Two metallic sphere of radii `1 cm` and `3 cm` are given charges of `4xx10^(-2)C` and respectively. If these are connected by a counducting wire, the final charge on the bigger sphere is

Options

A. `2xx10^(-2)C`
B. `3xx10^(-2)C`
C. `4xx10^(-2)C`
D. `1xx10^(-2)C`

Explanation

To find the final charge on the bigger sphere when two metallic spheres of different radii are connected by a conducting wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of the first sphere, \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Radius of the second sphere, \( r_2 = 3 \, \text{cm} = 0.03 \, \text{m} \)
- Initial charge on the first sphere, \( Q_1 = 4 \times 10^{-2} \, \text{C} \)
- Initial charge on the second sphere, \( Q_2 = 0 \, \text{C} \) (since it is not mentioned, we assume it starts with no charge).

2. **Understand the Concept of Equal Potential:**
- When the two spheres are connected by a conducting wire, charge will flow until the electric potential on both spheres is equal.
- The potential \( V \) of a sphere is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( r \) is the radius of the sphere.

3. **Set Up the Equation for Equal Potentials:**
- Let \( q_1 \) be the final charge on the first sphere and \( q_2 \) be the final charge on the second sphere.
- Since the potentials are equal:
\[ \frac{q_1}{r_1} = \frac{q_2}{r_2} \]

4. **Express \( q_2 \) in Terms of \( q_1 \):**
- Rearranging the equation gives:
\[ q_2 = \frac{r_2}{r_1} \cdot q_1 \]
- Substituting the values of \( r_1 \) and \( r_2 \):
\[ q_2 = \frac{0.03}{0.01} \cdot q_1 = 3 \cdot q_1 \]

5. **Use the Conservation of Charge:**
- The total charge before connecting the spheres must equal the total charge after they are connected:
\[ Q_1 + Q_2 = q_1 + q_2 \]
- Substituting the known values:
\[ 4 \times 10^{-2} + 0 = q_1 + 3 \cdot q_1 \]
- This simplifies to:
\[ 4 \times 10^{-2} = 4 \cdot q_1 \]

6. **Solve for \( q_1 \):**
- Dividing both sides by 4:
\[ q_1 = \frac{4 \times 10^{-2}}{4} = 1 \times 10^{-2} \, \text{C} \]

7. **Find \( q_2 \):**
- Now, substituting \( q_1 \) back into the equation for \( q_2 \):
\[ q_2 = 3 \cdot q_1 = 3 \cdot (1 \times 10^{-2}) = 3 \times 10^{-2} \, \text{C} \]

### Final Answer:
The final charge on the bigger sphere (radius 3 cm) is:
\[ \boxed{3 \times 10^{-2} \, \text{C}} \]