A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be :

Solution

Physics sir · 4 steps

The total volume stays the same, the surface area shrinks, and the lost surface energy turns into heat.

1. Volume conserved. If n small drops merge, n · 4/3π r³ = 4/3π R³, so n r³ = R³. The volume of water is V = 4/3π R³, both before and after.
2. Energy released. Surface energy is T × area. The area falls from n · 4π r² to 4π R², so the energy released is W = 4π T (n r² - R²). Put n = R³/r³ to get W = 4π T R³(1/r - 1/R).
3. Per unit volume. Divide W by V = 4/3π R³. The 4π R³ cancels and leaves a factor of 3. Dividing by J converts the work into heat.

   W/JV = 3T/J(1/r - 1/R)

4. So. The rise in heat energy per unit volume is 3T/J(1/r - 1/R), which is option (b).

The idea to keep. When drops merge or a drop splits, keep the volume fixed and track only the change in surface area. Per unit volume, a sphere gives the factor 3/r, because area divided by volume is 3/r.

Answer from the NTA final answer key. Solution by PrepWiser.