Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer The temperature coefficient of resistance of a   wire of \[0.00125/{}^\circ C\]. At 300 K, its resistance is 1                 \[\Omega \]. The resistance of the wire will be 2\[\Omega \], at               following temperature

    A)                  1127K                                  

    B)  1128K

    C)                  600 K                                   

    D)  1400 K

    Correct Answer: A

    Solution :

                    According to definition of thermal resistance coefficient                 \[\alpha =\frac{{{\text{R}}_{2}}-{{R}_{1}}}{R{{ & }_{1}}{{t}_{2}}-{{R}_{2}}{{t}_{1}}}\]                 \[0.00125=\frac{2-1}{1\times {{t}_{2}}-2\times {{27}^{0}}}\]                 \[0.00125=\frac{1}{{{t}_{2}}2-{{54}^{0}}}\]                 or \[{{t}_{2}}={{854}^{0}}C\]                           Hence,       \[\begin{align}   & {{t}_{2}}=854+273 \\  & \,\,\,\,\,=1127K \\ \end{align}\]

adversite


You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos