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Step-by-Step Solution
Step 1: Understand the Problem
We have a moving coil galvanometer (resistance $G = 50\,\Omega$) that reaches full-scale deflection when a current of $4\,\text{mA}$ flows through it. By adding a $5\,\text{k}\Omega$ ($5000\,\Omega$) series resistor, the galvanometer is converted into a voltmeter. We need to find the maximum voltage that this voltmeter can measure (i.e., the voltage drop across both the galvanometer and the series resistor when the galvanometer current is at full-scale deflection).
Step 2: Identify the Relevant Parameters
Galvanometer resistance, $G = 50\,\Omega$.
Series resistor, $S = 5000\,\Omega$.
Full-scale deflection current of the galvanometer, $I_g = 4 \times 10^{-3}\,\text{A}$ (or $4\,\text{mA}$).
Step 3: Write the Formula for Maximum Voltage
The maximum voltage $V$ that can be measured is the total current (at full deflection) multiplied by the total resistance in series (the galvanometer and the series resistor). Therefore,
$V = I_g \times (G + S)$.
Step 4: Calculate the Maximum Voltage
Substitute the known values into the formula:
$V = 4 \times 10^{-3}\,\text{A} \times (50\,\Omega + 5000\,\Omega)$
$V = 4 \times 10^{-3}\,\text{A} \times 5050\,\Omega$
$V = 4 \times 10^{-3} \times 5050\,\text{V}$
$V = 20.2\,\text{V}$
This is approximately $20\,\text{V}.$
Step 5: Conclusion
The maximum voltage that can be measured with the given configuration is about $20\,\text{V}.$ Thus, the correct answer is close to $20\,\text{V}.$
