A galvanometer having a resistance of `20Omega` and 30 divisions on both sides has figure of merit. 0.005 ampere /division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 V , is :

Options

A. `120Omega`
B. `100Omega`
C. `80 Omega`
D. `125Omega`

Answer

To solve the problem, we need to determine the resistance that should be connected in series with the galvanometer so that it can be used as a voltmeter for a maximum voltage of 15 V.

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance of the galvanometer, \( R_g = 20 \, \Omega \)
- Figure of merit (sensitivity), \( F = 0.005 \, \text{A/division} \)
- Total divisions on both sides = 30 divisions on each side, so the total maximum current, \( I_{max} = 30 \times 0.005 \, \text{A} = 0.15 \, \text{A} \)
- Maximum voltage, \( V_{max} = 15 \, \text{V} \)

2. **Use Ohm's Law:**
The relationship between voltage, current, and resistance is given by Ohm's Law:
\[ V = I \cdot R \]
Here, the total resistance in the circuit when the galvanometer is connected in series with the additional resistance \( R \) is:
\[ R_{total} = R_g + R \]

3. **Set up the equation:**
We want the maximum voltage (15 V) to be across the total resistance when the maximum current (0.15 A) flows through it:
\[ V_{max} = I_{max} \cdot R_{total} \]
Substituting the known values:
\[ 15 = 0.15 \cdot (20 + R) \]

4. **Solve for \( R \):**
Rearranging the equation:
\[ 15 = 0.15 \cdot (20 + R) \]
Dividing both sides by 0.15:
\[ \frac{15}{0.15} = 20 + R \]
\[ 100 = 20 + R \]
Now, subtract 20 from both sides:
\[ R = 100 - 20 = 80 \, \Omega \]

5. **Conclusion:**
The resistance that should be connected in series with the galvanometer to use it as a voltmeter up to 15 V is \( R = 80 \, \Omega \).

### Final Answer:
The required resistance is \( 80 \, \Omega \).

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