Mechanical Properties Of Fluids Question 8
Question 8 - 01 February - Shift 1
A mercury drop of radius $10^{-3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 Nm^{-1}$. The gain in surface energy is:
(1) $2.26 \times 10^{-5} J$
(2) $28 \times 10^{-5} J$
(3) $17.5 \times 10^{-5} J$
(4) $5 \times 10^{-5} J$
Show Answer
Answer: (1)
Solution:
Formula: Surface Tension
Initial surface energy $=0.45 \times 4 \pi(10^{-3})^{2}$
$\frac{4}{3} \pi(10^{-3})^{3}=125 \times \frac{4 \pi}{3} R _{\text{new }}^{3}$
$\therefore \quad 10^{-3}=5 R _{\text{new }}$
$\therefore \quad R _{\text{new }}=\frac{10^{-3}}{5} m$
So, final surface energy $=0.45 \times 125 \times 4 \pi(\frac{10^{-3}}{5})^{2}$
Increase in energy $=0.45 \times 4 \pi \times(10^{-3})^{2}[\frac{125}{25}-1]$
$=4 \times 0.45 \times 4 \pi \times 10^{-6}$
$=2.26 \times 10^{-5} J$
