CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

  • question_answer
    The resistance of a wire at 300 K is found to be 0.3 Sl. If the temperature coefficient of resistance of wire is\[1.5\times {{10}^{-3}}{{K}^{-1}}\], the temperature at which the resistance becomes 0.6\[\Omega \] is

    A)  720 K          

    B)  345 K

    C)  993 K          

    D)  690 K

    Correct Answer: C

    Solution :

    Given, \[{{R}_{300}}=0.3\Omega ,{{R}_{t}}=0.6\Omega ,\] \[T=300K={{27}^{o}}C\] Temperature coefficient of resistance, \[\alpha =1.5\times {{10}^{-3}}{{K}^{-1}}\] \[\therefore \] \[{{R}_{300}}={{R}_{0}}(1+\alpha \times 27)\] \[0.3={{R}_{0}}(1+1.5\times {{10}^{-3}}\times 27)\] ?(i) Again,     \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\] \[0.6={{R}_{0}}(1+1.5\times {{10}^{-3}}\times t)\] Dividing Eq. (ii) by Eq. (i), we get \[\frac{0.6}{0.3}=\frac{1+1.5\times {{10}^{-3}}t}{1+1.5\times {{10}^{-3}}\times 27}\] \[\Rightarrow \]\[2(1+1.5\times {{10}^{-3}}\times 27)=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \]\[2+81\times {{10}^{-3}}=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \]\[2+0.081=1+1.5\times {{10}^{-3}}t\] \[\Rightarrow \]\[t=\frac{1.081}{1.5\times {{10}^{-3}}}={{720}^{o}}C=993K\]


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