Which of the following combinations should be used for better tuning…

The correct answer is Option (3) → \( R = 15\,\Omega,\ L = 3.5\,\mathrm{H},\ C = 30\,\mu\mathrm{F} \)

Tuning means picking out one frequency and rejecting its neighbours, so the circuit with the sharpest resonance tunes best. The sharpness of resonance of a series LCR circuit is measured by its quality factor \( Q \): the larger \( Q \), the sharper the resonance.

    \( \displaystyle Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} \)

Working out \( Q \) for each combination (\( C \) in farads):

    \( \displaystyle Q_1 = \frac{1}{25}\sqrt{\frac{1.5}{35\times10^{-6}}} = \frac{207.0}{25} \approx 8.3 \)

    \( \displaystyle Q_2 = \frac{1}{25}\sqrt{\frac{2.0}{45\times10^{-6}}} = \frac{210.8}{25} \approx 8.4 \)

    \( \displaystyle Q_3 = \frac{1}{15}\sqrt{\frac{3.5}{30\times10^{-6}}} = \frac{341.6}{15} \approx 22.8 \)

    \( \displaystyle Q_4 = \frac{1}{15}\sqrt{\frac{1.0}{40\times10^{-6}}} = \frac{158.1}{15} \approx 10.5 \)

Option (3) has the largest quality factor, so it gives the sharpest resonance and the best tuning.

Note: A similar question was also asked in CUET (UG) 2023 on 30-05-2023 (Slot 2).