Question

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

• A. 660 V
• B. 1320 V
• C. 1520 V
• D. 1980 V

Answer 1

The Correct Option is D

To find the potential of the bigger drop formed by the combination of 27 smaller drops, we can use concepts from electrostatics concerning charged spheres. Here's a step-by-step explanation.

1. Understanding the Charge-Potential Relationship:

The potential \( V \) of a charged spherical drop is given by \( V = \frac{kQ}{R} \), where \( k \) is the electrostatic constant, \( Q \) is the charge, and \( R \) is the radius of the drop.

1. Initial Conditions:

Each of the 27 smaller drops has a potential of \(220\, V\). Assume each drop has a charge \( q \) and radius \( r \).

1. Total Charge Conservation: 

The total charge before and after merging remains the same. Hence, \( Q = 27q \).

1. Volume Conservation:

The volume of the combined drop is equal to the sum of the volumes of the 27 smaller drops. Since volume \( V \propto R^3 \), we have:

\(27 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3\)

Solving for \( R \), we get \( R = 3r \).

1. Potential of the Combined Drop:

Using the potential formula for the bigger sphere:

\(V' = \frac{kQ}{R} = \frac{k \cdot 27q}{3r} = 9 \times \frac{kq}{r}\)

From the small drop's potential: \(220\, V = \frac{kq}{r}\)

Substitute for \( \frac{kq}{r} \):

\(V' = 9 \times 220 = 1980\, V\)

1. Conclusion:

Thus, the potential of the bigger drop is \(1980\, V\).