A galvanometer having a coil resistance 100 $$\Omega$$ gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?

We are told that the galvanometer coil has an internal resistance $$G = 100\ \Omega$$ and that it produces full-scale deflection when a current of $$I_g = 1\ \text{mA} = 1\times10^{-3}\ \text{A}$$ flows through it.

To convert this galvanometer into a voltmeter that reads full scale at a potential difference of $$V = 10\ \text{V}$$, we must connect an additional resistance $$R_s$$ in series with the galvanometer. In this series arrangement, the same current $$I_g$$ will still correspond to full-scale deflection, but now the total voltage across the series combination must be $$V$$.

Ohm’s law states that for a resistor, $$V = I R.$$ Applying it to the entire series combination, we have $$V = I_g\,(G + R_s).$$

We substitute the known values: $$10\ \text{V} = (1\times10^{-3}\ \text{A})\,(100\ \Omega + R_s).$$

Now we divide both sides by $$1\times10^{-3}\ \text{A}$$ to isolate the bracket: $$\frac{10}{1\times10^{-3}} = 100\ \Omega + R_s.$$

The left side simplifies because $$\frac{10}{1\times10^{-3}} = 10 \times 10^{3} = 10\,000.$$ So we get $$10\,000\ \Omega = 100\ \Omega + R_s.$$

Next, we subtract $$100\ \Omega$$ from both sides to solve for $$R_s$$: $$R_s = 10\,000\ \Omega - 100\ \Omega.$$

Simplifying the subtraction, $$R_s = 9\,900\ \Omega.$$

Expressing this resistance in kilohms: $$R_s = 9.9\ \text{k}\Omega.$$

Hence, the correct answer is Option D.