The correct option is B 2.5π×10−12 J
Given,
Number of water drops, n=1000
Radius of water drop,r=10−7 m
Total volume =n×43πr3
=1000×43π×(10−7)3 ... (1)
Volume of large drop if its radius is R
=43πR3 ... (2)
On equating (1) and (2) as volume will remain same,
43πR3=1000×43π×(10−7)3
⇒R=10−6 m
We know, surface energy, U=TΔA
U=TΔA=T(4πR2−n×4πr2)
U=7×10−2×4×π×[(10−6)2−1000×(10−7)2]
⇒U=−2.5π×10−12 J
(-ve sign indicates release of energy)