Question

Two mercury drops each of radius $r$ merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension of mercury, is

• A. ${\left(\sqrt[3]{32}\right)}^8\pi r^{2}T$
• B. $2^4\pi r^{2}T$
• C. $2\pi r^{2}T$
• D. ${\left(\sqrt[3]{32}\right)}^4\pi r^{2}T$

Answer 1

The correct option is A ${\left(\sqrt[3]{32}\right)}^8\pi r^{2}T$

Let, $R$ be the radius of the bigger drop, then Volume of bigger drop = $2\times$ volume of samller drop
$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3$
$R=\sqrt[3]{32}r$
Surface energy of bigger drop,
$E=\text{Surface area}\times \text{Surface tension}=4\pi R^2\times T=4\times \pi \times {\left(\sqrt[3]{32}r\right)}^2\times T={\left(\sqrt[3]{32}\right)}^8\pi r^{2}T$