Question

A container of width $2a$ is filled with a liquid. A thin wire of mass per unit length $\mu$ is gently placed along the middle of the surface as shown in the figure. As a result, the liquid surface is depressed by a distance $y$. The surface tension of the liquid is $\sigma$. Then choose the correct option(s).
[Assume angle of contact between the wire and liquid is acute]

• A. $\sigma =\frac{\mu g}{2\cos \theta }$
• B. $\sigma =\frac{\mu g}{2\sin \theta }$
• C. If $y<<a$, then $\sigma =\frac{\mu ga}{2y}$
• D. If $y<<a$, then $\sigma =\frac{\mu ga}{y}$

Answer 1

Answer: (B), (C)

If $y<<a$, then $\sigma =\frac{\mu ga}{2y}$

Consider the FBD of the wire. Surface tension $T$ acts along the liquid surface as shown.


Let $l$ be the length of the wire.
Then, weight of the wire $W=\mu lg$
Force due to surface tension $f=\sigma l$.
Horizontal components $f\cos \theta$ cancel each other under equilibrium,

In vertical direction, $2f\sin \theta =W$
$\Rightarrow 2\sigma l\sin \theta =\mu lg$
$\Rightarrow \sigma =\frac{\mu g}{2\sin \theta }$

Also, $\sin \theta =\frac{y}{\sqrt{a^2+y^2}}$
If $y<<a\Rightarrow \sqrt{a^2+y^2}\approx a$ [when $\theta$ is very small ]
Then, $\sin \theta =\frac{y}{a}$
$\Rightarrow \sigma =\frac{\mu ga}{2y}$