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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    The charge flowing through a resistance R varies with time t as \[Q=at-b{{t}^{2}},\] where a and b are positive constants. The total heat produced in R is:                                                                                          

    A)  \[\frac{{{a}^{3}}R}{6b}\]                                             

    B)  \[\frac{{{a}^{3}}R}{3b}\]

    C)   \[\frac{{{a}^{3}}R}{2b}\]                                            

    D)   \[\frac{{{a}^{3}}R}{b}\]

    Correct Answer: A

    Solution :

                     \[Q=at-b{{t}^{2}}\]                 \[i=a-2bt\]          {for \[i=0\Rightarrow \,t=\frac{a}{2b}\] }                 From joule's law of heating \[dH={{i}^{2}}Rdt\] \[H=\int\limits_{0}^{a/2b}{{{(a-2b)}^{2}}}Rdt\] \[H=\frac{{{(a-2b)}^{3}}R}{-3\times 2b}\left| _{0}^{\frac{a}{2b}} \right.=\frac{{{a}^{3}}R}{6b}\]


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