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. $\frac{l}{(l+\frac{p_{0}r}{4\gamma })}$
• B. $l(l-\frac{p_{0}r}{4\gamma })$
• C. $l(l-\frac{p_{0}r}{2\gamma })$
• D. $\frac{l}{(l+\frac{p_{0}r}{2\gamma })}$

Answer 1

The correct option is D $\frac{l}{(l+\frac{p_{0}r}{2\gamma })}$

For air inside capillary, $p_0\left(lA\right)=p^{\prime }(l-x)A$ where $p^{\prime }$ is pressure in capillary after being submerged
$\therefore p^{\prime }=\frac{p_{0}l}{l-x}$
Now since level of water inside capillary coincides with outside $p^{\prime }-p_0=\frac{2\gamma }{r}$
$\therefore \frac{p_{0}l}{l-x}-p_0=\frac{2\gamma }{r}\Rightarrow x=\frac{l}{(l+\frac{p_{0}r}{2\gamma })}$