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. `50""^(@)C`
B. `60""^(@)C`
C. `70""^(@)C`
D. `80""^(@)C`

Answer

d

Explanation

We have , `R_(t)=R_(0)(1+alphat)`
Here , `20=R_(0)(1+20alpha)`
`60=R_(0)(1+500alpha)`
`rArr(60)/(20)=(R_(0)(1+500alpha))/(R_(0)(1+20alpha))`
`3=(1+500alpha)/(1+20alpha)`
`rArr3(1+20alpha)=1+500alpha`
`rArr3+60alpha=1+500alpha`
`rArr440alpha=2`
`rArralpha=(2)/(440)=(1)/(220)l^(@)C`
Also given `R_(t)=25^(@)C,t=`?
Now , `(20)/(25)=(R_(0)(1+(20)/(220)))/(R_(0)(1+(t)/(220)))`
`4(1+(t)/(220))=5(1+(20)/(220))`
On solving ,
`t=80^(@)C`