27 mercury drops of equal radii and charges combine into one large drop. Find the ratio of the potential of the large drop to that of one original small drop.
Physics · Mixed Chapter
JEE Mains

Question

27 mercury drops of equal radii and charges combine into one large drop. Find the ratio of the potential of the large drop to that of one original small drop.

G1 Physics · Accepted Answer

Step 1 — Conservation of liquid volume: For 27 identical small drops of radius r forming one large drop of radius R: (4/3)πR³ = 27(4/3)πr³ ⇒ R = 3r. Step 2 — Conservation of electric charge: If each original drop has charge q, the big drop carries Q = 27q. Step 3 — Potential of a charged spherical drop: Surface potential V = (1/4πε₀)(charge/radius). Therefore V(big)/V(small) = (27q/3r)/(q/r) = 9. Answer: 9.