Correct Answer - A
For a pure inductance, the current lags behind the voltage by `(pi)/2`.
`:.` For the circuit, `E=E_(0)I_(0) sin omega t sin(omega t = (pi)/2)`
`= E_(0)I_(0) sin omega t (-cos omega t)`
`=E_(0)I_(0) sin omega t cos omega t`
`1/2 E_(0)I_(0)(sin 2 omegat)`
Thus the angular frequency of the instantaneous power is `2 omega`.