Surface Tension and Surface Energy – Rankers Physics
Topic: Solid and Fluids
Subtopic: Surface Tension and Viscosity

Surface Tension and Surface Energy

A rectangular film of liquid is extended from $(4 \text{ cm} \times 2 \text{ cm})$ to $(5 \text{ cm} \times 4 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$, the value of the surface tension of the liquid is:

(2016 - II)

$0.2 \text{ Nm}^{-1}$
$8.0 \text{ Nm}^{-1}$
$0.250 \text{ Nm}^{-1}$
$0.125 \text{ Nm}^{-1}$

Solution:

Work done in stretching a liquid film is $W = T \times 2\Delta A$ (since it has two surfaces). The change in area is $\Delta A = (5 \times 4) - (4 \times 2) = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$. Thus, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.

Leave a Reply

Your email address will not be published. Required fields are marked *