1. Understand the assertion

   A choke coil is an inductor with:

   • large inductance $L$
   • very small resistance

   It is used in AC circuits, such as fluorescent/mercury tube fittings, to limit current without much power loss.

   If a mercury tube is directly connected to household AC supply, excessive current may flow once the tube conducts, and the tube can be damaged.

   So, Assertion (A) is true.

2. Check the reason mathematically

   Consider a resistor $R$ (representing the tube in simplified form) in series with an inductor $L$ (the choke coil), connected to an AC source of angular frequency $\omega$.

   The total impedance is

   $$Z = \sqrt{R^2 + (\omega L)^2}.$$

   Hence the current is

   $$I = \frac{V}{\sqrt{R^2 + \omega^2 L^2}}.$$

   Voltage across the resistor is

   $$V_R = IR = V\,\frac{R}{\sqrt{R^2 + \omega^2 L^2}}.$$

   Therefore, the voltage across the tube/resistor is reduced by the factor

   $$\frac{R}{\sqrt{R^2 + \omega^2 L^2}}.$$

   If the choke coil were absent, then effectively $L=0$, so

   $$V_R = V.$$

   Thus Reason (R) is also true.

3. Does the reason explain the assertion?

   Yes. The choke coil reduces the effective voltage across the tube (and hence limits the current) due to its inductive reactance $\omega L$. This is exactly why direct connection without choke can damage the tube.

   So (R) is the correct explanation of (A).

4. Evaluate options

   • A: $(A)$ true, $(R)$ false $\rightarrow$ incorrect
   • B: Both true and $(R)$ correctly explains $(A)$ $\rightarrow$ correct
   • C: Both true but $(R)$ not correct explanation $\rightarrow$ incorrect
   • D: $(A)$ false, $(R)$ true $\rightarrow$ incorrect

Therefore, the correct option is B.