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The work done to break up a drop of a liquid of radius R and surface tension $\sigma$ into eight drops, all of the same size, is
Option: 1
$\mathrm{4 \pi \sigma R^2}$

Option: 2
$\mathrm{2 \pi \sigma R^2}$

Option: 3
$\mathrm{\frac{1}{2} \pi \sigma R^2}$

Option: 4
$\mathrm{\frac{1}{4} \pi \sigma R^2}$

Answers (1)

best_answer

Volume of big drop $\mathrm{=\frac{4}{3} \pi R^3}$ . If r is the radius of each tiny drop, the total volume of eight drops $\mathrm{= 8 \times \frac{4 \pi}{3} r^3=\frac{32}{3} \pi r^3}$ . Since the total volume remains unchanged, we have

$\mathrm{ \frac{32}{3} \pi r^3=\frac{4}{3} \pi R^3 }$

which gives $\mathrm{r=\frac{R}{2}}$ . Now, surface area of big drop $\mathrm{= 4 \pi R^2.}$

Total surface area of eight drops $\mathrm{ =32 \pi r^2=8 \pi R^2}$

$\mathrm{(\because r=R / 2).}$

$\mathrm{\therefore \quad}$ Increase in surface area $\mathrm{=8 \pi R^2-4 \pi R^2=4 \pi R^2.}$

$\mathrm{\therefore \quad}$ Work done = surface tension $\times$ increase in surface area

$\mathrm{=4 \pi \sigma R^2}$

Posted by

jitender.kumar

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