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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2010

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

    A)  1154 K                                 

    B)  1100 K

    C)  1400 K                                 

    D)  1127 K

    Correct Answer: D

    Solution :

    \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{(1+\alpha {{t}_{1}})}{(1+\alpha {{t}_{2}})}\] \[\Rightarrow \]               \[\frac{1}{2}=\frac{(1+0.00125\times 27)}{(1+0.00125\times t)}\] \[\Rightarrow \]               \[t={{854}^{o}}C\] \[\Rightarrow \]               \[T=1127K\]


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