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 determine the new series resistance required to reduce the galvanometer's deflection from 30 divisions to 20 divisions while connected to a 3V battery.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of galvanometer, \( R_g = 50 \, \Omega \)
- Initial series resistance, \( R_s = 2950 \, \Omega \)
- Voltage of the battery, \( V = 3 \, V \)
- Initial deflection (current), \( I_{initial} \) corresponds to 30 divisions.
- Final deflection (current), \( I_{final} \) corresponds to 20 divisions.

2. **Calculate Initial Current:**
The total resistance in the circuit when the galvanometer shows full scale deflection (30 divisions) is:
\[ R_{total, initial} = R_g + R_s = 50 \, \Omega + 2950 \, \Omega = 3000 \, \Omega \]
Using Ohm's law, the initial current \( I_{initial} \) can be calculated as:
\[ I_{initial} = \frac{V}{R_{total, initial}} = \frac{3 \, V}{3000 \, \Omega} = 0.001 \, A \, (or \, 1 \, mA) \]

3. **Calculate Final Current:**
The final current \( I_{final} \) when the deflection is reduced to 20 divisions can be expressed in terms of the new series resistance \( R \):
\[ I_{final} = \frac{V}{R + R_g} = \frac{3 \, V}{R + 50 \, \Omega} \]

4. **Set Up the Ratio of Currents:**
Since the deflections are proportional to the currents, we can set up the following ratio:
\[ \frac{I_{initial}}{I_{final}} = \frac{30}{20} = \frac{3}{2} \]
Substituting the expressions for current:
\[ \frac{0.001 \, A}{\frac{3 \, V}{R + 50}} = \frac{3}{2} \]

5. **Cross Multiply and Solve for R:**
Cross multiplying gives:
\[ 0.001 \cdot (R + 50) = \frac{3}{2} \cdot 3 \]
Simplifying:
\[ 0.001(R + 50) = 4.5 \]
\[ R + 50 = \frac{4.5}{0.001} = 4500 \]
\[ R = 4500 - 50 = 4450 \, \Omega \]

6. **Conclusion:**
The new series resistance required to reduce the galvanometer deflection to 20 divisions is:
\[ R = 4450 \, \Omega \]