27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be _____________ V.

Solution

Physics sir · 3 steps

Charge adds up 27 times, but the radius grows only 3 times, so the potential rises 9 times.

1. Volume is conserved. The total volume stays the same when the drops merge: 4/3π R³ = 27 × 4/3π r³. So R³ = 27r³, which gives R = 3r.
2. Charge is conserved. Each small drop carries charge q, so the big drop carries Q = 27q. For one small drop, v = kq/r = 22 V.

   V = kQ/R = k(27q)/3r = 9 × kq/r = 9v

3. So. V = 9 × 22 = 198 V, so the answer is 198.

The common slip: 594. Adding the 27 potentials directly, 27 × 22, gives 594. That ignores the bigger radius, which divides the potential by 3.

The idea to keep. When n identical drops merge, charge scales as n and radius as n^(1/3), so the potential scales as n^(2/3). Always conserve volume and charge first.

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