CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The ratio of the resistance of conductor at temperature\[15{}^\circ C\]to its resistance at temperature\[37.5{}^\circ C\]is\[4:5\]. The temperature coefficient of resistance the conductor is:

    A)  \[\frac{1}{25}{}^\circ {{C}^{-1}}\]            

    B)         \[\frac{1}{50}{}^\circ {{C}^{-1}}\]

    C)  \[\frac{1}{80}{}^\circ {{C}^{-1}}\]            

    D)         \[\frac{1}{75}{}^\circ {{C}^{-1}}\]

    E)  \[\frac{1}{40}{{\,}^{\text{o}}}{{C}^{-1}}\]

    Correct Answer: D

    Solution :

    The resistance at temperature t is \[{{R}_{t}}={{R}_{0}}(1+\alpha t)\]                 \[R_{t}^{}={{R}_{0}}(1+\alpha t)\] \[\Rightarrow \]               \[4=1+15\alpha \]                                         ...(1) \[5=1+37.5\alpha \]                                        ...(2) \[\Rightarrow \]               \[\frac{4}{5}=\frac{1+15\alpha }{1+37.5\alpha }\] \[\Rightarrow \]               \[4(1+37.5\alpha )=5(1+15\alpha )\] \[\Rightarrow \]               \[75\alpha =1\] \[\Rightarrow \]               \[\alpha =\frac{1}{75}{}^\circ {{C}^{-1}}\]


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