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**Hint:**The inductive reactance and the capacitive (condenser is another name for capacitor) reactance can be equated. Since they are in the same circuit, then the frequency of the current flowing through them is the same.

**Formula used:**In this solution we will be using the following formula;

$ {X_L} = \omega L $ where $ {X_L} $ is the inductive reactance of the inductor, $ \omega $ is the angular frequency, and $ L $ is the inductance value.

$ {X_C} = \dfrac{1}{{\omega C}} $ where $ {X_C} $ is the capacitive reactance, and $ C $ is the capacitance value of the capacitor or condenser.

**Complete step by step solution:**

An inductor and condenser are said to be connected to the same circuit in series, and then, it was said that the reactance of the condenser and the inductor are same. This implies that

$ {X_C} = {X_L} $

But $ {X_C} = \dfrac{1}{{\omega C}} $ where $ C $ is the capacitance value of the capacitor or condenser, and $ \omega $ is the angular frequency of the current flowing in the circuit and

$ {X_L} = \omega L $ , where $ L $ is the inductance value.

Hence,

$ \dfrac{1}{{\omega C}} = \omega L $

Hence, by multiplying both sides by $ \omega $ and dividing by $ L $ , we get

$ {\omega ^2} = \dfrac{1}{{LC}} $

$ \Rightarrow \omega = \dfrac{1}{{\sqrt {LC} }} $

By inserting known values, we have

$ \omega = \dfrac{1}{{\sqrt {{{10}^{ - 3}} \times \left( {10 \times {{10}^{ - 6}}} \right)} }} $ (since $ 1H = {10^{ - 3}}mH $ and $ 1F = {10^{ - 6}}\mu F $ )

Hence, by computation, we get that

$ \omega = \dfrac{1}{{\sqrt {{{10}^{ - 8}}} }} = \dfrac{1}{{{{10}^{ - 4}}}} = {10^4} $

Then, to calculate the inductive reactance, we have

$ {X_L} = \omega L = {10^4} \times {10^{ - 3}} $

By computation it gives us,

$ {X_L} = 10\Omega $

**Hence, the correct option is D.**

**Note:**

Alternatively, since the inductive reactance and capacitive reactance are equal we can decide to calculate the capacitive reactance as follows

$ {X_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{{{10}^4} \times \left( {10 \times {{10}^{ - 6}}} \right)}} $

Hence, by computation, we have that

$ {X_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{{{10}^{ - 1}}}} $

Hence,

$ {X_C} = 10\Omega $

Also, the angular frequency calculated above $ {\omega ^2} = \dfrac{1}{{\sqrt {LC} }} $ is also called the natural frequency or resonant frequency. This is the frequency of oscillation between the inductor and the capacitor. If a current of this frequency flows through the circuit, the oscillation will be set in resonance.

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