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The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $72\times10^{-3}\, N/m$) :
$5.98\times10^{-5}\, J$
$10.98\times10^{-5}\, J$
$16.95\times10^{-5}\, J$
$8.95\times10^{-5}\, J$
Solution
Radius of new droplet if be $\mathrm{r}$ then,
$10^{6} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times(0.001)^{3}$
$r^{3}=10^{-15} \Rightarrow r=10^{-5}$
Increase in surface area
$=\left[4 \pi \times\left(10^{-5}\right)^{2} \times 10^{6}\right]-\left[4 \pi \times\left(10^{-3}\right)^{2}\right]$
$\left[4 \pi \times 10^{-4}\right]-\left[4 \pi \times 10^{-6}\right]=4 \pi 10^{-6}[100-1]$
$=4 \pi \times 10^{-6} \times 99=4 \pi \times 10^{-6} \times 99$
Work done
$=$ surface tension $\times$ increase in surface area
$=72 \times 4 \pi \times 99 \times 10^{-6} \times 10^{-3}=8.95 \times 10^{-5} \mathrm{\,J}$
