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. $9.68 \times 10^{-4} J$

Hint:

When droplets combine to form a single larger droplet, the total surface area decreases, and the surface energy is released in the process.

Answer 1

The Correct Option is B

$1000\times \frac{4\pi }{3}(1{)}^3=\frac{4\pi }{3}{R}^3$
$R=10mm$
$T\times 1000\times 4\pi {\left(10^{-3}\right)}^2-T\times 4\pi {\left(10\times 10^{-3}\right)}^2=\Delta E$
$\Delta E=4\times \pi \times 7\times 10^{-2}[1000-100]\times 10^{-6}$
$\Delta E=7.92\times 10^{-4}J$

Answer 2

The surface energy is given by the formula:

\[ \text{Surface energy} = \Delta E = \text{Number of droplets} \times \left( \frac{4\pi}{3} \right) R^3 \]

The radius of each droplet is 1 mm = \( 10 \, \text{mm} \). Therefore,

\[ T \times 1000 \times \left( \frac{4\pi}{3} \right) (10^{-3})^3 - T \times \left( \frac{4\pi}{3} \right) (10 \times 10^{-3})^3 = \Delta E \]

\[ \Delta E = 4 \times 10^{-7} [1000 - 100] \times 10^{-6} \]

\[ \Delta E = 7.92 \times 10^{-4} \, \text{J} \]