JEE Advanced 2023 Paper 2 · Q11 · Spring-Block

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 $\dfrac{100}{101} V_0$ with the movable piston. Considering air as an ideal gas, the height $h$ of the liquid column in the capillary above the liquid level is ___ cm. [Given: surface tension $T = 0.075$ N·m$^{-1}$, atmospheric pressure $P_a = 10^5$ N·m$^{-2}$, $g = 10$ m·s$^{-2}$, density $\rho = 10^3$ kg·m$^{-3}$, contact angle $0$]

SolveKar AI · Accepted Answer

Compressed pressure: $P_f = P_0\cdot V_0/(100V_0/101) = (101/100) P_0 = 1.01 P_0$. Pressure balance at the bottom of capillary: $P_f = P_0 + \rho g h - 2T/r$ (since the meniscus has surface-tension contribution lowering the pressure inside). Rearranging: $\rho g h = P_f - P_0 + 2T/r = 0.01 P_0 + 2(0.075)/(10^{-4}) = 1000 + 1500 = 2500$. So $h = 2500/(\rho g) = 2500/10^4 = 0.25$ m = $25$ cm.

JEE Advanced 2023 Paper 2 · Q11 · Spring-Block

Solution