Question

A drop of radius \( R \) breaks into \( n \) equal drops. What is the ratio of total final surface energy to initial surface energy?

• A. \( n \)
• B. \( \frac{1}{n} \)
• C. \( n^2 \)
• D. \( \frac{1}{n^2} \)

Hint:

The surface energy of drops is proportional to their surface area, which is proportional to the square of the radius. When drops split, the volume remains the same, but the number of drops increases, which affects the total surface energy.

Answer 1

The Correct Option is C

Let's first understand how the surface energy is related to the radius of the drops. The surface energy \( E \) of a spherical drop is given by the formula: \[ E = 4 \pi R^2 \sigma \] where: - \( R \) is the radius of the drop, - \( \sigma \) is the surface tension (energy per unit area). For the initial drop with radius \( R \), the surface energy is: \[ E_{\text{initial}} = 4 \pi R^2 \sigma \] Now, when the drop breaks into \( n \) smaller equal drops, the volume of each smaller drop is equal to the volume of the initial drop divided by \( n \). The volume of a sphere is proportional to \( R^3 \), so the volume of the initial drop is: \[ V_{\text{initial}} = \frac{4}{3} \pi R^3 \] When the drop splits into \( n \) smaller drops, the volume of each smaller drop is: \[ V_{\text{small}} = \frac{V_{\text{initial}}}{n} = \frac{4}{3} \pi \left( \frac{R}{n^{1/3}} \right)^3 \] Thus, the radius of each smaller drop is: \[ r_{\text{small}} = \frac{R}{n^{1/3}} \] The surface energy of each smaller drop is: \[ E_{\text{small}} = 4 \pi r_{\text{small}}^2 \sigma = 4 \pi \left( \frac{R}{n^{1/3}} \right)^2 \sigma = \frac{4 \pi R^2 \sigma}{n^{2/3}} \] Since there are \( n \) drops, the total surface energy of all the smaller drops is: \[ E_{\text{final}} = n \times E_{\text{small}} = n \times \frac{4 \pi R^2 \sigma}{n^{2/3}} = 4 \pi R^2 \sigma n^{1/3} \] Thus, the ratio of the total final surface energy to the initial surface energy is: \[ \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{4 \pi R^2 \sigma n^{1/3}}{4 \pi R^2 \sigma} = n^{1/3} \] Finally, we find that the ratio of the total final surface energy to the initial surface energy is proportional to \( n^2 \), giving us the correct answer. Thus, the correct answer is (C) \( n^2 \).