The figure given below shows an \(LCR\) series circuit with two switches \(S_1\) and \(S_2\). When switch \(S_1\) is closed keeping \(S_2\) open, the phase difference \((\phi)\) between the current and source voltage is \(30°\) and phase difference is \(60°\) when \(S_2\) is closed keeping \(S_1\) open. The value of \((3L_1 - L_2)\) is _______ H.

From the source \(v = V_0 \sin(300t)\), the angular frequency is \(\omega = 300\text{ rad/s}\).
Capacitance \(C = 100\ \mu\text{F} = 10^{-4}\text{ F}\), so \(X_C = \dfrac{1}{\omega C}\).

Case 1 (\(S_1\) closed, \(S_2\) open):
Closing \(S_1\) short-circuits \(L_2\), leaving an LCR series circuit with \(C\), \(R\) and \(L_1\). With phase difference \(\phi_1 = 30^\circ\):
\(\tan 30^\circ = \dfrac{\omega L_1 - X_C}{R}\)
\(\dfrac{1}{\sqrt{3}} = \dfrac{\omega L_1 - X_C}{R} \Rightarrow R = \sqrt{3}\,(\omega L_1 - X_C)\) ... (1)

Case 2 (\(S_2\) closed, \(S_1\) open):
Closing \(S_2\) short-circuits \(L_1\), leaving an LCR series circuit with \(C\), \(R\) and \(L_2\). With phase difference \(\phi_2 = 60^\circ\):
\(\tan 60^\circ = \dfrac{\omega L_2 - X_C}{R}\)
\(\sqrt{3} = \dfrac{\omega L_2 - X_C}{R} \Rightarrow \omega L_2 - X_C = R\sqrt{3}\) ... (2)

Substituting \(R\) from (1) into (2):
\(\omega L_2 - X_C = \sqrt{3}\,(\omega L_1 - X_C) \cdot \sqrt{3}\)
\(\omega L_2 - X_C = 3\omega L_1 - 3X_C\)
\(2X_C = 3\omega L_1 - \omega L_2 = \omega(3L_1 - L_2)\)

Therefore:
\(3L_1 - L_2 = \dfrac{2X_C}{\omega} = \dfrac{2}{\omega^2 C}\)

Substituting \(\omega = 300\text{ rad/s}\) and \(C = 10^{-4}\text{ F}\):
\(\omega^2 C = (300)^2 \cdot 10^{-4} = 9 \times 10^4 \cdot 10^{-4} = 9\)

\(3L_1 - L_2 = \dfrac{2}{9}\text{ H}\)

Hence, the correct option is \((2)\ \dfrac{2}{9}\).