The resistance of a bulb filmanet is `100Omega` at a temperature of `100^@C`. If its temperature coefficient of resistance be 0.005 per `^@C`, its resistance will become `200 Omega` at a temperature of

Options

A. `300^@C`
B. `400^@C`
C. `500^@C`
D. `200^@C`

Answer

B

Explanation

(b) `R_1 = R_0 [1+alphaxx 100] = 100` ………(1)
`R_2 = R_0 [1+alphaxx T] = 200` ………(2)
On dividing we get
`200/100 = (1+alphaT)/(1+100alpha)rArr 2 = (1+ 0.005T)/(1+100 xx 0.005) rArr T = 400^@C`.