The reactance of a 25μF capacitor at the AC frequency of 4000Hz is
(A) $ \dfrac{5}{\pi }\Omega $
(B) $ \sqrt {\dfrac{5}{\pi }} \Omega $
(C) $ 10\Omega $
(D) $ \sqrt {10} \Omega $

Hint
The capacitor is a component which has the ability or “capacity” to store energy in the form of an electrical charge producing a potential difference (Static Voltage) across its plates, much like a small rechargeable battery.

Complete step by step answer
We know the reactance of a capacitor is represented as,
 $\Rightarrow {X_C} = \dfrac{1}{{\omega C}} $
Or, we can say,
 $\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}} $ [As , ω = 2πf]
We know the value of f = 4000 Hz and the capacitor C = 25 muF
Now, put the values on the above equation
 $ \Rightarrow {X_C} = \dfrac{1}{{2\pi \times 4000 \times 25 \times {{10}^{ - 6}}}} $
 $ \therefore {X_C} = \dfrac{5}{\pi }\Omega $.
Option (A) is correct.

Additional Information
A capacitor (originally known as a condenser) is a passive two-terminal electrical component used to store energy electrostatically in an electric field. The forms of practical capacitors vary widely, but all contain at least two electrical conductors (plates) separated by a dielectric (i.e., insulator).

Note
Capacitive Reactance is the complex impedance of a capacitor whose value changes with respect to the applied frequency. In the RC Network tutorial we saw that when a DC voltage is applied to a capacitor, the capacitor itself draws a charging current from the supply and charges up to a value equal to the applied voltage.