The amount of work done to break a big water drop of radius ' R ' into 27 small drops of equal radius is 10 J . The work done required to break the same big drop into 64 small drops of equal radius will be

Solution

Physics sir · 4 steps

Volume stays fixed, so find how much new surface each break creates, then compare the two cases.

1. Size of small drops. Breaking the drop does not change the total volume, so n · 4/3π r³ = 4/3π R³. This gives r = R/n^(1/3).
2. Extra surface area. The n small drops have total area n · 4π r² = 4π R² n^(1/3). Work done equals surface tension times the increase in area.

   W = T · 4π R² (n^(1/3) - 1)

3. Use the 27 case. For n = 27, n^(1/3) - 1 = 2, so 2 × T · 4π R² = 10 J and T · 4π R² = 5 J.
4. So. For n = 64, n^(1/3) - 1 = 3, so W = 3 × 5 J = 15 J. The answer is (d).

The idea to keep. Work against surface tension goes only into new area. Splitting a drop into n equal drops multiplies its area by n^(1/3), so the work is proportional to n^(1/3) - 1.

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