Question

Two mercury drops of radii $ r $ and $ 2r $ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is } $S$ and take $9^{2/3} = 4.326$)

• A. \( 1.6 \pi S \, \text{J} \)
• B. \( 3.2 \pi S \, \text{J} \)
• C. \( 17.1 \pi S \, \text{J} \)
• D. \( 2.7 \pi S \, \text{J} \)

Hint:

When solving problems involving the merging of droplets, remember to use the concept of surface energy and apply the appropriate geometric formulas for surface area.

Answer 1

The Correct Option is D

The surface energy of a drop of radius \( r \) is given by: \[ E = 4 \pi r^2 S \] Where:
\( r \) is the radius of the drop,
\( S \) is the surface tension of mercury.
Let the radius of the smaller drop be \( r \) and the radius of the larger drop be \( 2r \). The total surface energy before merging is: \[ E_{\text{initial}} = 4 \pi r^2 S + 4 \pi (2r)^2 S = 4 \pi r^2 S + 16 \pi r^2 S = 20 \pi r^2 S \] After merging, the radius of the larger drop becomes \( r_{\text{final}} = r + 2r = 3r \). The surface area of the new drop is: \[ E_{\text{final}} = 4 \pi (3r)^2 S = 36 \pi r^2 S \] The surface energy released during the process is: \[ \Delta E = E_{\text{final}} - E_{\text{initial}} = 36 \pi r^2 S - 20 \pi r^2 S = 16 \pi r^2 S \] Now, using the given value for the ratio \( \left( \frac{3^3}{2^3} = 4.326 \right) \), we find: \[ \Delta E \approx 2.7 \pi S \] Thus, the correct answer is: \[ \boxed{2.7 \pi S \, \text{J}} \]