27 identical drops, each charged to potential 10 volts, combined to form a bigger drop. Find potential of this bigger drop ?

Options

A. `148 V`
B. `90 V`
C. 180 V`
D. `127 V`

Answer

To find the potential of the bigger drop formed by combining 27 identical drops, each charged to a potential of 10 volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 27 identical small drops, each with a potential of 10 volts.
- We need to find the potential of a single larger drop formed by combining these 27 smaller drops.

2. **Volume Conservation**:
- The volume of a single small drop can be expressed as:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of a small drop.
- The total volume of 27 small drops is:
\[ V_{\text{total}} = 27 \times V_{\text{small}} = 27 \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \]
- Let the radius of the larger drop be \( R \). The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
- Setting the total volume of the small drops equal to the volume of the larger drop gives:
\[ \frac{4}{3} \pi R^3 = 36 \pi r^3 \]

3. **Solving for R**:
- Cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 27 r^3 \]
- Taking the cube root of both sides:
\[ R = 3r \]

4. **Charge Calculation**:
- The potential \( V \) of a charged drop is given by:
\[ V = \frac{kQ}{r} \]
where \( Q \) is the charge on the drop and \( k \) is Coulomb's constant.
- For a single small drop, we have:
\[ V_{\text{small}} = \frac{kQ}{r} = 10 \text{ volts} \]
- The charge \( Q \) on one small drop can be expressed as:
\[ Q = 10 \cdot r / k \]

5. **Total Charge on the Larger Drop**:
- The total charge \( Q_{\text{total}} \) on the larger drop, which is formed by combining 27 small drops, is:
\[ Q_{\text{total}} = 27Q = 27 \left( \frac{10 \cdot r}{k} \right) = \frac{270r}{k} \]

6. **Potential of the Larger Drop**:
- The potential \( V' \) of the larger drop is given by:
\[ V' = \frac{kQ_{\text{total}}}{R} \]
- Substituting \( Q_{\text{total}} \) and \( R \):
\[ V' = \frac{k \left( \frac{270r}{k} \right)}{3r} = \frac{270r}{3r} = 90 \text{ volts} \]

### Final Answer:
The potential of the bigger drop is **90 volts**.

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