When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 - 10 V is:

We are told that the galvanometer gives a full-scale deflection when a current of $$5 \text{ mA}$$ flows through it, and that the resistance of its moving coil is $$15\;\Omega.$$ A current of $$5 \text{ mA} = 0.005 \text{ A}$$ therefore represents the maximum current that can ever be allowed to pass through the galvanometer.

To convert this galvanometer into a voltmeter of range $$0 \text{ V} \rightarrow 10 \text{ V},$$ we have to put a suitable resistance $$R_s$$ in series with the galvanometer. When the voltmeter is connected across a source that gives the maximum reading (that is, $$10\text{ V}$$), the current through the galvanometer must still be its full-scale current $$I_g = 0.005 \text{ A}.$$

Ohm’s law tells us that

$$V = I \, R_{\text{total}},$$

where $$V$$ is the applied voltage, $$I$$ is the current through the circuit, and $$R_{\text{total}}$$ is the total resistance in that circuit. In the present case, at full-scale deflection we have

$$V_{\max} = I_g \bigl(R_g + R_s\bigr),$$

because the current $$I_g$$ flows through both the coil resistance $$R_g$$ and the series resistance $$R_s.$$ Substituting the known numerical values:

$$10\text{ V} = 0.005\text{ A} \left(15\;\Omega + R_s\right).$$

Now we solve step by step for $$R_s$$. First divide both sides by $$0.005\text{ A}:$$

$$\frac{10\text{ V}}{0.005\text{ A}} = 15\;\Omega + R_s.$$

Perform the division on the left:

$$2000\;\Omega = 15\;\Omega + R_s.$$

Next, subtract $$15\;\Omega$$ from both sides in order to isolate $$R_s$$:

$$R_s = 2000\;\Omega - 15\;\Omega.$$

This gives

$$R_s = 1985\;\Omega.$$

Since $$1985\;\Omega = 1.985 \times 10^{3}\;\Omega,$$ the required resistance value is

$$R_s = 1.985 \times 10^{3}\;\Omega.$$

Hence, the correct answer is Option B.