A spherical droplet having a potential of 2.5 volt is obtained as a result of merging of 125 identical droplets. Find the potential of the constituent droplet

Options

A. `0.4 V`
B. `0.5 V`
C. `62.5 V`
D. `0.1 V`

Answer

To solve the problem, we need to find the potential of the constituent droplets given that a larger droplet is formed by merging 125 identical smaller droplets. Let's break down the solution step by step.

### Step 1: Understand the relationship between charge and potential
The potential \( V \) of a spherical droplet is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the total charge, and \( R \) is the radius of the droplet.

### Step 2: Define the variables for the smaller droplets
Let:
- The charge of each smaller droplet be \( q \).
- The radius of each smaller droplet be \( r \).
- The potential of each smaller droplet be \( V_s \).

Thus, the potential of each smaller droplet can be expressed as:
\[ V_s = \frac{kq}{r} \]

### Step 3: Find the total charge of the larger droplet
When 125 identical droplets merge, the total charge \( Q \) of the larger droplet is:
\[ Q = 125q \]

### Step 4: Find the radius of the larger droplet
The volume of a single smaller droplet is:
\[ V_s = \frac{4}{3}\pi r^3 \]
The volume of the larger droplet formed by merging 125 smaller droplets is:
\[ V_b = 125 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]
From this, we can equate the volumes:
\[ \frac{4}{3}\pi R^3 = 125 \cdot \frac{4}{3}\pi r^3 \]
Cancelling \( \frac{4}{3}\pi \) from both sides gives:
\[ R^3 = 125r^3 \]
Taking the cube root of both sides:
\[ R = 5r \]

### Step 5: Find the potential of the larger droplet
The potential \( V_b \) of the larger droplet can be expressed as:
\[ V_b = \frac{kQ}{R} \]
Substituting \( Q = 125q \) and \( R = 5r \):
\[ V_b = \frac{k(125q)}{5r} = \frac{25kq}{r} \]

### Step 6: Set the potentials equal
We know the potential of the larger droplet is given as \( V_b = 2.5 \) volts. Therefore, we can set up the equation:
\[ \frac{25kq}{r} = 2.5 \]

### Step 7: Find the potential of the smaller droplet
Now we can express \( V_s \):
\[ V_s = \frac{kq}{r} \]
From the equation \( \frac{25kq}{r} = 2.5 \), we can isolate \( \frac{kq}{r} \):
\[ \frac{kq}{r} = \frac{2.5}{25} = 0.1 \]
Thus, the potential of the smaller droplet is:
\[ V_s = 0.1 \text{ volts} \]

### Final Answer
The potential of the constituent droplet is **0.1 volts**.

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