Question

A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, then

• A. energy = $ 4VT \bigg( \frac{1}{r} - \frac{1}{R} \bigg) $ is released
• B. energy = $ 3VT \bigg( \frac{1}{r} + \frac{1}{R} \bigg) $ is absorbed
• C. energy = $ 3VT \bigg( \frac{1}{r} - \frac{1}{R} \bigg) $ is released
• D. energy is neither released nor absorbed.

Answer 1

The Correct Option is C

To solve the problem of determining the energy change when a certain number of small spherical drops of liquid coalesce into a single larger drop, we need to understand the relationship between surface energy, volume, and surface tension.

1. The volume of a single small drop of radius $r$ is given by: \( v = \frac{4}{3} \pi r^3 \).
2. Let the number of small drops be $n$. Therefore, the total volume of all small drops is: \( n \times \frac{4}{3} \pi r^3 \).
3. When they coalesce to form a single large drop of radius $R$, the volume of this larger drop is: \( \frac{4}{3} \pi R^3 \).
4. Since the volume is conserved, equate the total volume of small drops to the volume of the large drop: \( n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \).
5. Simplifying, we find: \( n = \left( \frac{R}{r} \right)^3 \).
6. The initial surface area of all small drops is \( n \times 4 \pi r^2 = \left( \frac{R}{r} \right)^3 \times 4 \pi r^2 = 4 \pi R^2 \cdot \frac{R}{r} \).
7. The final surface area of the large drop is \( 4 \pi R^2 \).
8. The change in surface area is thus: \( 4 \pi R^2 \left( \frac{R}{r} - 1 \right) \).
9. Surface energy is proportional to surface area. The change in energy due to change in surface area when drops coalesce can be calculated using the formula: \( \Delta E = T \times \Delta A \), where $T$ is the surface tension.
10. Substituting the change in area into the energy formula gives: \( \Delta E = T \times 4 \pi R^2 \left( \frac{R}{r} - 1 \right) \).
11. Simplifying, considering $V = \frac{4}{3} \pi R^3$, we have: \( \Delta E = 3V T \left( \frac{1}{r} - \frac{1}{R} \right) \).
12. The energy is released as \(\Delta E\) is negative because smaller \(r\) means greater initial energy.

Therefore, the correct option is energy = $ 3VT \left( \frac{1}{r} - \frac{1}{R} \right) $ is released.