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  • A mercury drop of radius   is broken into 125 equal size droplets. Surface tension of mer

A mercury drop of radius 10^{-3}m  is broken into 125 equal size droplets.
Surface tension of mercury is  0.45Nm^{-1} . The gain in surface energy is:

Option: 1

28\times10^{-5}J


Option: 2

17.5\times10^{-5}J


Option: 3

5\times10^{-5}J


Option: 4

2.26\times10^{-5}J


Answers (1)

best_answer

[Volume of bigger drop] = [volume of smaller drop] × 125

\begin{aligned} & \frac{4}{3} \pi \mathrm{R}^3=125 \times \frac{4}{3} \pi \mathrm{r}^3 \\ & \mathrm{R}^3=125 \mathrm{r}^3 \\ & \therefore \mathrm{R}=5 \times \mathrm{r} \\ & \Rightarrow \text { Gain in sinface energy }=\mathrm{TdA} \\ & \quad=0.45 \times\left[\mathrm{A}_2-\mathrm{A}_1\right] \end{aligned}

\begin{aligned} & =0.45 \times\left[125 \times 4 \pi r^2-4 \pi R^2\right] \\ & =0.45 \times\left[125 \times 4 \pi\left(\frac{R}{5}\right)^2-4 \pi R^2\right] \\ & =0.45 \times\left[20 \pi R^2-4 \pi R^2\right] \\ & =0.45 \times 16 \pi R^2 \\ & =0.45 \times 16 \times 3.14 \times\left(10^{-3}\right)^2 \\ & =2.26 \times 10^{-5} \mathrm{~J} \end{aligned}

Posted by

Suraj Bhandari

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