When a number of liquid drops each of surface charge density `sigma ` and energy E combine, a large drop is formed. If the charge density of the large drop is 3 `sigma` , its energy is

Options

A. 81E
B. 3E
C. 27E
D. 243 E

Answer

To solve the problem step by step, we will analyze the situation of small liquid drops merging to form a larger drop, considering the conservation of charge and volume.

### Step 1: Define the parameters for small drops
Let:
- \( \sigma \) = surface charge density of small drops
- \( E \) = energy of one small drop
- \( r \) = radius of one small drop
- \( A \) = surface area of one small drop = \( 4\pi r^2 \)
- \( q \) = charge on one small drop = \( \sigma \cdot A = \sigma \cdot 4\pi r^2 \)

### Step 2: Define the parameters for the large drop
Let:
- \( R \) = radius of the large drop
- The surface charge density of the large drop is given as \( 3\sigma \).
- The surface area of the large drop = \( 4\pi R^2 \)
- The charge on the large drop = \( 3\sigma \cdot 4\pi R^2 \)

### Step 3: Conservation of charge
The total charge before merging (from \( n \) small drops) must equal the charge of the large drop:
\[ n \cdot q = 3\sigma \cdot 4\pi R^2 \]
Substituting \( q \):
\[ n \cdot (\sigma \cdot 4\pi r^2) = 3\sigma \cdot 4\pi R^2 \]
Canceling \( 4\pi \sigma \) from both sides:
\[ n r^2 = 3 R^2 \quad \text{(1)} \]

### Step 4: Conservation of volume
The volume of \( n \) small drops must equal the volume of the large drop:
\[ n \cdot \left(\frac{4}{3} \pi r^3\right) = \frac{4}{3} \pi R^3 \]
Canceling \( \frac{4}{3} \pi \):
\[ n r^3 = R^3 \quad \text{(2)} \]

### Step 5: Relate \( R \) and \( r \) using equations (1) and (2)
From equation (1):
\[ R^2 = \frac{n r^2}{3} \]
Substituting this into equation (2):
\[ n r^3 = \left(\frac{n r^2}{3}\right)^{3/2} \]
Simplifying:
\[ n r^3 = \frac{n^{3/2} r^3}{3^{3/2}} \]
Canceling \( r^3 \) (assuming \( r \neq 0 \)):
\[ n = \frac{n^{3/2}}{3^{3/2}} \]
Rearranging gives:
\[ n^{1/2} = 3^{3/2} \quad \Rightarrow \quad n = 27 \]

### Step 6: Calculate the energy of the large drop
The electrostatic potential energy \( U \) of a drop is given by:
\[ U = \frac{1}{4\pi \epsilon_0} \frac{q^2}{R} \]
For one small drop:
\[ E = \frac{1}{4\pi \epsilon_0} \frac{q^2}{r} \]
For the large drop:
\[ E_{large} = \frac{1}{4\pi \epsilon_0} \frac{(3q)^2}{R} \]
Substituting \( q = \sigma \cdot 4\pi r^2 \):
\[ E_{large} = \frac{1}{4\pi \epsilon_0} \frac{(3 \cdot \sigma \cdot 4\pi r^2)^2}{R} \]
Now, substituting \( R \) using \( n = 27 \):
\[ E_{large} = \frac{1}{4\pi \epsilon_0} \frac{(3q)^2}{\frac{n r^2}{3}} = \frac{1}{4\pi \epsilon_0} \frac{9q^2}{\frac{27r^2}{3}} = \frac{1}{4\pi \epsilon_0} \frac{27q^2}{r^2} \]
Thus:
\[ E_{large} = 27 \cdot E \]

### Final Step: Conclusion
The energy of the large drop is:
\[ E_{large} = 243E \]