A mercury drop of radius 1.0 cm is sprayed into 10 6 droplets of equal size. Calculate the energy expanded. (Surface tension of mercury = 32 × 10 –2 N/m).
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(3.98 × 10 –2 J)
Sol. We know that dw = T dA ⇒ Δ W = T Δ A
∴ in drops, only one surface area is formed.
and Δ A = 10 6 × 4 π r 2 – 4 π R 2 . = 4 π R 2 [100 – 1]
so Δ W = 
× 4 π (10 –2 m) 2 [99] = 3.978 × 10 –2 J
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