Correct Answer - B
Let resistance of bulb filament is `R_(0)` at `0^(@)C`, then from expression
`R=R_(0)(1+alphaDeltaT)`
`:. 100=R_(0)(1+0.005xx100)`
`200=R_(0)(1+0.005xxx)`
Where `x` is temperature in `.^(@)C` at which resistance becomes `200 Omega`
Dividing the above two equation
`200/100=(1+0.005 x)/(1+0.005xx100)implies x=400^(@)C`