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 Question:
Multiple small charged spherical drops coalesce into a single large drop. We are given the potential of the small drops and must calculate the potential of the resulting large drop.

Step 2: Key Formula or Approach:
1. Conservation of Volume: The volume of the big drop equals the sum of the volumes of all small drops.
$$\frac{4}{3}\pi R^3 = n \left( \frac{4}{3}\pi r^3 \right) \implies R = n^{1/3} r$$
2. Conservation of Charge: The total charge on the big drop equals the sum of the charges of the small drops.
$$Q = n \cdot q$$
3. Potential of a Sphere: $V = \frac{kQ}{R}$.

Step 3: Detailed Explanation:
Let $n = 27$ be the number of drops.
Let $v = 20 \text{ V}$ be the potential of one small drop, where $v = \frac{kq}{r}$.
First, find the radius $R$ of the large drop:
$$R = (27)^{1/3} r = 3r$$
Next, find the total charge $Q$ of the large drop:
$$Q = 27q$$
Now, calculate the potential $V_{big}$ of the large drop:
$$V_{big} = \frac{kQ}{R}$$
Substitute $Q = 27q$ and $R = 3r$:
$$V_{big} = \frac{k(27q)}{3r} = 9 \left( \frac{kq}{r} \right)$$
Notice that the term in the parenthesis is simply the potential of a small drop ($v$):
$$V_{big} = 9 \times v$$
Substitute the given value $v = 20 \text{ V}$:
$$V_{big} = 9 \times 20 = 180 \text{ V}$$

Step 4: Final Answer:
The potential of the big drop is 180 V, matching option (b).