A) 2
B) \[\sqrt{2}\]
C) 1
D) zero
Correct Answer: D
Solution :
Given \[OP=OQ=OR=\sqrt{2}m\] The gravitational force on mass 2 kg due to mass 1 kg at P. \[{{F}_{OP}}=G\frac{2\times 1}{{{(\sqrt{2})}^{2}}}=G\] along OP Similarly, and \[{{F}_{OR}}=G\frac{2\times 1}{{{(\sqrt{2})}^{2}}}=G\] along OR \[{{F}_{OQ}}=\cos {{30}^{o}}\] and \[{{F}_{OR}}=\cos {{30}^{o}}\] are equal and acting in opposite directions, then cancel out each other. Then the resultant force on the mass 2 kg at 0 \[F={{F}_{OP}}-({{F}_{OQ}}\sin {{30}^{o}}+{{F}_{OR}}\sin {{30}^{o}})\] \[F=G-\left( \frac{G}{2}+\frac{G}{2} \right)\] F = 0 (zero)You need to login to perform this action.
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