27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop.

Solution

Physics sir · 4 steps

Get the new radius from volume, add up the charges, then use U = kQ²/2R for the ratio.

1. New radius. Mercury does not compress, so the volume is conserved: 4/3π R³ = 27 × 4/3π r³. This gives R³ = 27r³, so R = 3r.
2. New charge. Charge is conserved too, so the big drop carries Q = 27q. Its potential is V = kQ/R = 27/3 × kq/r = 9 × 10 V = 90 V.

   U = 1/2QV = kQ²/2R ⇒ U_(big)/Uₛₘₐₗₗ = (Q/q)² r/R = 27²/3

3. Put in numbers. 729/3 = 243. As a check, U = 1/2QV grows by 27 × 9 = 243 as well.
4. So. The big drop has 243 times the potential energy of one small drop.

The common slip: 729. Scaling the energy by Q² alone and forgetting that the radius also grows to 3r gives 729, not 243.

The idea to keep. When n drops merge, R = n^(1/3)r, Q = nq, V becomes n^(2/3) times and U becomes n^(5/3) times. Get the radius from volume first, then scale everything else.

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