Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is :

<p>The volume of a sphere is given by $\frac{4}{3}\pi R^3$.</p> <p>So the volume of the two small mercury drops each of radius $R$ is $2\times \frac{4}{3}\pi R^3$.</p> <p>When they coalesce to form a larger drop, the volume is conserved. So, the volume of the larger drop is also $2\times \frac{4}{3}\pi R^3$.</p> <p>Let&#39;s denote the radius of this large drop as $R&#39;$.</p> <p>Therefore, $2\times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi {R&#39;}^3$.</p> <p>Solving for $R&#39;$, we get $R&#39; = 2^{1/3}R$.</p> <p>Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$.</p> <p>So, the ratio of total surface energy before and after the change is:</p> <p>$$\frac{{2\times 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1$$</p>