Question

A uniform capillary tube of length $l$ and inner radius $r$ with its upper end sealed is submerged vertically into water. The outside pressure is $p_0$ and surface tension of water is $\gamma$. When a length $x$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of $x$ is

• A. $\frac{l}{\left(l+\frac{p_{o}r}{4\gamma}\right)}$
• B. $l\left(l-\frac{p_{o}r}{4\gamma}\right)$
• C. $l\left(l-\frac{p_{o}r}{2\gamma}\right)$
• D. $\frac{l}{\left(l+\frac{p_{o}r}{2\gamma }\right)}$

Answer 1

Answer: (D)

$\frac{l}{\left(l+\frac{p_{o}r}{2\gamma }\right)}$

Answer 2

For air inside capillary, $p_{0} \left(\ell A\right) = p' \left(\ell-x\right)A$ where p' is pressure in capillary after being submerged
$\therefore p' = \frac{p_{0}\ell}{\ell-x}$
Now since level of water inside capillary coincides with outside, $\therefore p' - p_{0} = \frac{2\gamma}{r}$
$\therefore \frac{p_{0}\ell}{\ell-x} - p_{0} = \frac{2\gamma}{r}\Rightarrow x = \frac{\ell}{\left(1+\frac{p_{0}r}{2\gamma}\right)}$