Two metal spheres having radii in the ratio 3:4 are put into contact and a charge `14muC` is given o the system. Then they are seperated so that there is no mutual force between them. The potential due to the larger sphere at a distance of 3 m is

Options

A. 72KV
B. 36KV
C. 24KV
D. 12KV

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
We have two metal spheres with radii in the ratio of 3:4. They are given a total charge of \(14 \mu C\) and then separated. We need to find the potential due to the larger sphere at a distance of 3 meters.

### Step 2: Define the Radii
Let the radius of the smaller sphere be \(r_1 = 3x\) and the radius of the larger sphere be \(r_2 = 4x\) for some \(x\).

### Step 3: Calculate the Capacities of the Spheres
The capacitance \(C\) of a sphere is given by:
\[ C = 4\pi \epsilon_0 r \]
Thus, the capacitance of the smaller sphere \(C_1\) and the larger sphere \(C_2\) are:
\[ C_1 = 4\pi \epsilon_0 (3x) = 12\pi \epsilon_0 x \]
\[ C_2 = 4\pi \epsilon_0 (4x) = 16\pi \epsilon_0 x \]

### Step 4: Calculate the Potentials
When the spheres are in contact, the charge will redistribute until the potentials are equal. Let the charge on the smaller sphere be \(q_1\) and on the larger sphere be \(q_2\). We know:
\[ q_1 + q_2 = 14 \mu C \]

The potentials \(V_1\) and \(V_2\) are given by:
\[ V_1 = \frac{q_1}{C_1} = \frac{q_1}{12\pi \epsilon_0 x} \]
\[ V_2 = \frac{q_2}{C_2} = \frac{q_2}{16\pi \epsilon_0 x} \]

Setting \(V_1 = V_2\):
\[ \frac{q_1}{12\pi \epsilon_0 x} = \frac{q_2}{16\pi \epsilon_0 x} \]

### Step 5: Solve for Charges
Cross-multiplying gives:
\[ 16q_1 = 12q_2 \]
Substituting \(q_2 = 14 \mu C - q_1\) into the equation:
\[ 16q_1 = 12(14 \mu C - q_1) \]
\[ 16q_1 = 168 \mu C - 12q_1 \]
\[ 28q_1 = 168 \mu C \]
\[ q_1 = 6 \mu C \]
Thus, \(q_2 = 14 \mu C - 6 \mu C = 8 \mu C\).

### Step 6: Calculate the Potential due to the Larger Sphere
The potential \(V\) due to the larger sphere at a distance \(r = 3 m\) is given by:
\[ V = \frac{kq_2}{r} \]
Where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) and \(q_2 = 8 \times 10^{-6} \, C\):
\[ V = \frac{9 \times 10^9 \cdot 8 \times 10^{-6}}{3} \]
\[ V = \frac{72 \times 10^3}{3} = 24 \times 10^3 \, V = 24 \, kV \]

### Final Answer
The potential due to the larger sphere at a distance of 3 meters is **24 kV**.

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