2. Solved Papers
  3. CET Karnataka Engineering

CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

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


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