Given V = V₀ sin $$\omega$$t

At resonant frequency, current and voltage will be in same phase i.e., $$\omega L = {1 \over {\omega C}}$$

$$\therefore$$ Resonant frequency $$\omega = {1 \over {\sqrt {LC} }}$$ ...... (i)

Here, L = 1 $$\mu$$H = 10^$$-$$6 H, C = 10^$$-$$6 F

$$\therefore$$ $$\omega = {1 \over {\sqrt {{{10}^{ - 12}}} }} = {10^6}$$ rad s^$$-$$1

So, option (a) is incorrect.

Since, $${X_C} = {1 \over {\omega C}},\,{X_L} = \omega L$$ ....... (ii)

For large $$\omega$$, X_C $$\sim$$ 0, so circuit will behave like a inductive circuit.

So, option (d) is incorrect.

From Eqn. (i)

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

At resonant frequency or the frequency at which the current will be in phase with the voltage is independent of R.

$$\therefore$$ option (b) is correct.

From Eqn. (ii), at $$\omega$$ $$\sim$$ 0, X_C $$\sim$$ $$\infty$$ and X_L = 0

$$\therefore$$ The current flowing through the circuit becomes nearly zero.

So, option (c) is correct.