If n identical drops of mercury are combined to form a bigger drop then find the capacity of bigger drop, if capacity of each drop of mercury is C.

Options

A. `n^(1//3)`C
B. `n^(2//3)`C
C. `n^(1//4)` C
D. nC

Explanation

To solve the problem of finding the capacity of a bigger drop formed by combining n identical drops of mercury, we can follow these steps:

### Step 1: Understand the relationship between charge, capacity, and voltage
The capacity (C) of a drop is defined as the charge (Q) stored per unit voltage (V). Thus, we can express this relationship as:
\[ Q = C \cdot V \]

### Step 2: Determine the volume of the drops
The volume of a single small drop of mercury can be expressed as:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the small drop.

For \( n \) identical drops, the total volume is:
\[ V_{\text{total}} = n \cdot V_{\text{small}} = n \cdot \frac{4}{3} \pi r^3 \]

### Step 3: Relate the volume of the bigger drop
When these \( n \) drops combine to form a bigger drop, the volume of the bigger drop can be expressed as:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the bigger drop.

Since the volume is conserved, we can equate the two volumes:
\[ n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 4: Simplify the equation
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ n \cdot r^3 = R^3 \]

### Step 5: Solve for the radius of the bigger drop
From the equation above, we can solve for \( R \):
\[ R = n^{1/3} \cdot r \]

### Step 6: Find the capacity of the bigger drop
The capacity of a spherical drop is given by:
\[ C = 4 \pi \epsilon_0 R \]
where \( \epsilon_0 \) is the permittivity of free space.

Substituting \( R \) into the capacity formula for the bigger drop:
\[ C_{\text{big}} = 4 \pi \epsilon_0 (n^{1/3} \cdot r) \]

### Step 7: Relate the capacity of the bigger drop to the smaller drop
Since the capacity of the small drop is given as \( C = 4 \pi \epsilon_0 r \), we can express the capacity of the bigger drop in terms of the capacity of the small drop:
\[ C_{\text{big}} = n^{1/3} \cdot C \]

### Final Answer
Thus, the capacity of the bigger drop formed by combining \( n \) identical drops of mercury is:
\[ C_{\text{big}} = n^{1/3} \cdot C \]

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