Question

A big oil droplet of radius 10 cm is broken into a thousand equal droplets. What will be the gain in surface energy? (Surface tension of the oil is 0.1 N/m)

• A. 5 J
• B. 10 J
• C. 0.11 J
• D. 0.25 J

Hint:

When breaking a large droplet into smaller droplets, the surface area increases due to the relationship between volume and surface area in spherical objects. The total surface energy increase can be calculated using surface tension and the change in surface area.

Answer 1

The Correct Option is C

To determine the gain in surface energy when a big oil droplet is broken into smaller droplets, follow these steps: 
Given:
Radius of the big droplet (\( R \)): \( 10 \, {cm} = 0.1 \, {m} \)
Number of small droplets (\( n \)): \( 1000 \)
Surface tension of oil (\( T \)): \( 0.1 \, {N/m} \)
Step 1: Calculate the Volume of the Big Droplet
The volume \( V \) of the big droplet is:
\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (0.1)^3 = \frac{4}{3} \pi \times 10^{-3} \, {m}^3 \]
Step 2: Volume of Each Small Droplet
Since the big droplet breaks into \( 1000 \) equal droplets, the volume of each small droplet \( v \) is:
\[ v = \frac{V}{1000} = \frac{\frac{4}{3} \pi \times 10^{-3}}{1000} = \frac{4}{3} \pi \times 10^{-6} \, {m}^3 \]
Let the radius of each small droplet be \( r \). Then:
\[ \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 10^{-6} \]
\[ r^3 = 10^{-6} \Rightarrow r = 10^{-2} \, {m} = 1 \, {cm} \]
Step 3: Calculate the Surface Areas
Surface area of the big droplet (\( A_{{big}} \)):
\[ A_{{big}} = 4 \pi R^2 = 4 \pi (0.1)^2 = 4 \pi \times 10^{-2} \, {m}^2 \]
Total surface area of the small droplets (\( A_{{small}} \)):
\[ A_{{small}} = 1000 \times 4 \pi r^2 = 1000 \times 4 \pi (10^{-2})^2 = 1000 \times 4 \pi \times 10^{-4} = 4 \pi \times 10^{-1} \, {m}^2 \]
Step 4: Determine the Change in Surface Area
\[ \Delta A = A_{{small}} - A_{{big}} = 4 \pi \times 10^{-1} - 4 \pi \times 10^{-2} = 4 \pi (0.1 - 0.01) = 4 \pi \times 0.09 = 0.36 \pi \, {m}^2 \]
Step 5: Calculate the Gain in Surface Energy
The surface energy \( E \) is given by:
\[ E = T \times \Delta A = 0.1 \times 0.36 \pi = 0.036 \pi \, {J} \]
Approximating \( \pi \approx 3.14 \):
\[ E \approx 0.036 \times 3.14 \approx 0.11 \, {J} \]