Question

Two spherical liquid droplets coalesce to form one large spherical droplet. The initial radius of each droplet is r, the surface tension of the liquid is s, and its density is d. The energy released in this process depends on:

(a) r²
(b) r³
(c) r⁴
(d) r⁵

Answer 1

The energy released when two liquid droplets coalesce is related to the change in surface area and is proportional to the square of the initial radius of the droplets, r².

Step-by-step explanation:

The process of two spherical liquid droplets coalescing into one larger spherical droplet can be explained by the principles of surface tension and conservation of energy. Surface tension, caused by cohesive forces, drives the liquid to minimize surface area relative to volume, forming a sphere. When two droplets with radius r merge, they release some of the energy that was originally required to maintain their separate surfaces.

The initial total surface area of two droplets is proportional to r2, as the area of a sphere is 4πr2. After coalescing, the new droplet's volume is the sum of the volumes of the two smaller droplets, which is proportional to r3. To find the radius of the new droplet, we take the cube root of the sum of the volumes and then apply the surface area formula again to get the surface area proportional to the new radius squared. The released surface energy can be calculated as the difference between the initial combined surface energy and the final surface energy of the larger droplet.

Therefore, the energy released during the coalescence of the two droplets is related to the surface area change and is proportional to the square of the initial radius (r2). So, the correct answer to the given question is (a) r2.