The resistance of a heating is `99Omega` at room temperature. What is the temperature of the element if the resistance is found to be `11 Omega`
(Temperature coefficient of the material of the resistor is `1.7xx10^(-4)^(2).C^(-1)`)

Options

A. `999.9^(@)C`
B. `1005.3^(@)C`
C. `1020.2^(@)C`
D. `1037.1^(@)C`

Answer

D

Explanation

`Here, R_(0)=99Omega, T_(0)=27^(2)C`
`R_(r)=116Omega, alpha=1.7 10^(-4) .^(@)C^(-1)`
`therefore R_(T)=R_(0)(1+alpha(T-T_(0))]`
`therefore (R_(T))/(R_(0))-1=alpha (T-T_(0)) Rightarrow (116)/(99)-1=alpha(T-T_(0))`
`T-T_(0)=(1)/(alpha)[(116-99)/(99)]=(17)/(99alpha)=(1)/(1.7xx10^(-4))xx(17)/(99)`
`therefore T-T_(0)=(10^(5))/(99)=1010.10.^(@)C`
`Rightarrow T=1010.1+T_(0)=1010.1+27=1037.1.^(@)C`