Question

Twenty seven drops of same size are charged at $220\text{ V}$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.}

• A. $1320\text{ V}$
• B. $1520\text{ V}$
• C. $1980\text{ V}$
• D. $660\text{ V}$

Hint:

For $n$ identical droplets combining, the potential of the big drop is always $V_{big} = n^{2/3} \times V_{small}$. Here, $27^{2/3} = (3^3)^{2/3} = 3^2 = 9$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When droplets coalesce, charge and volume are conserved. The potential changes because the radius of the combined drop is not simply the sum of individual radii.
Step 2: Detailed Explanation:
Let \( n = 27 \).
Volume of big drop \( = n \times \) Volume of small drop \( \implies R^3 = n r^3 \implies R = n^{1/3} r = 27^{1/3} r = 3r \).
Charge of big drop \( Q = nq = 27q \).
Potential of small drop \( v = kq/r = 220 \, \text{V} \).
Potential of big drop \( V = kQ/R = k(27q)/(3r) = 9 (kq/r) = 9 \times v \).
\( V = 9 \times 220 = 1980 \, \text{V} \).
Step 3: Final Answer:
The potential of the bigger drop is \( 1980 \, \text{V} \).