}^{- 2}$ , acceleration due to gravity $(g)$ is $10 {ms}^{- 2}$ , density of the liquid is $10^{3} {kg m}^{- 3}$ and contact angle of capillary surface with the liquid is zero] [2023]

Smart Achievers · Accepted Answer

(25)

$h_{0} = \frac{2 T \cos θ}{ρ g r} = \frac{2 T \cos 0 °}{ρ g r}$

$= \frac{2 \times 0.075 \times 1}{10^{3} \times 10 \times 10^{- 4}} = 15 cm$

$P_{0} V_{0} = P (\frac{100 V_{0}}{101}) \Rightarrow P = \frac{101}{100} P_{0}$

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

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

$∴$ $h = 15 cm + \frac{10^{5} \times 100}{100 \times 10^{3} \times 10} = 25 cm$

Solution

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Solution

Ans. (25) h0=2Tcosθρgr=2Tcos0°ρgr =2×0.075×1103×10×10-4=15 cm P0V0=P(100V0101)⇒P=101100P0 P0-2Tcosθr+ρgh=P=101100P0 ⇒-ρgh0+ρgh=P0100⇒h=h0+P0100ρg ∴ h=15 cm+105×100100×103×10=25 cm

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