The resistance of the bulb filament is 100 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

Options

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

Answer

A

Explanation

As the resistance of the bulb filament at `T^(@)C` is
`R_(T) = R_(0) (1+alphaT)`
where `R_(0)` is its resistance at `0^(@)C` and `alpha ` is the temperature coefficient of resistance .
` :. R_(T_(1)) = R_(0) (1+alphaT_(1)) " "...(i)`
and `R_(T_(2)) = R_(0) (1+alphaT_(2))" " ....(ii)`
Dividing eqn . (i) by eqn . (ii) , we get
` (R_(T_(1)))/(R_(T_(2))) = (1+alphaT_(1))/(1+alphaT_(2))`
Here , `R_(T_(1)) = 100 Omega , T_(1) = 100^(@)C`
`R_(T_(2)) = 200 Omega , T_(2) = ? `
`alpha = 0.005^(@)C^(-1)`
`:. " " (100 Omega )/(200 Omega) = (1+(0.005^(@)C^(-1))(100^(@)C))/(1+(0.005^(@)C^(-1))T_(2))`
` 1/2 = (1+0.5)/(1+(0.005^(@)C^(-1))T_(2))`
`2(1+0.5) = 1 + (0.005 ^(@)C^(-1))T_(2)`
` 3 = 1 + (0.005 ^(@)C^(-1))T_(2)`
or `(0.005 ^(@)C^(-1))T_(2)=3-1=2 `
`T_(2) = 2/((0.005 ^(@)C^(-1))) = 400^(@)C`