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Work of $3.0\times10^{-4}$ joule is required to be done in increasing the size of a soap film from $10\, cm\times6\, cm$ to $10\, cm\times11\, cm$. The surface tension of the film is
$5\times10^{-2}\, N/m$
$3\times10^{-2}\, N/m$
$1.5\times10^{-2}\, N/m$
$1.2\times10^{-2}\, N/m$
Solution
Area increased $=(10 \times 11)-(10 \times 6) \mathrm{\,cm}^{2}$
$=110-60=50 \mathrm{\,cm}^{2}$
Since film has $2$ sides
$\therefore$ total increased area $=50 \times 2=100 \mathrm{cm}^{2}$
work done $=$ surface tension $\times$ increase in surface area
$\Rightarrow$ Surface tension
$=\frac{\text { Work done }}{\text { increasein surface area }}$
$=\frac{3 \times 10^{-4}}{100 \mathrm{\,cm}^{2}}=\frac{3 \times 10^{-4}}{100 \times 10^{-4} \mathrm{\,m}^{2}}$
$=0.03 \mathrm{\,N} / \mathrm{m}=3 \times 10^{-2} \mathrm{\,N} / \mathrm{m}$
