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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    The resistance of an electrical toaster has a temperature dependence given by \[R\,\,(T)={{R}_{0}}[1+\alpha (T-{{T}_{0}}]\]in its range of operation. At \[{{T}_{0}}=300\,K,\,R=100\Omega \]. And at\[T=500\,K,\,R=120\Omega \]. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate form300 to 500 K in 30s. The total work done in raising the temperature is:   JEE Main Online Paper (Held On 10 April 2016)

    A) \[400\,In\frac{1.5}{1.3}J\]                               

    B)  300 J

    C) \[200\,In\frac{2}{3}J\]                      

    D) \[400\,In\frac{5}{6}J\]

    Correct Answer: D

    Solution :

                 Bonus              \[\frac{{{(200)}^{2}}}{{{R}_{0}}(1+\alpha (T-{{T}_{0}})}\] T \[\to \]?temperature at 't' T0 \[\to \]?temperature at t = 300 K T\[T-{{T}_{0}}=\frac{500-300}{30}(t)\]              \[T-{{T}_{0}}=\frac{200}{30}t\]              \[T-{{T}_{0}}=\frac{20t}{3}\] \[\int\limits_{0}^{30}{\frac{{{(200)}^{2}}}{100(1+\alpha \frac{20t}{3})}dt=\frac{200\times 200}{100}\,\,\,\int\limits_{0}^{30}{\frac{dt}{1+\frac{20\alpha }{3}}t}}\] \[=\frac{400\times 3}{20\alpha }\ell n\left( \frac{\frac{1+20\alpha }{3}\times 30}{1} \right)\] \[120=100\,(1+\alpha (200))\] \[1+(200)\alpha =\frac{6}{5}\] \[(200\alpha )=\frac{1}{5}\]\[\alpha =\frac{1}{1000}\] \[=60,000\,\ell n\left( \frac{6}{5} \right)\]


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