Two water drops each of radius ' $r$ ' coalesce to form a bigger drop. If ' $T$ ' is the surface tension, the surface energy released in this process is :

<p>To determine the surface energy released when two identical water drops coalesce to form a larger drop, let's go through the process step by step:</p> <p><p><strong>Volume Conservation</strong>: </p> <p>When two droplets, each with a radius $ R $, merge to form a larger droplet, the total volume is conserved. Therefore:</p> <p>$ 2 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 $</p> <p>Solving for $ r $, the radius of the larger drop, we find:</p> <p>$ r = 2^{1/3} R $</p></p> <p><p><strong>Initial Surface Energy ($ U_i $)</strong>: </p> <p>The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension $ T $:</p> <p>$ U_i = 2 \times 4 \pi R^2 T $</p></p> <p><p><strong>Final Surface Energy ($ U_f $)</strong>: </p> <p>The surface energy of the larger droplet is given by:</p> <p>$ U_f = 4 \pi r^2 T = 4 \pi R^2 T (2^{2/3}) $</p></p> <p><p><strong>Energy Released</strong>: </p> <p>The energy released in the process is the difference between the initial and final surface energy:</p> <p>$ \text{Heat lost} = U_i - U_f = 4 \pi R^2 T \left[2 - 2^{2/3}\right] $</p></p> <p>This expression gives us the amount of surface energy that is released when two identical water drops combine to form a single larger drop.</p>