The resistance of a wire at `20^(@)C` is `20Omega` and at `500^(@)C` is `60Omega`. At which temperature its resistance will be `25Omega`?

Options

A. `160^(@)C`
B. `250^(@)C`
C. `100^(@)C`
D. `80^(@)C`

Answer

D

Explanation

Here, `R_(20)=20Omega,R_(500)=60Omega,R_(T)=25Omega`
`R_(T)=R_(0)(1+alphaT)`
where `alpha` is the temperature coefficient of resistance.
`therefore R_(20)=R_(0)(1+alphaxx20)`
or `20=R_(0)(1+alphaxx20)" "...(i)`
`R_(500)=R_(0)(1+alphaxx500)`
or `60=R_(0)(1+alphaxx500)" "...(ii)`
Divide (ii) by (i), we get
`(60)/(20)=(1+alphaxx500)/(1+alphaxx20)or3+60alpha=1+500alpha`
or `alpha=(2)/(440)=(1)/(220)""^(@)C^(-1)`
Again, `R_(20)=R_(0)(1+alphaxx20)or20=R_(0)(1+(1)/(220)xx20)...(iii)`
`R_(T)=R_(0)(1+alphaT)or25=R_(0)(1+(1)/(220)xxT)" "...(iv)`
Divide (iv) by (iii), we get
`(25)/(20)=((1+(1)/(220)xxT))/((1+(1)/(220)xx20))or4+(4T)/(220)=5+(100)/(220)`
or `(4T)/(220)=1+(5)/(11)=(16)/(11)orT=(16)/(11)xx(220)/(4)=80^(@)C`