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, we need to analyze the situation involving a charged conductor enclosed by a hollow charged conducting shell, which are connected by a conducting wire.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charged inner conductor with charge \( Q_1 \).
- There is a hollow conducting shell surrounding it with charge \( Q_2 \).
- Both conductors are connected by a conducting wire.

2. **Potential of Conductors**:
- In electrostatics, the potential inside a conductor is constant. When two conductors are connected by a wire, they must be at the same electric potential.
- Let \( V_1 \) be the potential of the inner conductor and \( V_2 \) be the potential of the outer shell.

3. **Equating Potentials**:
- Since the conductors are connected, we have \( V_1 = V_2 \).
- This implies that the potential difference \( V_1 - V_2 = 0 \).

4. **Calculating Potentials**:
- The potential \( V_1 \) of the inner conductor can be expressed as:
\[ V_1 = \frac{kQ_1}{r_1} \]
- The potential \( V_2 \) of the outer shell can be expressed as:
\[ V_2 = \frac{-kQ_2}{r_2} + \frac{Q_1 + Q_2}{4\pi \epsilon_0 r_2} \]
- Here, \( k \) is Coulomb's constant, \( r_1 \) is the radius of the inner conductor, and \( r_2 \) is the radius of the outer shell.

5. **Charge Redistribution**:
- When the wire connects the two conductors, charge will redistribute until equilibrium is reached.
- The total charge remains conserved: \( Q_{\text{total}} = Q_1 + Q_2 \).

6. **Final Charge Distribution**:
- If \( Q_1 \) is transferred to the outer shell, the final charge on the outer conductor becomes \( Q_2 + Q_1 \).
- If \( Q_1 \) is positive and \( Q_2 \) is negative or zero, the inner conductor may lose its charge to the outer shell.

7. **Conclusion**:
- The potential difference between the two conductors becomes zero when they are connected.
- The charge on the inner conductor can be completely transferred to the outer conductor, depending on their initial charges.

### Summary of Correct Statements:
1. The potential difference \( V_1 - V_2 = 0 \) (Correct).
2. The final charge on the outer conductor will be \( Q_2 + Q_1 \) (Correct).
3. The charge on the inner conductor can be transferred to the outer conductor (Correct).
4. If the inner conductor has charge \( Q \) and the outer conductor has charge \( 0 \), the final charge on each conductor will not be \( Q/2 \) (Incorrect).