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

<p>To convert a galvanometer into a voltmeter to measure higher voltages, you need to add a series resistance to it. Let's figure out the required resistance value.</p> <p>First, let's find the maximum voltage that can be directly measured by the galvanometer without any additional resistance. We know the maximum current, $I$, that the galvanometer can safely measure is $5 \text{ mA}$, and the resistance of the galvanometer, $R_g$, is $50 \Omega$.</p> <p>Using Ohm's law $ V = I \times R $, the maximum voltage $V_g$ the galvanometer can measure is:</p> <p>$ V_g = I \times R_g $</p> <p>$ V_g = 5 \times 10^{-3} \text{ A} \times 50 \Omega $</p> <p>$ V_g = 0.25 \text{ V} $</p> <p>Next, to measure up to $100 \text{ V}$, we need to add a series resistor $R_s$ so that the total voltage drop when the maximum current is flowing is $100 \text{ V}$. The voltage drop across the additional resistor $R_s$ when the maximum current flows would be the total voltage minus the voltage across the galvanometer:</p> <p>$ V_s = V_{\text{total}} - V_g $</p> <p>$ V_s = 100 \text{ V} - 0.25 \text{ V} $</p> <p>$ V_s = 99.75 \text{ V} $</p> <p>Applying Ohm's law to the series resistor to find its resistance value:</p> <p>$ R_s = \frac{V_s}{I} $</p> <p>$ R_s = \frac{99.75 \text{ V}}{5 \times 10^{-3} \text{ A}} $</p> <p>$ R_s = 19950 \Omega $</p> <p>Therefore, the resistance of the series resistor required to convert the galvanometer into a voltmeter that can measure up to $100 \text{ V}$ is $19950 \Omega$.</p> <p>The correct option is D: $19950 \Omega$.</p>