Question

A film of water is formed between two straight parallel wires each $10\, cm$ long and at separation $0.5\, cm$. The work required to increase $1 mm$ distance between the wires is $p \times 10^{-5}\, J$. Find $p$.
(Surface tension $=72 \times 10^{-3} N / m$ )

Answer 1

The process is shown in the figure. As we have to produce 2 films, so $\Delta W=2(\Delta A) T$ $=2[10 \,cm \times 0.6\, cm -10\, cm \times 0.5\, cm ] \times 72 \times 10^{-3} \,N / m$ $=2 \times\left(10 \times 10^{-2} m \right)\left(0.1 \times 10^{-2} m \right) \times 72 \times 10^{-3} N / m$ $=1.44 \times 10^{-5} J$
$=2[10\, cm \times 0.6 \,cm -10\, cm \times 0.5 \,cm ] \times 72 \times 10^{-3} N / m$
$=2 \times\left(10 \times 10^{-2} m \right)\left(0.1 \times 10^{-2} m \right) \times 72 \times 10^{-3} N / m $
$=1.44 \times 10^{-5}\, J$