Question

Assuming the drops to be spherical, 27 identical drops of mercury are charged simultaneously to the same potential of 20 volt. If all the charged drops are made to combine to form one big drop, then potential of big drop will be ______.

• A. 90 V
• B. 180 V
• C. 270 V
• D. 360 V

Hint:

Memorize these coalescing drop shortcut multipliers for $n$ drops:
- Radius: $R = n^{1/3}r$
- Capacitance: $C = n^{1/3}c$
- Potential: $V = n^{2/3}v$
- Energy: $E = n^{5/3}u$
Here, $V = (27)^{2/3} \times 20 = (3)^2 \times 20 = 9 \times 20 = 180 \text{ V}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When $n$ drops combine, volume is conserved, and total charge is conserved. $V_{drop} = \frac{kq}{r}$.

Step 2: Formula Application:
Volume conservation: $\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi r^3 \implies R = 3r$.
Charge conservation: $Q = 27q$.

Step 3: Explanation:
$V_{big} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 \left( \frac{kq}{r} \right)$.
Since $V_{small} = 20$ V, $V_{big} = 9 \times 20 = 180$ V.

Step 4: Final Answer:
The potential of the big drop is 180 V.