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**Hint:**Using the formula for voltage of LCR circuits and applying the boundary conditions of resonance, we will be able to calculate the frequency at which voltage drop across $R$ is maximum.

**Formula Used:**

Formula for voltage of LCR circuits: ${V_R} = R\dfrac{V}{{\sqrt {({R^2} + {{(\omega L - \dfrac{1}{{\omega C}})}^2})} }}$.

Where $R$ is the voltage drop across resistor and is expressed in Volt $(V)$, $V$ is the voltage supplied and is expressed in Volt $(V)$, $L$ inductance and is expressed in Henry $(H)$, $R$ is the resistance value and is expressed in Ohms $(\Omega )$, $C$ is the capacitance value and is expressed in microFarads $(\mu F)$ and $\omega $ is the resonant frequency value and is expressed in Hertz $(Hz)$.

**Complete step by step answer:**

Relation between frequency and $\omega = 2\pi f$

Where $f$ is the frequency value in any condition and is expressed in Hertz $(Hz)$ $\omega $ is the resonant frequency value and is expressed in Hertz $(Hz)$.

Step by step solution: When the switch is closed, current flows in the circuit due to the battery connected to it. A constant $rms$ value $V$ and a variable frequency $f$is generated.

The voltage drop across the inductor is expressed as ${V_R} = R\dfrac{V}{{\sqrt {({R^2} + {{(\omega L - \dfrac{1}{{\omega C}})}^2})} }}$.

This value of ${V_R}$ will be maximum, that is, equal to the supplied voltage $V$ at resonance condition. In this condition the inductive and capacitive reactances are equal and have a cancelling effect on each other due to their $180^\circ $ phase difference.

Hence, in the above equation, ${V_R}$ will be maximum when $\omega L - \dfrac{1}{{\omega C}} = 0$.

Applying this deduction we get,

$

\omega L - \dfrac{1}{{\omega C}} = 0 \\

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

$

Applying square root on both sides,

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

Substituting the necessary values in resonance condition we get,

$

\omega = \dfrac{1}{{\sqrt {\dfrac{1}{\pi } \times {{10}^{ - 6}} \times \dfrac{1}{\pi }} }} \\

\Rightarrow \omega = \pi \times {10^3}rad \\

\\

$

We know that $\omega = 2\pi f$

Therefore, substituting the value of $\omega $ from above we get,

$

\omega = 2\pi f \\

\Rightarrow f = \dfrac{\omega }{{2\pi }} = \dfrac{{\pi \times {{10}^3}}}{{2\pi }} \\

\Rightarrow f = 500Hz \\

$

**In conclusion, the correct option is B.**

**Note:**At the boundary condition, that is, resonance the inductive and capacitive reactances are equal and have a cancellation effect on each other due to their $180^\circ $ phase difference. Therefore, they are considered $\dfrac{1}{\pi }$ and not simply $\pi $.

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