Question

A large number of liquid drops each of radius \( r \) coalesce to form a single drop of radius \( R \). The energy released in the process is converted into kinetic energy of the big drop \( M \) is given by (given, surface tension of liquid \( \sigma \)):

• A. \( \dfrac{T}{\rho} \left( 1 - \dfrac{1}{R} \right) \)
• B. \( \dfrac{2T}{\rho} \left( 1 - \dfrac{1}{R} \right) \)
• C. \( 2T \left( 1 - \dfrac{1}{R} \right) \)
• D. \( \dfrac{T}{\rho} \left( 1 - \dfrac{1}{R^2} \right) \)

Hint:

When liquid drops coalesce, the surface area decreases and the released energy is proportional to the change in surface area and the surface tension.

Answer 1

The Correct Option is A

Step 1: When drops of liquid coalesce, the total surface area decreases, and this results in the release of energy equal to the change in surface energy.
Step 2: The energy released is proportional to the change in surface area, and using the relationship between surface tension \( \sigma \), radius, and volume, the energy released is: \[ E = \dfrac{T}{\rho} \left( 1 - \dfrac{1}{R} \right). \]
Final Answer: \[ \boxed{\dfrac{T}{\rho} \left( 1 - \dfrac{1}{R} \right)} \]