Question

27 water droplets of radius 3 mm each and charged with the same charge are coalesced to form a big water drop. The ratio of surface charge density of the small drop and large drop will be

• A. 1 : 1
• B. 1 : 3
• C. 3 : 1
• D. 1 : 27

Hint:

When multiple droplets coalesce into one large droplet, the surface charge density decreases with the increase in the radius of the larger drop.

Answer 1

The Correct Option is C

Step 1: Understanding the concept of surface charge density.
Surface charge density (\( \sigma \)) is the charge per unit surface area. It is
given by:
\[ \sigma = \frac{Q}{A} \] where \( Q \) is the charge and \( A \) is the surface area of the droplet.
Step 2: Total charge conservation.
Since all 27 droplets have the same charge, and they coalesce to form a single larger droplet, the total charge remains the same.
Step 3: Relating the radius of the droplets.
The total volume is conserved in the process. The volume of a sphere is
given by:
\[ V = \frac{4}{3} \pi r^3 \] For 27 small droplets of radius \( r = 3 \, \text{mm} \), the total volume of the small droplets is: \[ V_{\text{small}} = 27 \times \frac{4}{3} \pi r^3 \] For the large droplet, the radius is \( R \), and the volume is: \[ V_{\text{large}} = \frac{4}{3} \pi R^3 \] Since the total volume is conserved, we have: \[ V_{\text{small}} = V_{\text{large}} \quad \Rightarrow \quad 27 \times r^3 = R^3 \] This gives: \[ R = 3r \]
Step 4: Finding the ratio of surface charge densities.
The surface area of a sphere is
given by:
\[ A = 4 \pi r^2 \] Thus, the surface area of the small droplet is \( A_{\text{small}} = 4 \pi r^2 \) and the surface area of the large droplet is \( A_{\text{large}} = 4 \pi (3r)^2 = 36 \pi r^2 \). Now, the surface charge densities are: \[ \sigma_{\text{small}} = \frac{Q}{A_{\text{small}}} = \frac{Q}{4 \pi r^2}, \quad \sigma_{\text{large}} = \frac{Q}{A_{\text{large}}} = \frac{Q}{36 \pi r^2} \] Thus, the ratio of surface charge densities is: \[ \frac{\sigma_{\text{small}}}{\sigma_{\text{large}}} = \frac{\frac{Q}{4 \pi r^2}}{\frac{Q}{36 \pi r^2}} = 9 \] Therefore, the ratio is \( \text{3:1} \), not \( 9:1 \). Thus, the correct answer is
(C) 3 : 1.