512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.

Solution

Physics sir · 3 steps

Charge adds up 512 times, but the radius grows only 8 times, so the potential rises by 512 over 8.

1. Volume is conserved. The 512 small drops of radius r merge into one drop of radius R, and the total volume stays the same. So 4/3π R³ = 512 × 4/3π r³, which gives R = 8r.
2. Charge is conserved. Each small drop carries charge q, so the big drop carries Q = 512q. For one small drop, kq/r = 2 V.

   V = kQ/R = k(512q)/8r = 64 × kq/r = n^(2/3) v

3. So. V = 64 × 2 V = 128 V, so the answer is 128.

The common slip: 1024. Adding the potentials directly, 512 × 2, gives 1024. That treats potential like charge and ignores that the radius also grows 8 times.

The idea to keep. When n identical drops merge, charge scales by n and radius by n^(1/3). So the potential becomes n^(2/3) times the potential of one drop.

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