question_answer

Two fixed point masses of 10 kg and 20 kg are located at (0 m, 0 m) and (5 m, 0 m). Where should a third point mass of 30 kg be kept so that the net force of attraction on 30 kg mass due to point masses 10 kg and 20 kg is zero?

A) At 2.05 m from 10 kg mass

B) At 2.05 m from 20 kg mass

C) At 2.95 m from 10 kg mass

D) At 3 m from 20 kg mass

Correct Answer: A

Solution :

Force on 30 kg mass due to 10 kg mass. \[{{F}_{1}}=6.67\times {{10}^{-11}}\times \frac{10\times 30}{{{x}^{2}}}\] towards left Force on 30 kg mass due to 20 kg mass \[{{F}_{2}}=6.67\times {{10}^{-11}}\times \frac{30\times 20}{{{(5-x)}^{2}}}\]towards right Since the net force of 30 kg mass is zero, \[\therefore \] \[{{\text{F}}_{\text{1}}}\text{=}{{\text{F}}_{\text{2}}}\] \[\therefore 6.67\times {{10}^{-11}}\times \frac{10\times 30}{{{x}^{2}}}\]             \[=6.67\times {{10}^{-11}}\times \frac{30\times 20}{{{(5-x)}^{2}}}\] Or \[{{(5-x)}^{2}}=2{{x}^{2}}\] or \[25+{{x}^{2}}-10x=2{{x}^{2}}\] Or \[{{x}^{2}}+10x-25=0\] \[x=2.05m\]