Bulb A rated as (60 W, 120 V) and bulb B rated as (100 W, 220 V), are connected in series to electric mains supply of 220 V, then  Only one correct answerA.Bulb A will glow brighterB.Bulb B will glow brighterC.Both bulbs will glow with same brightnessD.Both bulbs will be fused

To answer this question, we first need to understand the concept of resistance and power in a series circuit.

1. In a series circuit, the total voltage is divided among the components based on their resistance. The bulb with the higher resistance will get a larger share of the total voltage.

2. The power of a bulb is given by the formula P = V^2/R, where P is power, V is voltage, and R is resistance.

3. The resistance of each bulb can be calculated using Ohm's law, R = V/I, where V is voltage and I is current. But since we don't have the current, we can rearrange the power formula to P = V^2/R to find R = V^2/P.

4. For bulb A, the resistance R_A = (120V)^2 / 60W = 240 ohms. For bulb B, the resistance R_B = (220V)^2 / 100W = 484 ohms.

5. Since bulb B has a higher resistance, it will get a larger share of the total voltage when connected in series.

6. Therefore, bulb B will glow brighter than bulb A when connected in series to an electric mains supply of 220 V.

So, the correct answer is B. Bulb B will glow brighter.