CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    The temperature coefficient of resistance of a wire is \[0.00125/{}^\circ C\]. Its resistance is 1 ohm at 300 K. Its resistance will be 2 ohm at:

    A)  1127 K

    B)  1400 K

    C)  1154 K

    D)  1100 K

    Correct Answer: A

    Solution :

    Using relation \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] where \[{{R}_{0}}\] = resistance at \[{{0}^{o}}C\] \[{{R}_{t}}\]= resistance at \[{{t}^{o}}C\] \[\alpha \]= temperature coefficient = 0.00125 Here:  \[{{t}_{1}}=300-273={{27}^{o}}C\] \[\therefore \] \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{1+\alpha {{t}_{2}}}{1+\alpha {{t}_{1}}}\]             \[2=\frac{1+0.00125\times {{t}_{2}}}{1+0.00125\times {{27}^{o}}}\] \[\Rightarrow \]       \[0.00125{{t}_{2}}=2.0675-1\] or          \[{{t}_{2}}=\frac{1.0675}{0.00125}={{854}^{o}}C=1127K\]


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