The resistance of a wire at 300 K is found to be `0.3Omega`. If the temperature co-efficent of resistance of wire is `1.5xx10^(-3)K^(-1)`, the temperature at which the resistance becomes `0.6Omega` is

Options

A. 720 K
B. 345 K
C. 993 K
D. 690 K

Answer

C

Explanation

Here `R_(300)=0.3Omega, R_(t)=0.6Omega`
temperature coefficient of resistance
`alpha=1.5xx10^(-3)K^(-1)`
`:.R_(300)=R_(0)(1+alpha+27)`
`0.3=R_(0)(1+1.5xx10^(-3)xx27)`…………….i
`R_(t)=R_(0)(1+alpha t)`
`0.6=R_(0)(1+1.5xx10^(-3)xxt)` ................ii
Divide ii by i, we get
`0.6/0.3=(1+1.5xx10^(-3)t)/(1+1.5xx10^(-3)xx27)`
`implies2(1+1.5xx10^(-3)xx27)=1+1.5xx10^(-3)t`
`=2+81xx10^(-3)=1+1.5xx10^(-3)t`
`implies2+0.081+1+1.5xx10^(-3)t`
`impliest=1.081/(1.5xx10^(-3))=720^(@)C`
`TK=t^(@)C+273K`
`=(720+273)K=993K`