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$\text{Hint:} \ P = V_{\text{rms}} i_{\text{rms}} \cos \phi$ $\text{Step 1: Find the total impedance} \ (Z) \ \text{in the circuit.}$ $\text{The angular frequency} \ (\omega) \ \text{is given by:}$ $\Rightarrow \omega = 2\pi f$ $\text{Substituting the values we get:}$ $\Rightarrow \omega = 2 \times \pi \times 750 = 4712.39 \ \text{rad/s}$ $\text{The inductive reactance} \ (X_L) \ \text{is given by:}$ $X_L = \omega L = 4712.39 \times 0.1803 \approx 850.74 \ \Omega$ $\text{The capacitive reactance} \ (X_C) \ \text{is given by:}$ $X_C = \frac{1}{\omega C} = \frac{1}{4712.39 \times 10 \times 10^{-6}} \approx 21.24 \ \Omega$ $\text{The total impedance} \ (Z) \ \text{in the circuit is given by:}$ $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\text{Substituting the values we get:}$ $Z = \sqrt{100^2 + (850.74 - 21.24)^2} \approx 838.5 \ \Omega$ $\text{Step 2: Find the time in which the resistance will get heated by} \ 10^\circ\text{C}$ $\text{The RMS current} \ (I_{\text{RMS}}) \ \text{can be calculated using Ohm's law:}$ $I_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{20}{838.5} \approx 0.0239 \ \text{A}$ $\text{The power} \ (P) \ \text{dissipated in the resistor is given by:}$ $P = I_{\text{RMS}}^2 \times R = (0.0239)^2 \times 100 \approx 0.057 \ \text{W}$ $\text{The heat} \ (Q) \ \text{required to raise the temperature of the resistor by} \ \Delta \theta \ \text{is given by:}$ $\Rightarrow Q = S \times \Delta \theta = 2 \times 10 = 20 \ \text{J}$ $P = \frac{Q}{T} \Rightarrow T = \frac{Q}{P}$ $\text{Substituting the values we get:}$ $\Rightarrow T = \frac{20}{0.057} \approx 350.88 \ \text{s}$ $\text{Hence, option} \ (4) \ \text{is the correct answer.}$