Correct Answer - a
Resistance of one bulb, `R_(1) = (220)^(2)//25` ohm
Resistance of other bulb, `R_(2) = (220)^(2)//100` ohm
Total resistance when two bulb are in series will be `R = (220)^(2)//25 + (220)^(2)//100 = (220)^(2)//20 Omega`
Given `I = (V)/(R ) = (440)/((220)^(2)//20) = (2)/(11)` Amp
Potential difference across `25W` bulb
`= IR_(1) = ((220)^(2))/(25) xx (2)/(11) = 352V`
Potential difference across `100W` bulb
`= ((220)^(2))/(100) xx (2)/(11) = 88 V`
The bulb of `25` watt will be fixed become if can tolerate only `220V` while the voltage across it is `352V`