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**Hint:**It is mentioned in the question that (self-inductance) \[L = \dfrac{1}{2}\] and (inductive reactance) \[{X_L} = 3140\Omega \] . First, we will calculate angular frequency using ${X_L} = \omega L$ . Then, using this value of \[\omega \] in $f = \dfrac{\omega }{{2\pi }}$ , we will calculate the frequency of the inductor.

**Complete step by step solution:**

Self-inductance (L)

It is the capacity of the inductor with the help it opposes the direction of the flow of current. It is because of emf produced in wire on its own. Emf is produced in the opposite direction of applied voltage. It is denoted by L and its S.I. The unit is Henry.

Inductive reactance \[\left( {{X_L}} \right)\] :

Electrical field is produced when the current is passed through the wire. This field is known as an induced field. The process by which it gets induced is known as inductance. The resistance in the flow of current in an inductor is known as Inductive reactance.

${X_L} = \omega L$

$\omega = \dfrac{{{X_L}}}{L}$ …(1)

Putting value of self-inductance and inductive reactance in equation (1), we get

$\omega = \dfrac{{3140}}{{\dfrac{1}{2}}}$

\[\omega = 6140\] radian/sec

$f\, = \dfrac{\omega }{{2\pi }}$ …(2)

Putting value of angular frequency and \[\pi \] as 3.14 in equation (2), we get

$f\, = \dfrac{{6140}}{{2 \times 3.14}}$

$f = 1000Hz$

**Thus option (C) is correct.**

**Note:**

If we have not used the word simple ‘frequency’ instead of ‘angular frequency’ then we might not reach the results. If we have used the value of $\pi $ as \[\dfrac{{22}}{7}\] and not 3.14 and we are provided with different options then we might end up with results as 999.49 instead of 1000Hz. Do not use approximate values of L as 1 otherwise solution will vary.

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