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

Answer 1

The Correct Option is A

Let's solve the problem step-by-step to find the potential of the bigger drop formed by combining 27 identical drops, each charged at 220 V.

Step 1: Calculate the charge on one small drop

The potential \( V \) of a drop is given by the formula: 

\(V = \frac{k \cdot q}{r}\)

where \( k \) is Coulomb's constant, \( q \) is the charge on the drop, and \( r \) is the radius of the drop.

Rearranging, we find the charge \( q \) on one drop:

\(q = \frac{V \cdot r}{k}\)

Since all 27 drops are identical, each has the same charge \( q \).

Step 2: Combine the charges

When the drops combine to form a bigger drop, their total charge \( Q \) is the sum of the charges of the 27 smaller drops:

\(Q = 27 \cdot q\)

Step 3: Relate the volumes of the drops

Since volume is proportional to the cube of the radius, we have:

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

where \( R \) is the radius of the bigger drop.

This implies:

\(R = 3r\)

Step 4: Calculate the potential of the bigger drop

The potential \( V' \) of the bigger drop is given by:

\(V' = \frac{k \cdot Q}{R}\)

Substituting \( Q = 27q \) and \( R = 3r \):

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

Since each small drop has a potential of \( 220 \, \text{V} \):

\(V' = 9 \cdot 220 = 1980 \, \text{V}\)

Thus, the potential of the bigger drop is 1980 V.

The correct answer is therefore \(1980 V\).