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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    A uniform thin rod AB of length L has linear mass density \[\mu \] \[(x)=a+\frac{bx}{L},\]where x is measured from A. If the CM of the rod lies at a distance of \[\left( \frac{7}{12} \right)L\] from A, then a and b are related as:

    A) \[a=2b\]                       

    B) \[2a=b\]

    C) \[a=b\]                

    D)        \[3a=2b\]

    Correct Answer: B

    Solution :

    Centre of mass of the rod is given by: \[{{x}_{cm}}=\frac{\int\limits_{0}^{L}{(ax+\frac{b{{x}^{2}}}{L})dx}}{\int\limits_{0}^{L}{(a+\frac{bx}{L})dx}}\] \[=\frac{\frac{a{{L}^{2}}}{2}+\frac{b{{L}^{2}}}{3}}{aL+\frac{bL}{2}}=\frac{L\left( \frac{a}{2}+\frac{b}{3} \right)}{a+\frac{b}{2}}\] Now      \[\frac{7L}{12}=\frac{\frac{a}{2}+\frac{b}{3}}{a+\frac{b}{2}}\] On solving we get, \[b=2a\]  


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