Question

A large number of liquid drops each of radius \( r \) coalesce to form a big drop of radius \( R \). The energy released in the process is converted into kinetic energy of the big drop. The speed of the big drop is: (T = surface tension of liquid, \( \rho = \) density of liquid)

• A. \( \frac{6T}{\rho} \left( \frac{1}{r} + \frac{1}{R} \right)^{\frac{1}{2}} \)
• B. \( \frac{3T}{\rho} \left( \frac{1}{r} + \frac{1}{R} \right)^{\frac{1}{2}} \)
• C. \( \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)^{\frac{1}{2}} \)
• D. \( \frac{3T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)^{\frac{1}{2}} \)

Hint:

When many small drops coalesce to form a larger drop, the energy released due to surface area change is converted into the kinetic energy of the large drop.

Answer 1

The Correct Option is C

Step 1: Energy Considerations.
The total energy released in the process of coalescence is the difference between the total energy of the small drops and the total energy of the big drop. This difference in energy is converted into the kinetic energy of the big drop. The energy of a single drop is related to its surface area and surface tension, and for the big drop, we have: \[ E_{\text{big}} = 4 \pi R^2 T, \quad E_{\text{small}} = 4 \pi r^2 T \] The total energy released is proportional to the difference in the radii of the small and big drops, leading to the final velocity of the big drop.
Step 2: Final Answer.
Thus, the speed of the big drop is given by \( \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)^{\frac{1}{2}} \).