A galvanometer of resistance `50 Omega ` is connected to a battery of `3 V` along with resistance of `2950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 division the above series resistance should be

Options

A. `4450 Omega`
B. `5050 Omega`
C. `5550 Omega`
D. `6050Omega`

Explanation

To solve the problem, we need to find the new series resistance \( R' \) that will reduce the galvanometer's deflection from 30 divisions to 20 divisions.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
The galvanometer has a resistance \( G = 50 \, \Omega \) and is connected in series with a resistor \( R = 2950 \, \Omega \) and a battery of voltage \( V = 3 \, V \).

2. **Current Calculation for Full Scale Deflection**:
The current through the galvanometer when it shows full scale deflection (30 divisions) can be calculated using Ohm's law:
\[ I_1 = \frac{V}{G + R} = \frac{3}{50 + 2950} = \frac{3}{3000} = 0.001 \, A \]

3. **Deflection Proportionality**:
Since the deflection is proportional to the current, we can express this relationship as:
\[ I_1 \propto 30 \quad \text{(for 30 divisions)} \]

4. **Current Calculation for Reduced Deflection**:
We want to find the new current \( I_2 \) that corresponds to a deflection of 20 divisions:
\[ I_2 \propto 20 \]

5. **Setting Up the Equations**:
We can set up the following equations based on the proportionality of current to deflection:
\[ \frac{I_1}{I_2} = \frac{30}{20} \implies I_2 = \frac{2}{3} I_1 \]

6. **Substituting for \( I_2 \)**:
Substitute \( I_1 \) into the equation:
\[ I_2 = \frac{2}{3} \times 0.001 = \frac{2}{3000} = 0.0006667 \, A \]

7. **Finding the New Resistance \( R' \)**:
Now we can write the equation for the new current \( I_2 \) with the new resistance \( R' \):
\[ I_2 = \frac{V}{G + R'} \implies 0.0006667 = \frac{3}{50 + R'} \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 50 + R' = \frac{3}{0.0006667} \implies 50 + R' = 4500 \]

9. **Solving for \( R' \)**:
\[ R' = 4500 - 50 = 4450 \, \Omega \]

### Final Answer:
The resistance that should be connected in series to reduce the deflection to 20 divisions is \( R' = 4450 \, \Omega \).