If two bulbs of wattage 60 W and 100 W respectively each rated at 110 V are connected in series with the supply of 220 V, which bulb will fuse ?

To answer this question, we need to understand how the bulbs work in a series circuit and how the voltage is distributed.

1. In a series circuit, the total voltage is the sum of the voltages across each component. In this case, the total supply voltage is 220V, which is divided between the two bulbs.

2. The voltage drop across each bulb is inversely proportional to their resistances. The resistance of a bulb can be calculated using Ohm's law (R = V/I), but since we don't have the current, we can use the power rating and voltage to calculate the resistance using the formula P = V^2/R.

3. Calculate the resistance of each bulb:

   • For the 60W bulb: R1 = (110V)^2 / 60W = 202.3 ohms
   • For the 100W bulb: R2 = (110V)^2 / 100W = 121.0 ohms

4. In a series circuit, the bulb with the higher resistance gets a higher voltage drop. So, the 60W bulb (with higher resistance) will have a higher voltage drop.

5. If the voltage drop across the 60W bulb exceeds its rated voltage of 110V, it will fuse.

So, in this case, the 60W bulb will fuse when connected in series with a 100W bulb to a 220V supply.