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…

From the source $v = V_0 \sin(300t)$v=V0​sin(300t), the angular frequency is $\omega = 300\text{ rad/s}$ω=300 rad/s.

Capacitance $C = 100\ \mu\text{F} = 10^{-4}\text{ F}$C=100 μF=10−4 F, so $X_C = \dfrac{1}{\omega C}$XC​=ωC1​.

Case 1 ($S_1$S1​ closed, $S_2$S2​ open):

Closing $S_1$S1​ short-circuits $L_2$L2​, leaving an LCR series circuit with $C$C, $R$R and $L_1$L1​. With phase difference $\phi_1 = 30^\circ$ϕ1​=30∘:

$$\tan 30^\circ = \dfrac{\omega L_1 - X_C}{R}$$tan30∘=RωL1​−XC​​

$\dfrac{1}{\sqrt{3}} = \dfrac{\omega L_1 - X_C}{R} \Rightarrow R = \sqrt{3}\,(\omega L_1 - X_C)$3​1​=RωL1​−XC​​⇒R=3​(ωL1​−XC​) ... (1)

Case 2 ($S_2$S2​ closed, $S_1$S1​ open):

Closing $S_2$S2​ short-circuits $L_1$L1​, leaving an LCR series circuit with $C$C, $R$R and $L_2$L2​. With phase difference $\phi_2 = 60^\circ$ϕ2​=60∘:

$$\tan 60^\circ = \dfrac{\omega L_2 - X_C}{R}$$tan60∘=RωL2​−XC​​

$\sqrt{3} = \dfrac{\omega L_2 - X_C}{R} \Rightarrow \omega L_2 - X_C = R\sqrt{3}$3​=RωL2​−XC​​⇒ωL2​−XC​=R3​ ... (2)

Substituting $R$R from (1) into (2):

$$\omega L_2 - X_C = \sqrt{3}\,(\omega L_1 - X_C) \cdot \sqrt{3}$$ωL2​−XC​=3​(ωL1​−XC​)⋅3​

$$\omega L_2 - X_C = 3\omega L_1 - 3X_C$$ωL2​−XC​=3ωL1​−3XC​

$$2X_C = 3\omega L_1 - \omega L_2 = \omega(3L_1 - L_2)$$2XC​=3ωL1​−ωL2​=ω(3L1​−L2​)

Therefore:

$$3L_1 - L_2 = \dfrac{2X_C}{\omega} = \dfrac{2}{\omega^2 C}$$3L1​−L2​=ω2XC​​=ω2C2​

Substituting $\omega = 300\text{ rad/s}$ω=300 rad/s and $C = 10^{-4}\text{ F}$C=10−4 F:

$$\omega^2 C = (300)^2 \cdot 10^{-4} = 9 \times 10^4 \cdot 10^{-4} = 9$$ω2C=(300)2⋅10−4=9×104⋅10−4=9

$$3L_1 - L_2 = \dfrac{2}{9}\text{ H}$$3L1​−L2​=92​ H

Hence, the correct option is $(2)\ \dfrac{2}{9}$(2) 92​.