JEE Main 2026 Physics question
A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α × 10⁻⁶ J. The value of α is . (Take surface tension of liquid = 0.08 N/m)
Answer: (B)
Solution
512 droplets have radius r/8 = 0.125 mm. ΔE = T·4π(512r′² − r²) = 0.08 × 4π × (8 − 1) × 10⁻⁶ ≈ 7.0 × 10⁻⁶ J.