A solid conducting sphere of radius a is enclosed in a concentric shell of radius `b gt a`. The charge on the inner sphere is `3muC` and on the conducting shell is `-10muC`. The potential difference between the conductors is V. If `+10muC` additional charge is given to the outer shell, the new potential difference is

A. V
B. V/2
C. 2V
D. zero

Options

A. V
B. V/2
C. 2V
D. zero

Explanation

To solve the problem step by step, we will analyze the situation involving the solid conducting sphere and the conducting shell.

### Step 1: Understand the Configuration
We have a solid conducting sphere of radius \( a \) with a charge of \( +3 \, \mu C \). This sphere is enclosed by a concentric conducting shell of radius \( b \) (where \( b > a \)), which initially has a charge of \( -10 \, \mu C \).

### Step 2: Determine the Initial Potential Difference
The potential difference \( V \) between the inner sphere and the conducting shell can be calculated using the formula:
\[ V = V_{\text{inner}} - V_{\text{outer}} \]
Where:
- \( V_{\text{inner}} \) is the potential at the surface of the inner sphere.
- \( V_{\text{outer}} \) is the potential at the inner surface of the conducting shell.

The potential \( V_{\text{inner}} \) at the surface of the sphere is given by:
\[ V_{\text{inner}} = \frac{Q_{\text{inner}}}{4 \pi \epsilon_0 a} \]
Substituting \( Q_{\text{inner}} = 3 \, \mu C \):
\[ V_{\text{inner}} = \frac{3 \times 10^{-6}}{4 \pi \epsilon_0 a} \]

The potential \( V_{\text{outer}} \) at the inner surface of the shell can be calculated considering the total charge enclosed by the shell:
\[ V_{\text{outer}} = \frac{Q_{\text{inner}} + Q_{\text{shell}}}{4 \pi \epsilon_0 b} \]
Where \( Q_{\text{shell}} = -10 \, \mu C \):
\[ V_{\text{outer}} = \frac{3 \times 10^{-6} - 10 \times 10^{-6}}{4 \pi \epsilon_0 b} = \frac{-7 \times 10^{-6}}{4 \pi \epsilon_0 b} \]

Thus, the potential difference \( V \) is:
\[ V = \frac{3 \times 10^{-6}}{4 \pi \epsilon_0 a} - \frac{-7 \times 10^{-6}}{4 \pi \epsilon_0 b} \]

### Step 3: Adding Additional Charge
Now, we add an additional charge of \( +10 \, \mu C \) to the outer shell. The new charge on the shell becomes:
\[ Q_{\text{new shell}} = -10 \, \mu C + 10 \, \mu C = 0 \, \mu C \]

### Step 4: Determine the New Potential Difference
With the new charge on the shell being \( 0 \, \mu C \), the potential at the inner surface of the shell becomes:
\[ V_{\text{outer new}} = \frac{3 \times 10^{-6}}{4 \pi \epsilon_0 b} \]

Now, the potential difference \( V_{\text{new}} \) becomes:
\[ V_{\text{new}} = V_{\text{inner}} - V_{\text{outer new}} \]
Substituting the values:
\[ V_{\text{new}} = \frac{3 \times 10^{-6}}{4 \pi \epsilon_0 a} - \frac{3 \times 10^{-6}}{4 \pi \epsilon_0 b} \]

### Step 5: Factor Out Common Terms
Factoring out \( \frac{3 \times 10^{-6}}{4 \pi \epsilon_0} \):
\[ V_{\text{new}} = \frac{3 \times 10^{-6}}{4 \pi \epsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) \]
This shows that the new potential difference is still proportional to the original potential difference \( V \).

### Conclusion
Since the potential difference depends only on the charges inside and the geometry, and the outer charge does not affect the potential difference, we conclude that the new potential difference remains the same:
\[ V_{\text{new}} = V \]

Thus, the correct answer is **A. V**.