A galvanometer has a resistance of $$50\ \Omega$$ and it allows maximum current of $$5$$ mA. It can be converted into voltmeter to measure upto $$100$$ V by connecting in series a resistor of resistance.

Convert a galvanometer (resistance 50 $$\Omega$$, max current 5 mA) into a voltmeter reading up to 100 V.

A galvanometer measures current. To use it as a voltmeter, we connect a high resistance in series so that when the full voltage is applied, only the maximum safe current flows through the galvanometer.

By Ohm’s law, when $$V = 100$$ V is applied and maximum current $$I_g = 5$$ mA = $$5 \times 10^{-3}$$ A flows:

$$ R_{\text{total}} = \frac{V}{I_g} = \frac{100}{5 \times 10^{-3}} = \frac{100}{0.005} = 20000 \text{ } \Omega $$

Subtracting the galvanometer resistance of 50 $$\Omega$$ gives the required series resistance:

$$ R_{\text{series}} = R_{\text{total}} - R_g = 20000 - 50 = 19950 \text{ } \Omega $$

The correct answer is Option C: 19950 $$\Omega$$.