For a series resonant circuit, what happens to the Q–factor when the capacitance of the circuit is increased three times and the frequency is slashed by four times?

Correct option is D

​$$Q = \frac{1}{\omega_0 C R} \\[6pt]Q_1 = \frac{1}{2 \pi f_0 C R} \hspace{20pt} \text{..................(i)} \\[6pt]\text{According to the question} \\[6pt]Q = \frac{1}{2 \pi \frac{f_0}{4} \times 3 C R} = \frac{4}{6 \pi f_0 C R} \hspace{20pt} \text{..................(ii)} \\[6pt]\text{From equation (i) and (ii):} \\[6pt]\frac{Q_2}{Q_1} = \frac{4}{6 \pi f_0 C R} \times \frac{2 \pi f_0 C R}{1} \\[6pt]Q_2 = 1.33 Q_1$$​​