Question

Calculate the amount of energy evolved when 8 drops of mercury of radius 1mm each ,combine to form 1drop (surface tension of Hg = .55/m)

Answer 1

Answer:

T = 0.55 N/m  Amount of energy evolved = W = T ∙ ΔA  ∴ W = 055 × ΔA   (1) ΔA = (n × 4πr2) – (1 × 4πR2) where r is radius of small drop. R is radius of big drop. = (4π × 8 × r2) – 4πR2 ΔA = 4π(8r2 – R2)    (2) Also as mass remains same hence ρ ∙ (4/3)πr3 × 8 = ρ × (4/3)πR3

Explanation:

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