A galvanometer of resistance `10kOmega` can measure a maximum current 1 mA. A non-ideal battery of emf 6 V is connected across this galvanometer. A resistance `1Omega` is also connected across the galvanometer. If the galvanometer reads a current 0.2 mA, the internal resistance of the battery is:

Options

A. `0.5Omega`
B. `1Omega`
C. `2 Omega`
D. `4 Omega`

Explanation

To solve the problem, we need to find the internal resistance of the battery given the parameters of the galvanometer and the current it reads.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance of the galvanometer, \( R_g = 10 \, k\Omega = 10,000 \, \Omega \)
- Maximum current the galvanometer can measure, \( I_{max} = 1 \, mA = 0.001 \, A \)
- Current reading of the galvanometer, \( I_g = 0.2 \, mA = 0.0002 \, A \)
- Resistance connected across the galvanometer, \( R = 1 \, \Omega \)
- EMF of the battery, \( E = 6 \, V \)

2. **Calculate the Total Resistance in the Circuit:**
The total resistance \( R_{total} \) in the circuit is the sum of the galvanometer resistance and the additional resistance:
\[ R_{total} = R_g + R = 10,000 \, \Omega + 1 \, \Omega = 10,001 \, \Omega \]

3. **Calculate the Potential Difference Across the Galvanometer:**
The potential difference \( V_g \) across the galvanometer can be calculated using Ohm's law:
\[ V_g = I_g \times R_g = 0.0002 \, A \times 10,000 \, \Omega = 2 \, V \]

4. **Determine the Total Voltage Drop in the Circuit:**
The total voltage drop in the circuit is equal to the EMF of the battery minus the voltage drop across the internal resistance \( r \) of the battery:
\[ E = V_g + I_g \times r \]
Rearranging gives:
\[ r = \frac{E - V_g}{I_g} \]

5. **Substitute the Known Values:**
Substituting the known values into the equation:
\[ r = \frac{6 \, V - 2 \, V}{0.0002 \, A} = \frac{4 \, V}{0.0002 \, A} = 20,000 \, \Omega \]

6. **Final Answer:**
The internal resistance of the battery is:
\[ r = 20,000 \, \Omega \]