Mechanical Properties Of Fluids Question 7
Question 7 - 31 January - Shift 1
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 $.\pi=\frac{22}{7})$
(1) $7.92 \times 10^{-6} J$
(2) $7.92 \times 10^{-4} J$
(3) $9.68 \times 10^{-4} J$
(4) $8.8 \times 10^{-5} J$
Show Answer
Answer: (2)
Solution:
Formula: Surface Tension
$1000 \times \frac{4 \pi}{3}(1)^{3}=\frac{4 \pi}{3} R^{3}$
$R=10 mm$
$T \times 1000 \times 4 \pi(10^{-3})^{2}-T \times 4 \pi(10 \times 10^{-3})^{2}=\Delta E$
$\Delta E=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$
$\Delta E=7.92 \times 10^{-4} J$
Option 2.
