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JEE Advanced 2023 · Thermodynamics

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 mm0.1\,\text{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\frac{100}{101}V_0 with the movable piston. Considering air as an ideal gas, the height (hh) 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\,\text{N}\,\text{m}^{-1}, atmospheric pressure is 105 N m−210^{5}\,\text{N}\,\text{m}^{-2}, acceleration due to gravity (g) is 10 m s−210\,\text{m}\,\text{s}^{-2}, density of the liquid is 103 kg m−310^{3}\,\text{kg}\,\text{m}^{-3} and contact angle of capillary surface with the liquid is zero]

Answer: 2525

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Written solution

Step 1: Identify the given parameters.

Convert the given physical quantities into SI units.

r=10−4 mr = 10^{-4}\,\text{m} T=0.075 N m−1T = 0.075\,\text{N}\,\text{m}^{-1} Patm=105 N m−2P_{\text{atm}} = 10^{5}\,\text{N}\,\text{m}^{-2} ρ=103 kg m−3\rho = 10^{3}\,\text{kg}\,\text{m}^{-3} g=10 m s−2g = 10\,\text{m}\,\text{s}^{-2}

Step 2: Determine the pressure of the compressed air.

Apply Boyle's Law for the isothermal compression of the air inside the container.

PV=PatmV0P V = P_{\text{atm}} V_0 P=Patm(V0100101V0)P = P_{\text{atm}} \left(\frac{V_0}{\frac{100}{101}V_0}\right) P=101100PatmP = \frac{101}{100} P_{\text{atm}}

Step 3: Calculate the excess pressure inside the container.

Find the pressure difference relative to the atmosphere.

ΔP=P−Patm\Delta P = P - P_{\text{atm}} ΔP=1100Patm\Delta P = \frac{1}{100} P_{\text{atm}} ΔP=103 Pa\Delta P = 10^{3}\,\text{Pa}

Step 4: Establish the pressure balance at the meniscus.

Equate the pressure at the liquid level inside the container to the pressure at the same level inside the capillary.

P=Patm−2Tr+ρghP = P_{\text{atm}} - \frac{2T}{r} + \rho g h ρgh=(P−Patm)+2Tr\rho g h = \left(P - P_{\text{atm}}\right) + \frac{2T}{r} ρgh=ΔP+2Tr\rho g h = \Delta P + \frac{2T}{r}

Step 5: Calculate the surface tension pressure term.

2Tr=2×0.07510−4\frac{2T}{r} = \frac{2 \times 0.075}{10^{-4}} 2Tr=1.5×103 Pa\frac{2T}{r} = 1.5 \times 10^{3}\,\text{Pa}

Step 6: Solve for the height of the liquid column.

Substitute the calculated values into the pressure balance equation.

103×10×h=103+1.5×10310^{3} \times 10 \times h = 10^{3} + 1.5 \times 10^{3} 104h=2.5×10310^{4} h = 2.5 \times 10^{3} h=0.25 mh = 0.25\,\text{m} h=25 cmh = 25\,\text{cm}

Answer: 25

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