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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
    A straight rod of length L extends from x=a to \[x=L+a.\] The gravitational force it exerts on a point mass m at x=0, if the mass per unit length of the rod is \[A+B{{x}^{2}},\], is given by [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) \[Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)-BL \right]\]

    B) \[Gm\left[ A\left( \frac{1}{a+L}-\frac{1}{a} \right)-BL \right]\]

    C) \[Gm\left[ A\left( \frac{1}{a+L}-\frac{1}{a} \right)+BL \right]\]

    D) \[Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)+BL \right]\]

    Correct Answer: D

    Solution :

    \[dF=\frac{Gm(\mu dx)}{{{x}^{2}}}\] \[F=Gm\int\limits_{x=a}^{x=(a+L)}{\frac{(A+B{{x}^{2}})}{{{x}^{2}}}}dx\] \[F=Gm\left( A\int\limits_{x=a}^{x=a+L}{{{x}^{-2}}dx+}\int\limits_{x=a}^{x=a+L}{B}.dx \right)\] \[F=Gm\left( A\left[ \frac{-1}{x} \right]_{a}^{a+L}+B[x]_{a}^{a+L} \right)\] \[F=Gm\left( -A\left[ \frac{1}{a+L}-\frac{1}{a} \right]+B[a+L-a] \right)\] \[F=Gm\left( A\left[ \frac{1}{a}-\frac{1}{a+L} \right]+BL \right)\]


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