Step-by-Step Solution:

1. Understand the Given Information:

   • The circuit contains a metal filament lamp (acting as a resistor, R) and a capacitor (C) in series.
   • Supply voltage (rms), $V_s = 200$ V.
   • Supply frequency, $f = 50$ Hz.
   • Power consumed by the lamp, $P_L = 500$ W.
   • Voltage drop across the lamp (rms), $V_R = 100$ V.
   • The circuit is a series RC circuit.
   • We need to find the magnitude of the phase angle, $\varphi$, between the supply voltage and the current.

2. Calculate the Current in the Circuit:

   • The power consumed in an AC circuit is only dissipated by the resistive component. The lamp acts as the resistor.
   • The power consumed by the lamp is given by the formula $P_L = V_R \times I$, where $I$ is the rms current in the circuit.
   • Since the components are in series, the current is the same throughout the circuit.
   • Rearranging the formula to find the current: $I = \frac{P_L}{V_R} = \frac{500 \text{ W}}{100 \text{ V}} = 5 \text{ A}$

3. Analyze the Voltages using a Phasor Diagram:

   • In a series RC circuit, the voltage across the resistor ($V_R$) is in phase with the current ($I$).
   • The voltage across the capacitor ($V_C$) lags the current by $90^\circ$.
   • The supply voltage ($V_s$) is the vector sum of $V_R$ and $V_C$.
   • The relationship between the magnitudes of the voltages is given by the Pythagorean theorem: $V_s^2 = V_R^2 + V_C^2$
   • The phase angle $\varphi$ is the angle between the supply voltage $V_s$ and the current $I$ (or $V_R$).

4. Calculate the Phase Angle ($\varphi$):

   • From the phasor diagram, the power factor of the circuit is given by $\cos \varphi$.
   • The relationship is: $\cos \varphi = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{V_R}{V_s}$
   • We are given $V_R = 100$ V and $V_s = 200$ V.
   • Substituting these values: $\cos \varphi = \frac{100 \text{ V}}{200 \text{ V}} = \frac{1}{2}$
   • Now, we find the angle $\varphi$ whose cosine is $1/2$: $\varphi = \arccos\left(\frac{1}{2}\right) = 60^\circ$

5. Conclusion:

   • The magnitude of the phase angle between the current and the supply voltage is $60^\circ$.
   • The information $\pi\sqrt 3 \approx 5$ and the capacitance value C are not needed to find the phase angle, but could be used to calculate the capacitance itself.

Alternative Method (using Impedance):

1. Calculate Resistance (R) and Impedance (Z):

   • Resistance of the lamp: $R = \frac{V_R}{I} = \frac{100 \text{ V}}{5 \text{ A}} = 20 \, \Omega$.
   • Total impedance of the circuit: $Z = \frac{V_s}{I} = \frac{200 \text{ V}}{5 \text{ A}} = 40 \, \Omega$.

2. Calculate the Phase Angle from the Impedance Triangle:

   • The power factor is also given by the ratio of resistance to impedance: $\cos \varphi = \frac{R}{Z}$
   • Substituting the calculated values: $\cos \varphi = \frac{20 \, \Omega}{40 \, \Omega} = \frac{1}{2}$
   • This again gives: $\varphi = 60^\circ$

The final answer is 60.