Potential Drop Across Rated Bulb in Resistor Circuit

To determine the potential drop across the bulb, we first calculate the resistance of the bulb using its rated voltage and power: $R_b = \frac{V^2}{P} = \frac{(200)^2}{100} = \frac{40000}{100} = 400\ \Omega$

The bulb ($R_b = 400\ \Omega$) is connected in parallel with the $400\ \Omega$ resistor. The equivalent resistance of this parallel combination ($R_p$) is: $R_p = \frac{400 \times 400}{400 + 400} = 200\ \Omega$

This parallel combination is connected in series with the $200\ \Omega$ resistor and a $100\text{ V}$ battery. The total equivalent resistance of the circuit ($R_{\text{total}}$) is: $R_{\text{total}} = 200\ \Omega + R_p = 200\ \Omega + 200\ \Omega = 400\ \Omega$

The total current supplied by the battery is: $I = \frac{V}{R_{\text{total}}} = \frac{100\text{ V}}{400\ \Omega} = 0.25\text{ A}$

The potential drop across the parallel combination, which is equal to the potential drop across the bulb ($V_{\text{bulb}}$), is: $V_{\text{bulb}} = I \times R_p = 0.25\text{ A} \times 200\ \Omega = 50\text{ V}$

Thus, the potential drop across the bulb is $50\text{ V}$, which corresponds to option B.