Question

If 1000 droplets of water of surface tension 0.07 N/m , having same radius 1 mm each, combine to from a single drop In the process the released surface energy is - (Take π= $\frac {22}{7}$)

• A. $8.8 \times 10^{-5} J$
• B. $7.92 \times 10^{-4} J$
• C. $7.92 \times 10^{-6} J$
• D. $9.68 \times 10^{-4} J$

Answer 1

Answer: (B)

$7.92 \times 10^{-4} J$

Answer 2

1000×34π​(1)3=34π​R3
R=10mm
T×1000×4π(10−3)2−T×4π(10×10−3)2=ΔE
ΔE=4×π×7×10−2[1000−100]×10−6
ΔE=7.92×10−4J