Question

A circular film having radius $2\text{cm}$ is extended such that its radius becomes $4\text{cm}$. If the value of surface tension of the liquid is $0.25\text{N/m}$, then the work done in extending the circular film is

• A. $3\pi \times {10}^{-4}\text{J}$
• B. $4\pi \times {10}^{-4}\text{J}$
• C. $6\pi \times {10}^{-4}\text{J}$
• D. $8\pi \times {10}^{-4}\text{J}$

Answer 1

The correct option is C $6\pi \times {10}^{-4}\text{J}$

The circular film of liquid has two surfaces.
Initial area $A_i=\pi r_i^2=4\pi {\text{cm}}^2$
Final area $A_j=\pi r_j^2=16\pi {\text{cm}}^2$
$\therefore \Delta A=2(A_j-A_i)$
$=2(16-4)\pi {\text{cm}}^2$
$=24\pi {\text{cm}}^2$

We know,
$W=T.\Delta A$
$=0.25\times 24\pi \times {10}^{-4}\text{J}$
$=6\pi \times {10}^{-4}\text{J}$