64 identical drops each charged upto potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be __________ mV.

Solution

Physics sir · 3 steps

Volume fixes the new radius, charge just adds up; then V = kQ/R grows as n^(2/3).

1. Radius of big drop. Volume is conserved: 4/3π R³ = 64 × 4/3π r³, so R³ = 64r³ and R = 4r.
2. Charge of big drop. Charge is conserved too, so the big drop carries Q = 64q. For a single small drop, v = kq/r = 10 mV.

   V = kQ/R = k(64q)/4r = 16 kq/r = n^(2/3) v

3. So. V = 16 × 10 mV = 160 mV, so the answer is 160.

The common slip: 640. Treating the big drop's radius as 64r, or forgetting that the radius grows at all, gives 10 or 640 mV. Using R = 4r with Q = 64q is what gives 160.

The idea to keep. When n identical drops merge, charge scales as n but radius only as n^(1/3). So potential goes up by n^(2/3) and charge density by n^(1/3).

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