The deflection in a galnometer falls from 50 divisions to 20 divisions, when a `12Omega` shunt is applied. The galvanometer resistance (in `Omega`) is

Options

A. `18Omega`
B. `24 Omega`
C. `30Omega`
D. `36 Omega`

Explanation

To solve the problem, we need to determine the resistance of the galvanometer (R) using the information provided about the deflections and the shunt resistance.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The initial deflection of the galvanometer (without shunt) is 50 divisions.
- After applying the shunt of resistance 12 Ω, the deflection falls to 20 divisions.

2. **Determine the Current through the Galvanometer**:
- Let \( I_G \) be the current through the galvanometer when it shows 50 divisions.
- Let \( I_S \) be the current through the shunt when the galvanometer shows 20 divisions.
- The change in deflection indicates that the current through the galvanometer has decreased.
- The current through the galvanometer after applying the shunt is proportional to the new deflection:
\[ I_G = 50 \text{ divisions} \quad \text{(initial)} \]
\[ I_G' = 20 \text{ divisions} \quad \text{(after shunt)} \]
- The current through the shunt can be calculated as:
\[ I_S = I_G - I_G' = 50 - 20 = 30 \text{ divisions} \]

3. **Apply the Concept of Equal Potential**:
- The potential drop across the galvanometer (V_G) and the shunt (V_S) must be equal:
\[ V_G = V_S \]
- The potential drop across the galvanometer is given by:
\[ V_G = I_G' \times R = 20 \times R \]
- The potential drop across the shunt is given by:
\[ V_S = I_S \times R_S = 30 \times 12 \]

4. **Set Up the Equation**:
- Equating the two potential drops:
\[ 20R = 30 \times 12 \]

5. **Solve for R**:
- Calculate the right side:
\[ 20R = 360 \]
- Now, divide both sides by 20:
\[ R = \frac{360}{20} = 18 \, \Omega \]

### Final Answer:
The resistance of the galvanometer is \( R = 18 \, \Omega \).