JEE Main 2026 Physics question
Eight mercury drops, each of radius r, coalesce to form a bigger drop. The surface energy released in this process is . (S is the surface tension of mercury)
Answer: (B)
Solution
Volume conservation gives R=2r. Energy released = S(8×4πr²−4πR²) = S×4π(8r²−4r²) = 16πr²S.