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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    A solid sphere of mass M and of uniform density and radius 4R is located with its centre at origin, O. Two spheres of equal radii R with their centres at A (2R, 0, 0) and B(0, 2R, 0) respectively are taken out of the solid sphere leaving spherical cavities. The gravitational force at the origin will be (\[R=1\] Unit)

    A)  \[\frac{G{{M}^{2}}}{64}\]                           

    B)  \[\frac{G{{M}^{2}}}{128\sqrt{2}}\]

    C)  zero                                     

    D)  \[\frac{G{{M}^{2}}}{64\sqrt{2}}\]

    Correct Answer: B

    Solution :

    \[{{\vec{F}}_{A}}+{{\vec{F}}_{B}}\,+{{\vec{F}}_{R}}=0\]             \[M=\frac{4}{3}\,\pi {{(4R)}^{3}}\rho \]             \[{{M}_{A}}={{M}_{B}}=\frac{4}{3}\pi ({{R}^{3}})\rho \,=\frac{M}{64}\]             \[{{\vec{F}}_{A}}\,=\frac{G(M/64)m}{{{2}^{2}}}=\frac{GMm}{64\times 4}\]             \[\because \,\,2R=2\times 1=2\] \[|{{\vec{F}}_{A}}\,+{{\vec{F}}_{B}}|\,=\sqrt{\vec{F}_{A}^{2}+\vec{F}_{B}^{2}}\,=\frac{GMm\sqrt{2}}{64\times 4}\] \[|{{\vec{F}}_{R}}\,|\,=|{{\vec{F}}_{A}}+{{\vec{F}}_{B}}|\,=\frac{GMm}{128\sqrt{2}}\]           Here m = M, so \[|{{\vec{F}}_{R}}|\,=\frac{G{{M}^{2}}}{128\sqrt{2}}\]


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