Question

64 identical drops each of capacity 5 µF combine to form a big drop. What will be the capacity of the big drop?

• A. 25 µF
• B. 4 µF
• C. 164 µF
• D. 20 µF

Hint:

For \(n\) identical drops combining, the new radius is \(R = n^{1/3}r\), the new capacitance is \(C' = n^{1/3}C\), and if the drops are charged, the new potential is \(V' = n^{2/3}V\). Memorizing these relations can save time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When small charged conducting drops combine to form a larger drop, the total volume is conserved. The capacitance of a spherical conductor is proportional to its radius. We can find the radius of the big drop in terms of the small drops' radius and then find its capacitance.
Step 2: Key Formula or Approach:
Let \(r\) be the radius and \(C\) be the capacity of each small drop.
Let \(R\) be the radius and \(C'\) be the capacity of the big drop.
The capacity of a spherical conductor is given by \(C = 4\pi\epsilon_0 r\).
The volume of \(n\) small drops equals the volume of the big drop: \[ n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \] This simplifies to \(R = n^{1/3} r\).
Step 3: Detailed Explanation:
Given:
Number of drops, \(n = 64\).
Capacity of each small drop, \(C = 5 \, \mu F\).
First, find the radius of the big drop (\(R\)) in terms of the small drop radius (\(r\)): \[ R = (64)^{1/3} r = 4r \] The radius of the big drop is 4 times the radius of a small drop.
Now, find the capacity of the big drop (\(C'\)): \[ C' = 4\pi\epsilon_0 R \] Substitute \(R = 4r\): \[ C' = 4\pi\epsilon_0 (4r) = 4 \times (4\pi\epsilon_0 r) \] Since \(C = 4\pi\epsilon_0 r\), we have: \[ C' = 4 \times C \] Finally, substitute the given value of \(C\): \[ C' = 4 \times 5 \, \mu F = 20 \, \mu F \] Step 4: Final Answer:
The capacity of the big drop will be 20 µF. Therefore, option (D) is correct.