An alternating voltage source \(V =260 \sin (628\,t )\) is connected across a pure inductor of \(5\,mH\). Inductive reactance in the circuit is \(......... \Omega\)

Explanation:
Let us calculate the inductive reactance step-by-step based on the Alternating Current chapter of the NCERT Class 12 Physics textbook:

Step 1: Find the angular frequency (\(\omega\))
The standard equation for an alternating voltage source is given by:
\(V = V_m\sin(\omega t)\)
Comparing this standard expression with the given equation \(V = 260\sin(628t)\), we get:
\(\omega = 628\text{ rad s}^{-1}\)

Step 2: Convert the inductance (\(L\)) into standard SI units
The given inductance value is:
\(L = 5\text{ mH} = 5 \times 10^{-3}\text{ H}\)

Step 3: Calculate the inductive reactance (\(X_L\))
The formula for inductive reactance is:
\(X_L = \omega L\)
Substituting the values into the formula:
\(X_L = 628 \times (5 \times 10^{-3})\)
\(X_L = 3140 \times 10^{-3} = 3.14\ \Omega\)

Hence, Option A is the correct choice.