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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The resistance of a bulb filament is\[100\text{ }\Omega \].at a temperature of\[100{}^\circ C\]. If its temperature coefficient of resistance is\[0.005/{}^\circ C,\]its resistance will become\[200\,\Omega \]. at a temperature of     AIEEE  Solved  Paper-2006

    A) \[300{}^\circ C\]              

    B)        \[400{}^\circ C\]              

    C)        \[500{}^\circ C\]              

    D)        \[200{}^\circ C\]

    Correct Answer: B

    Solution :

    Let resistance of bulb filament be\[{{R}_{0}}\]at\[0{}^\circ C,\] then from expression\[R={{R}_{0}}[1+\alpha \Delta \theta ],\]we have \[100={{R}_{0}}[1+0.005\times 100]\]         ...(i) and       \[200={{R}_{0}}[1+0.005\times \times ]\]       ...(ii) where,\[x\]is temperature in\[{}^\circ C\]at which resistance become\[200\,\Omega \]. Dividing Eq. (ii) by Eq. (i), we get \[\frac{200}{100}=\frac{1+0.005x}{1+0.005\times 100}\] \[\Rightarrow \] \[x=400{}^\circ C\]


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