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QJEEAdv23Ph6688: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph6688 Solution: An incompressible liquid is kept in a container having a…

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 \mathrm{~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 (h)(h) of the liquid column in the capillary above the liquid level in cm\mathrm{cm} is _______.

[Given: Surface tension of the liquid is 0.075 N m−10.075 \mathrm{~N} \mathrm{~m}^{-1} , atmospheric pressure is 105 N m−210^5 \mathrm{~N} \mathrm{~m}^{-2} , acceleration due to gravity (g)(\mathrm{g}) is 10 m s−210 \mathrm{~m} \mathrm{~s}^{-2} , density of the liquid is 103 kg m−310^3 \mathrm{~kg} \mathrm{~m}^{-3} and contact angle of capillary surface with the liquid is zero] JEE Advanced 2023 Paper 2 Online Physics - Properties of Matter Question 3 English

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JEE Advanced 2023 Paper 2 Online Physics - Properties of Matter Question 3 English Explanation

amp;h0=2 Tcos⁡θρgr=2×0.075×1103×10×10−4=15 cmamp;P0 V0=P100 V0101⇒P=101100P0\begin{aligned}& \mathrm{h}_0=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 0.075 \times 1}{10^3 \times 10 \times 10^{-4}}=15 \mathrm{~cm} \\\\& \mathrm{P}_0 \mathrm{~V}_0=\mathrm{P} \frac{100 \mathrm{~V}_0}{101} \Rightarrow \mathrm{P}=\frac{101}{100} \mathrm{P}_0\end{aligned}

amp;P0−2 Tcos⁡θr+ρgh=P=101100P0amp;⇒−ρgh0+ρgh=P0100\begin{aligned}& \mathrm{P}_0-\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r}}+\rho \mathrm{gh}=\mathrm{P}=\frac{101}{100} \mathrm{P}_0 \\\\& \Rightarrow-\rho \mathrm{gh}_0+\rho \mathrm{gh}=\frac{\mathrm{P}_0}{100}\end{aligned}

amp;⇒h=h0+P0100ρgamp;=15 cm+105100×103×10=25 cm\begin{aligned}& \Rightarrow \mathrm{h}=\mathrm{h}_0+\frac{\mathrm{P}_0}{100 \rho \mathrm{g}} \\\\& =15 \mathrm{~cm}+\frac{10^5}{100 \times 10^3 \times 10}=25 \mathrm{~cm}\end{aligned}

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