An incompressible liquid is kept in a container having a weightless...

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius $0 . 1mm$ 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 \left(h\right) $of the liquid column in the capillary above the liquid level in$ cm $is _____.</p><p>[Given: Surface tension of the liquid is$ 0 . 075N{m}^{- 1} $, atmospheric pressure is$ {10}^{5}N{m}^{- 2} $, acceleration due to gravity \left(g\right)$ is $10m{s}^{- 2}$ , density of the liquid is ${10}^{3}kg{m}^{- 3}$ and contact angle of capillary surface with the liquid is zero]

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner

Correct answer

Step-by-step solution

Let ${P}_{f}$ be the air pressure.
As the process is isothermal, we can write
${P}_{0}{V}_{0} = {P}_{f}{V}_{f}$
$⇒ {P}_{0}{V}_{0} = {P}_{f}\left(\dfrac{100}{101}\right){V}_{0}$
$⇒ {P}_{f} = 101 \times {10}^{3}Pa\left(∵ {P}_{0} = {10}^{5}N{m}^{- 2}\right)$
Now, consider the $4$ points shown in diagram
${P}_{d} - {P}_{c} = \dfrac{2T}{R}\left(∵ {P}_{d} = {P}_{0}\right)$
$∴ {P}_{c} = {P}_{0} - \dfrac{2T}{R}$
Now,
${P}_{a} = {P}_{b}$ (also, ${P}_{a} = {P}_{f}$ )
$⇒ {P}_{f} = ρgh + {P}_{c}$
$⇒ 101 \times {10}^{3} = \left({10}^{3} \times 10 \times h\right) + \left({10}^{5} - \dfrac{2 \times 0 . 075}{0 . 1 \times {10}^{- 3}}\right)$
$⇒ h = \dfrac{1}{4}m = 25cm$

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