A moving coil galvanometer has resistance `50 Omega` and it indicates full deflection at 4 m A current. A voltmeter is made using this galvanometer and a `5k Omega` resistance. The maximum voltmeter, will be close be :

Options

A. `40 V`
B. `15 V`
C. `20 V`
D. `10 V`

Answer

To solve the problem, we need to determine the maximum reading of a voltmeter made using a moving coil galvanometer with given specifications. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer (R_g) = 50 Ω
- Full-scale deflection current (I_max) = 4 mA = 4 × 10^(-3) A
- Resistance used in the voltmeter (R) = 5000 Ω

2. **Understanding the Circuit:**
- The voltmeter is constructed by connecting the galvanometer in series with an additional resistor (R).
- The total resistance in the circuit when the voltmeter is connected is the sum of the galvanometer resistance and the additional resistor.

3. **Calculate Total Resistance:**
\[ R_{total} = R + R_g = 5000 \, \Omega + 50 \, \Omega = 5050 \, \Omega \]

4. **Calculate the Maximum Voltage (V_max):**
- The maximum voltage that can be measured by the voltmeter can be calculated using Ohm's Law, which states that Voltage (V) = Current (I) × Resistance (R).
- Here, we will use the full-scale deflection current (I_max) to find the maximum voltage.
\[ V_{max} = I_{max} \times R_{total} \]
\[ V_{max} = (4 \times 10^{-3} \, A) \times (5050 \, \Omega) \]

5. **Perform the Calculation:**
\[ V_{max} = 4 \times 10^{-3} \times 5050 = 20.2 \, V \]

6. **Final Result:**
- The maximum reading of the voltmeter will be approximately 20 V.

### Conclusion:
The maximum voltmeter reading will be close to **20 V**.