8 equally charged drops are combined to form a big drop. If the potential on each drop is 10 V, then the potential of the big drop will be

Options

A. 40 V
B. 30 V
C. 25 V
D. 20 V

Explanation

To solve the problem, we need to find the potential of a big drop formed by combining 8 equally charged small drops, each having a potential of 10 V.

### Step-by-Step Solution:

**Step 1: Understand the relationship between volume and radius.**
The volume \( V \) of a sphere (small drop) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the small drop.

**Step 2: Calculate the total volume of 8 small drops.**
If we have 8 small drops, the total volume \( V_{\text{total}} \) is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

**Step 3: Relate the volume of the big drop to its radius.**
Let \( R \) be the radius of the big drop. The volume of the big drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]

**Step 4: Set the volumes equal.**
Since the total volume of the small drops equals the volume of the big drop, we have:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

**Step 5: Simplify the equation.**
Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ 8r^3 = R^3 \]

**Step 6: Find the relationship between the radii.**
Taking the cube root of both sides, we find:
\[ R = 2r \]

**Step 7: Use the formula for potential.**
The potential \( V \) of a charged sphere is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]
where \( Q \) is the charge.

**Step 8: Relate charge to radius and potential.**
The charge \( Q \) of a small drop can be expressed in terms of its potential:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \implies Q = 4 \pi \epsilon_0 V r \]
For the small drop, substituting \( V = 10 \, \text{V} \):
\[ Q = 4 \pi \epsilon_0 \cdot 10 \cdot r \]

**Step 9: Substitute \( Q \) into the potential formula for the big drop.**
The potential of the big drop becomes:
\[ V_{\text{big}} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} = \frac{1}{4 \pi \epsilon_0} \frac{4 \pi \epsilon_0 \cdot 10 \cdot r}{2r} = \frac{10}{2} = 5 \, \text{V} \]

### Final Answer:
The potential of the big drop is **5 V**.