Question 73: Properties of Matter

Question

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

Accepted Answer

Initial surface energy $=0.45 	imes 4 \pi\left(10^{-3}ight)^2$
$$
\begin{aligned}
& \frac{4}{3} \pi\left(10^{-3}ight)^3=125 	imes \frac{4 \pi}{3} R_{	ext {new }}^3 \\
	herefore & 10^{-3}=5 R_{	ext {new }} \\
	herefore & R_{	ext {new }}=\frac{10^{-3}}{5} \mathrm{~m}
\end{aligned}
$$
So, final surface energy $=0.45 	imes 125 	imes 4 \pi\left(\frac{10^{-3}}{5}ight)^2$
Increase in energy $=0.45 	imes 4 \pi 	imes\left(10^{-3}ight)^2\left[\frac{125}{25}-1ight]$
$$
\begin{aligned}
& =4 	imes 0.45 	imes 4 \pi 	imes 10^{-6} \\
& =2.26 	imes 10^{-5} \mathrm{~J}
\end{aligned}
$$

Answer

$$28	imes10^{-5}~\mathrm{J}$$

Answer

$$17.5	imes10^{-5}~\mathrm{J}$$

Answer

$$5	imes10^{-5}~\mathrm{J}$$

Answer

$$2.26	imes10^{-5}~\mathrm{J}$$

Properties of Matter

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

Choose the correct answer: