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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceMedium2 minPYQ_2020

PhysicsMediumsingle choice

Three particles each of mass ' $m_{1}$ ' are placed at the corners of an equilateral triangle of side $\frac{L}{3}$ '. A particle of mass 'm $_{2}$ ' is placed at the mid point of any one side of triangle. Due to the system of particles the force acting on ' $m_{2}$ ' is ( G = Universal constant of gravitation )

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Solution:

Forces on mass $m_{2}$ due to masses at $B$ and $C$ will be equal and opposite and cancel each other. $h = \frac{L}{3} cos 3 0^{\circ} = \frac{L}{3} \frac{3}{2} = \frac{L}{2 3}$ Force on $m_{2}$ due to mass $m_{1}$ at $A$ is given by $F = G \frac{m _{1} m _{2}}{( \frac{L}{2 3} ) ^{2}} = \frac{12 G m _{1} m _{2}}{L ^{2}}$

Stream:MHT_CETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2020

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