Question

A water film is formed between two straight parallel wires of $10\, cm$ length $0.5\, cm$ apart. If the distance between wires is increased by $1 \,mm$. What will be the work done?
(surface tension of water $=72\, dyne / cm$ )

• A. $36\, erg$
• B. $288\, erg$
• C. $144 \,erg$
• D. $72\, erg$

Answer 1

Answer: (C)

Work done $=$ Surface tension $\times$ increase in area of the film
$W = S \times \Delta A$
Increase in area $=$ Final area $-$ initial area
$=10 \times(0.5+0.1)-10 \times 0.5=1\, cm ^2$
$\therefore W =72 \times 2 \times 1=144 \,erg$
[ $\because$ There are $2$ free surfaces; $\therefore \Delta A =2 \times 1]$.