JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    The resistance of a wire at 300 K is found to be\[0.3\] \[\Omega \]. If the temperature coefficient of resistance of wire is \[1.5\times {{10}^{-3}}{{K}^{-1}}\], the temperature at which the resistance becomes \[0.5\,\Omega \]. is

    A)  720 K                          

    B)  345 K

    C)  993 K                          

    D)  690 K

    Correct Answer: C

    Solution :

    Given \[{{R}_{300}}\,=0.3\,\Omega ,\,\,{{R}_{\tau }}\,=0.6\,\Omega \] \[T=300\,K={{27}^{o}}C\] Temperature coefficient of resistance \[\alpha =1.5\,\times {{10}^{-3}}\,{{K}^{-1}}\] \[{{R}_{300}}\,={{R}_{0}}(1+\alpha \times 27)\] (where \[{{R}_{0}}\] is resistance at \[0{}^\circ C\] or \[273{}^\circ K\].) \[0.3={{R}_{0}}(1+1.5\,\times {{10}^{-3}}\,\times 27)\]            ?(i) Again \[{{R}_{t}}={{R}_{0}}\,(1+\alpha t)\] \[0.6={{R}_{0}}(1+{{10}^{-3}}\times t)\,\]                  ?(ii) Dividing eq. (ii) by eq(i) we get \[\frac{0.6}{0.3}=\frac{1+1.5\times {{10}^{-3}}t}{1+1.5\,\times {{10}^{-3}}\,\times 27}\] \[2(1+1.5\times {{10}^{-3}}\,\times 27\,)\,=1+2.5\,\times {{10}^{-3}}t\] \[t=\frac{1.081}{1.5\,\times {{10}^{-3}}}\,={{720}^{0}}\,C\,=993K\]

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