Question

27 drops of mercury coalesce to form a bigger drop. What is the relative increase in surface energy?

• A. 3/2
• B. 2/3
• C. -2/3
• D. 8

Answer 1

The Correct Option is C

Given: 27 drops of mercury coalesce to form a bigger drop.
Let the radius of each small drop be \( r \). The volume of one drop is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \] The total volume of 27 small drops is:
\[ V_{\text{total}} = 27 \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \] Now, let the radius of the bigger drop formed by coalescing all 27 smaller drops be \( R \). The volume of the bigger drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \] Since the total volume is conserved, we have:
\[ V_{\text{total}} = V_{\text{big}} \quad \Rightarrow \quad 36 \pi r^3 = \frac{4}{3} \pi R^3 \] Simplifying:
\[ 36 r^3 = \frac{4}{3} R^3 \quad \Rightarrow \quad R^3 = 27 r^3 \quad \Rightarrow \quad R = 3r \] Next, we calculate the surface energy of the small drops and the big drop. The surface area of one small drop is:
\[ A_{\text{small}} = 4 \pi r^2 \] The total surface area of 27 small drops is:
\[ A_{\text{total small}} = 27 \times 4 \pi r^2 = 108 \pi r^2 \] The surface area of the bigger drop is:
\[ A_{\text{big}} = 4 \pi R^2 = 4 \pi (3r)^2 = 36 \pi r^2 \] The surface energy is proportional to the surface area. The relative increase in surface energy is:
\[ \frac{A_{\text{big}} - A_{\text{total small}}}{A_{\text{total small}}} = \frac{36 \pi r^2 - 108 \pi r^2}{108 \pi r^2} = \frac{-72 \pi r^2}{108 \pi r^2} = -\frac{2}{3} \] Final Answer: The relative increase in surface energy is -2/3.