Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.

Solution

Physics sir · 4 steps

Merging conserves charge and volume, so find the new charge and radius first; the energy follows.

1. Volume fixes radius. Mercury does not compress, so volume is conserved: 4/3π R³ = 8 × 4/3π r³. This gives R³ = 8r³, so R = 2r.
2. Charge adds up. Charge is conserved, so the big drop carries Q = 8q. Its potential is V' = kQ/R = (k × 8q)/2r = 4V, which is 48 V.

   U = 1/2QV' = 1/2(8q)(4V) = 32 × 1/2qV

3. Compare with one drop. A single small drop has E = 1/2qV. So the big drop has U = 32E. The value 12 V is not needed for the ratio.
4. So. The potential energy of the big drop is 32 E.

The common slip: 64. Keeping the radius the same and only squaring the charge gives 8² = 64, not 32. The radius doubles, which halves the energy.

The idea to keep. For n identical drops merging: R = n^(1/3)r, Q = nq, V → n^(2/3)V, and U = Q²/(8πε₀ R) → n^(5/3)U. With n = 8 that is 2⁵ = 32.

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