Consider two concentric metalic shell's of radii R and 2R. The inner shell is having charge Q and outer shell is uncharged. If they are connected with a conducting wire. Then,

Options

A. Q amount of change will flow from inner to outer shell
B. Q/e number of electrons will flow from outer to inner shell, where 'e' is the charge on an electron
C. `(KQ^(2))/(4R)` amount of heat is produced in the process
D. `(KQ^(2))/(2R)` amount of heat is produced in the process

Answer

To solve the problem step by step, we will analyze the situation of two concentric metallic shells with given charges and radii.

### Step 1: Understand the Setup
We have two concentric metallic shells:
- Inner shell with radius \( R \) and charge \( Q \).
- Outer shell with radius \( 2R \) and initially uncharged (charge = 0).

### Step 2: Connect the Shells with a Wire
When the two shells are connected by a conducting wire, they will reach the same electric potential because conductors in electrostatic equilibrium must have the same potential.

### Step 3: Calculate the Potential of Each Shell
The potential \( V \) at the surface of a charged spherical shell is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius.

- For the inner shell (radius \( R \)):
\[ V_{\text{inner}} = \frac{kQ}{R} \]

- For the outer shell (radius \( 2R \)):
\[ V_{\text{outer}} = \frac{kQ_{\text{outer}}}{2R} \]
where \( Q_{\text{outer}} \) is the charge on the outer shell after they are connected.

### Step 4: Set the Potentials Equal
Since both shells are at the same potential after connection:
\[ \frac{kQ}{R} = \frac{kQ_{\text{outer}}}{2R} \]

### Step 5: Simplify the Equation
Cancelling \( k \) and \( R \) from both sides:
\[ Q = \frac{Q_{\text{outer}}}{2} \]
Thus, we can express \( Q_{\text{outer}} \) in terms of \( Q \):
\[ Q_{\text{outer}} = 2Q \]

### Step 6: Charge Conservation
Since the total charge must be conserved, the charge on the outer shell after connection must equal the initial charge on the inner shell:
\[ Q_{\text{outer}} + Q_{\text{inner}} = Q + 0 \]
Substituting \( Q_{\text{outer}} = 2Q \) and \( Q_{\text{inner}} = 0 \):
\[ 2Q + 0 = Q \]
This indicates that the charge \( Q \) flows from the inner shell to the outer shell.

### Step 7: Calculate the Number of Electrons Transferred
The number of electrons \( n \) that flow from the outer shell to the inner shell can be calculated using:
\[ Q = n \cdot e \]
where \( e \) is the charge of an electron. Rearranging gives:
\[ n = \frac{Q}{e} \]

### Step 8: Calculate the Change in Potential Energy
The initial potential energy \( U_i \) of the system (with charge \( Q \) on the inner shell) is given by:
\[ U_i = \frac{kQ^2}{2R} \]
The final potential energy \( U_f \) (with charge \( 2Q \) on the outer shell) is:
\[ U_f = \frac{k(2Q)^2}{2(2R)} = \frac{2kQ^2}{2R} = \frac{kQ^2}{R} \]
The change in potential energy \( \Delta U \) is:
\[ \Delta U = U_f - U_i = \frac{kQ^2}{R} - \frac{kQ^2}{2R} = \frac{kQ^2}{2R} \]

### Step 9: Determine the Heat Produced
The energy lost in the process will be dissipated as heat, which is equal to the change in potential energy:
\[ \text{Heat produced} = -\Delta U = -\frac{kQ^2}{2R} \]

### Summary of Results
1. Charge \( Q \) flows from the inner shell to the outer shell.
2. The number of electrons that flow from the outer shell to the inner shell is \( \frac{Q}{e} \).
3. The amount of heat produced in the process is \( -\frac{kQ^2}{2R} \).