A spherical conductor of radius `2 m` is charged to a potential of `120 V`. It is now placed inside another hollow spherical conductor of radius `6 m`. Calculate the potential to which the bigger sphere would be raised

Options

A. 20 V
B. 60 V
C. 80 V
D. 40 V

Answer

To solve the problem, we need to calculate the potential to which the bigger spherical conductor will be raised when the smaller charged spherical conductor is placed inside it.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the inner sphere (R1) = 2 m
- Potential of the inner sphere (V1) = 120 V
- Radius of the outer sphere (R2) = 6 m

2. **Calculate the Charge on the Inner Sphere:**
The potential (V) of a charged sphere is given by the formula:
\[ V = \frac{k \cdot Q}{R} \]
where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(R\) is the radius of the sphere.

Rearranging the formula to find the charge \(Q\):
\[ Q = \frac{V \cdot R}{k} \]
Substituting the values for the inner sphere:
\[ Q_1 = V_1 \cdot R_1 = 120 \, \text{V} \cdot 2 \, \text{m} = 240 \, \text{C} \]

3. **Calculate the Potential of the Outer Sphere:**
When the inner sphere is placed inside the outer sphere, it will influence the potential of the outer sphere. The potential (V2) at the surface of the outer sphere due to the charge \(Q_1\) on the inner sphere is given by:
\[ V_2 = \frac{k \cdot Q_1}{R_2} \]
Substituting the values:
\[ V_2 = \frac{k \cdot 240 \, \text{C}}{6 \, \text{m}} \]

4. **Calculate the Potential Increase:**
Since \(k\) is a constant, we can simplify the equation. The potential increase at the outer sphere due to the inner sphere is:
\[ V_2 = \frac{240}{6} = 40 \, \text{V} \]

5. **Final Potential of the Outer Sphere:**
The potential of the outer sphere will now be raised by this amount. Therefore, the total potential of the outer sphere becomes:
\[ V_{\text{total}} = V_{\text{initial}} + V_2 \]
Since the outer sphere was initially uncharged, its initial potential can be considered as 0 V. Thus:
\[ V_{\text{total}} = 0 + 40 \, \text{V} = 40 \, \text{V} \]

### Conclusion:
The potential to which the bigger sphere would be raised is **40 V**.