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 \[|{{\vec{F}}_{R}}|\,=\frac{G{{M}^{2}}}{128\sqrt{2}}\]You need to login to perform this action.
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