A big drop is formed by coalescing 1000 small identical drops of water. If E₁ be the total surface energy of 1000 small drops of water and E₂ be the surface energy of single big drop of water, then E₁: E₂ is x: 1 where x= ________.

Solution

Physics sir · 3 steps

Volume is conserved when drops merge, so first find the big radius, then compare the surface areas.

1. Volume stays same. Let each small drop have radius r and the big drop radius R. The water is only rearranged, so 1000 × 4/3π r³ = 4/3π R³. This gives R³ = 1000 r³, so R = 10r.
2. Energy follows area. Surface energy is surface tension times surface area, E = T × 4π r². With the same T for both, the ratio of energies is the ratio of total surface areas.

   E₁/E₂ = (1000 × 4π r² T)/(4π (10r)² T) = (1000 r²)/(100 r²) = 10

3. So. E₁ : E₂ = 10 : 1, so x = 10. The extra surface energy of the small drops is released when they merge.

The common slip: 100. Stopping at R²/r² = 100 gives 100. That compares the big drop with only one small drop, but E₁ counts all 1000 of them.

The idea to keep. When n identical drops merge, volume is conserved, so R = n^(1/3) r. The total surface area, and so the surface energy, falls by a factor of n^(1/3).

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