An ac source of emf V(t) = V_0 sinω t is put across a pure capacitor.…

The correct answer is Option (3) → \( 2\omega \)

In a pure capacitor the current leads the voltage by \( \pi/2 \). The instantaneous power is the product of the instantaneous voltage and current.

    \( \displaystyle I(t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) = I_0 \cos\omega t \)

    \( \displaystyle P(t) = V(t)\,I(t) = V_0 I_0 \sin\omega t\,\cos\omega t \)

Using \( 2\sin\theta\cos\theta = \sin 2\theta \):

    \( \displaystyle P(t) = \frac{V_0 I_0}{2}\,\sin 2\omega t \)

So the instantaneous power oscillates with angular frequency \( 2\omega \). (Its average over a cycle is zero, which is why a pure capacitor consumes no power.)

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… The instantaneous power supplied to the capacitor is

    \( \displaystyle p_c = i\,v = i_m \cos(\omega t)\, v_m \sin(\omega t) = i_m v_m \cos(\omega t)\sin(\omega t) = \frac{i_m v_m}{2}\sin(2\omega t) \qquad (7.19) \)”