If a charged conductor is enclosed by a hollow charged conducting shell (assumed concentric and spherical in shape), and they are connected by a conducting wire, then which of the following statement (s) would be correct ?

Options

A. Potential difference between the two conductors becomes zero.
B. If charge on the inner conductor is `q` and on the outer conductor is `2 q`, then finally charge on the outer conductor will be `3q`.
C. The charge on the inner conductor is totally transferred to the outer conductor.
D. If charge on the inner conductor is `q` and charge on the outer conductor is zero, then finally the charge on each conductor will be `q//2`.

Explanation

To solve the problem regarding the charged conductor enclosed by a hollow charged conducting shell, we will analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a charged conductor (let's call it Conductor 1) with charge \( Q_1 \).
- This conductor is enclosed by a hollow conducting shell (Conductor 2) with charge \( Q_2 \).
- Both conductors are connected by a conducting wire.

2. **Connecting the Conductors**:
- When the two conductors are connected by a wire, they will reach the same electric potential because they are conductors.
- This means that the potential difference \( V_1 - V_2 = 0 \).

3. **Potential of Conductors**:
- The potential \( V_1 \) of the inner conductor (Conductor 1) is given by:
\[ V_1 = \frac{Q_1}{4 \pi \epsilon_0 R_1} \]
- The potential \( V_2 \) of the outer conductor (Conductor 2) is given by:
\[ V_2 = \frac{Q_2}{4 \pi \epsilon_0 R_2} \]

4. **Equating Potentials**:
- Since \( V_1 = V_2 \), we can set the equations equal:
\[ \frac{Q_1}{4 \pi \epsilon_0 R_1} = \frac{Q_2}{4 \pi \epsilon_0 R_2} \]
- This implies:
\[ Q_1 R_2 = Q_2 R_1 \]

5. **Charge Redistribution**:
- When the conductors are connected, charge will flow until the potentials are equal.
- If \( Q_1 \) is the initial charge on the inner conductor and \( Q_2 \) is the initial charge on the outer conductor, charge will redistribute between them.

6. **Final Charge Distribution**:
- If the inner conductor has charge \( Q \) and the outer conductor has charge \( 2Q \), after redistribution, the inner conductor will lose its charge and the outer conductor will gain it.
- Thus, the final charge on the outer conductor will be \( 3Q \) and the inner conductor will have \( 0 \) charge.

7. **Conclusion**:
- The potential difference becomes zero when the total charge from the inner conductor is transferred to the outer conductor.
- Therefore, the correct statements are:
- If the charge on the inner conductor is \( Q \) and on the outer conductor is \( 2Q \), then finally the charge on the outer conductor will be \( 3Q \).
- The charge on the inner conductor will be \( 0 \) after redistribution.