Question

A very light, rectangular wire-frame of dimensions $7\,cm\times 5\,cm$ hangs just above the free surface of a liquid of surface tension $T$ , with its plane parallel to the free surface. The wire-frame is just brought in contact with the liquid surface and then, lifted up. If the force required to lift the wire-frame is $3.36\,N$ , then what is the value of $T$ (in $Nm^{- 1}$ )?

Answer 1

The downward force of surface tension acting on the wire frame is
$F_{\text{surface tension}}=T\times l$ ,
where $l$ is the total length of the wire which is in contact with the liquid.
Now since it is a wire frame, the free surface of the liquid will be in contact with the inner as well as the outer perimeter of the wire.
Hence, $l=2\times 2\times \left(7 + 5\right)=48\,cm$
$T\times \frac{48}{100}=3.36\,N$
$\Rightarrow T=7\,N\,m^{- 1}$