Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Fluids2 Marks
Question
A liquid drop of radius 4mm breaks into 1000 identical drops. Find the change in surface energy. $S=0.07 \mathrm{Nm}^{-1}$.
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Answer
Volume of 1000 small drops = Volume of a large drop$1000\times\frac43\pi\text{r}^3=\frac43\pi\text{R}^3$
$\text{r}=\frac{\text{R}}{10}$
Surface area of large drop $=4\pi\text{R}^2$ Surface area of 1000 drop $4\pi\times1000\text{r}^2=40\pi\text{R}^2$$\therefore$ Increase in surface area $=(40-4)\pi\text{R}^2=36\pi\text{R}^2$
The increase in surface energy = Surface tension × increase in suraface area$=36\pi\text{R}^2\times0.07=36\times3.14\times(4\times10^{-3})^2\times0.07$
$=1.26\times10^{-4}\text{J}$
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