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(A) The radius of the large drop is $R = 2\, mm = 2 	imes 10^{-3}\, m$.The number of small droplets formed is $n = 8$.The surface tension of mercury

Vedclass · Accepted Answer

$23.4$

Vedclass · Answer

(A) The radius of the large drop is $R = 2\, mm = 2 	imes 10^{-3}\, m$.The number of small droplets formed is $n = 8$.The surface tension of mercury is $T = 0.465\, J/m^2$.When a large drop of radius $R$ is split into $n$ identical small droplets of radius $r$,the volume remains constant:$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3 \Rightarrow r = \frac{R}{n^{1/3}}$.For $n = 8$,$r = \frac{R}{8^{1/3}} = \frac{R}{2}$.The increase in surface energy $\Delta U$ is given by the difference in surface areas multiplied by surface tension:$\Delta U = T 	imes (A_{final} - A_{initial}) = T 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$.Substituting $r = R/n^{1/3}$:$\Delta U = 4\pi R^2 T (n^{1/3} - 1)$.Substituting the values:$\Delta U = 4 	imes 3.14159 	imes (2 	imes 10^{-3})^2 	imes 0.465 	imes (8^{1/3} - 1)$,$\Delta U = 4 	imes 3.14159 	imes 4 	imes 10^{-6} 	imes 0.465 	imes (2 - 1)$,$\Delta U = 16 	imes 3.14159 	imes 0.465 	imes 10^{-6} \approx 23.37 	imes 10^{-6}\, J = 23.4\, \mu J$.

$A$ drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy in $\mu J$. (Surface tension of mercury is $0.465\;J/m^2$)

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