Question

A ring of radius $1m$ is lying on the surface of liquid. It is lifted from the liquid surface by a force of $4N$ in such a way that the liquid film in it remains intact. What is the surface tension of liquid?

Answer 1

Step 1. Given data:

The radius of the ring = $1m$

Lifting force from the liquid surface = $4N$

Step 2. Formula used:

Mathematically surface tension can be expressed as,

$T=\raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$L$}\right.$

Where,

$F$ is the force per unit length

$L$ is the length in which force act

$T$ is the surface tension of the liquid

Step 3. Calculations:

We assume the inside and outside perimeter same because the thickness of the ring is negligible and the liquid film in it remains intact or the surface tension in the ring is constant throughout.

Perimeter of the inside surface = $2\mathrm{\pi r}$

Perimeter of the outside surface = $2\mathrm{\pi r}$

Total length = $2\mathrm{\pi r}+2\mathrm{\pi r}=4\mathrm{\pi r}$,

As $r=1m,$Total length = $4\mathrm{\pi }$

Surface tension $\left(T\right)=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$4\mathrm{\pi }$}\right.=\frac{1}{\mathrm{\pi }}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m$}\right..$

Thus, the surface tension of liquid will be $\frac{1}{\mathrm{\pi }}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m$}\right..$