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_2` is

Options

A. Q
B. `Q(a/b)`
C. `-4Q`
D. zero

Explanation

To solve the problem step by step, we will analyze the situation involving the solid metallic sphere and the conducting spherical shell, considering the charge distributions and potentials at each stage.

### Step 1: Initial Setup
- We have a solid metallic sphere of radius \( a \) with a charge \( Q \).
- It is surrounded by 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 shell is \( V_b \).

### Step 2: Charge on the Shell
- The student then puts a charge of \( -4Q \) on the shell.
- This means the total charge on the shell is now \( -4Q \).

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

### Step 4: Earthing the Shell
- The outer shell is then connected to the earth. When a conductor is earthed, its potential becomes zero.
- Therefore, the charge on the outer shell will adjust such that its potential becomes zero. This will cause the charge on the outer shell to be \( -Q \) (as derived from the potential equation).

### Step 5: Charge Distribution After Earthing
- The inner surface of the shell will have a charge of \( -Q \) (to balance the charge \( Q \) on the inner sphere).
- The outer surface of the shell will have a charge of \( +Q \) (since the total charge on the shell must remain constant).

### Step 6: Earthing the Inner Sphere
- The inner solid sphere is then earthed. When this happens, the potential of the inner sphere becomes zero.
- This means that the charge on the inner sphere will adjust to maintain zero potential. The charge on the inner sphere becomes \( q_2 \).

### Step 7: Charge on the Inner Sphere
- The charge \( q_2 \) on the inner sphere can be calculated using the formula for potential:
\[ V = \frac{kQ}{a} + \frac{kq_2}{a} = 0 \]
From this, we can derive:
\[ q_2 = -\frac{Q \cdot a}{b} \]

### Step 8: Connecting the Two Spheres
- Finally, the inner sphere and the outer shell are connected by a conducting wire.
- The charges will redistribute until both spheres reach the same potential.

### Step 9: Final Charge on the Outer Shell
- After the connection, the charge on the outer shell becomes \( q_3 \).
- The final charge distribution will depend on the total charge and the radii of the spheres.

### Final Answer
The charge on the inner solid sphere, \( q_2 \), is given by:
\[ q_2 = -\frac{Q \cdot a}{b} \]