A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius `b(bgt a)`. The solid sphere is given a charge Q. A student measures the potential at the surface of the solid sphere as V and the potential at the surface of spherical shell as `V_b`. After taking these readings, he decides . to put charge of `-4Q` on the shell. He then noted the readings of the potential of solid sphere and the shell and found that the potential difference is `/_\V` .
He then connected the outer spherical shell to the earth by a conducting wire and found that the charge on the outer surface of the shell as He then decides to remove the earthing connection from the shell and earthed the inner solid sphere. Connecting the inner sphere with the earth he observes the charge on the solid sphere as `q_2`. He then wanted to check what happens if the two are connected by the conducting wire. So he removed the earthing connection and connected a conducting wire between the solid sphere and the spherical shelll. After the connections were made he found the charge on the outer shell as `q_3`.
`q_3` is

Options

A. `(Q(a+b))/(a-b)`
B. `(Qa^2)/b`
C. `(Q(a-b))/b`
D. `-(Qb)/a`

Explanation

To solve the problem step by step, we will analyze the situation involving a solid metallic sphere and a conducting spherical shell, and how the charges and potentials change throughout the process.

### Step-by-Step Solution:

1. **Initial Setup**:
- We have a solid metallic sphere of radius \( a \) with charge \( Q \).
- Surrounding it is a conducting spherical shell of radius \( b \) (where \( b > a \)).
- The potential at the surface of the solid sphere is \( V \) and at the surface of the spherical shell is \( V_b \).

2. **Charge on the Shell**:
- The student puts a charge of \( -4Q \) on the conducting shell.
- The total charge on the shell is now \( -4Q \).

3. **Potential Difference**:
- The potential difference between the solid sphere and the shell is measured as \( \Delta V = V - V_b \).

4. **Earthing the Shell**:
- The outer shell is connected to the earth. When this happens, the potential of the shell becomes zero, and the charge on the shell redistributes.
- The charge on the outer shell becomes \( 0 \) (since it is earthed).

5. **Charge Redistribution**:
- Since the shell is earthed, it will have no net charge. The charge \( -4Q \) will flow to the earth, leaving the shell with \( 0 \) charge.
- The charge on the inner solid sphere remains \( Q \).

6. **Earthing the Inner Sphere**:
- The student then decides to earth the inner solid sphere. When the solid sphere is earthed, it will also have a potential of \( 0 \).
- The charge on the inner sphere will also become \( 0 \) as it loses its charge to the earth.

7. **Connecting the Two**:
- After earthing both the shell and the sphere, the student connects the two with a conducting wire.
- When the two conductors are connected, they will share charge until they reach the same potential.

8. **Final Charge Distribution**:
- The total charge in the system before connecting the two is \( Q + (-4Q) = -3Q \).
- When the two conductors are connected, they will form a single conducting body, and the total charge \( -3Q \) will distribute over the outer surface of the shell.

9. **Charge on the Outer Shell**:
- Since the shell is the outer conductor, all the charge will reside on its outer surface.
- Therefore, the charge on the outer shell after the connection is \( q_3 = -3Q \).

### Final Answer:
The charge on the outer shell after connecting the inner sphere and the outer shell is:
\[ q_3 = -3Q \]