A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will be

Options

A. `500 Omega`
B. `900 Omega`
C. `1000 Omega`
D. `1800 Omega`

Answer

To solve the problem of determining the resistance required to be added to a galvanometer to convert it into a voltmeter with a specified range, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance of the galvanometer, \( R_g = 100 \, \Omega \)
- Full-scale deflection current, \( I_g = 30 \, \text{mA} = 30 \times 10^{-3} \, \text{A} \)
- Desired voltage range for the voltmeter, \( V = 30 \, \text{V} \)

2. **Understand the Configuration**:
- When converting a galvanometer into a voltmeter, a resistor \( R \) is added in series with the galvanometer. The total resistance of the voltmeter will be \( R_g + R \).

3. **Apply Ohm's Law**:
- According to Ohm's Law, the voltage across the voltmeter can be expressed as:
\[ V = I_g \times (R_g + R) \]

4. **Substitute Known Values**:
- We need to find \( R \) such that:
\[ 30 \, \text{V} = 30 \times 10^{-3} \, \text{A} \times (100 \, \Omega + R) \]

5. **Rearrange the Equation**:
- Dividing both sides by \( 30 \times 10^{-3} \):
\[ \frac{30}{30 \times 10^{-3}} = 100 + R \]
- Simplifying the left side:
\[ 1000 = 100 + R \]

6. **Solve for \( R \)**:
- Subtract \( 100 \) from both sides:
\[ R = 1000 - 100 = 900 \, \Omega \]

7. **Conclusion**:
- The resistance required to be added to the galvanometer to convert it into a voltmeter with a range of 30 V is \( R = 900 \, \Omega \).

### Final Answer:
The resistance required to be added is **900 Ω**.

---