The resistance of a thin silver wire is `1.0 Omega` at `20^@C`. The wire is placed in a liquid bath and its resistance rises to `1.2 Omega`. What is the temperature of the bath? `alpha` for silver is `3.8xx10^03//(.^(@) C)`

Explanation

`R(T)R_0[1+alpha(T-T_0)]`
Here, `R(T)=1.2Omega, R_0=1.0Omega, alpha=3.8xx10^-3//(.^@C) "and" T_0=2^@C`
substituting the values we have `1.2=1.0[1+3.8x10^-3(T - 20)]`
or `3.8xx10^-3(T-20)=0.2`
solving this we get `T=72.6^@C`