Three particles P , Q and R are placed on a straight line as shown in the figure. The masses of P , Q and R are 3 m , 3 m and m respectively. The gravitational force on a fourth particle S of mass m is equal to

Along the direction parallel to TU Net force = G 3 mm 1 2 d 2 cos 30 ∘ - Gm 2 4 d 2 cos 60 ∘ = Gm 2 8 d 2 - Gm 2 8 d 2 = 0 Along the direction perpendicular to TU Net force = G 3 m 2 1 2 d 2 cos 60 ∘ + G 3 m 2 3 d 2 + Gm 2 4 d 2 cos 30 ∘ = 3 Gm 2 2 4 d 2 + 3 Gm 2 3 d 2 + 3 Gm 2 8 d 2 = 3 Gm 2 d 2 1 + 8 + 3 2 4 = 3 Gm 2 2 d 2 along SQ

Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

Three particles $P$ , $Q$ and $R$ are placed on a straight line as shown in the figure. The masses of $P$ , $Q$ and $R$ are $\sqrt{3} m$ , $\sqrt{3} m$ and $m$ respectively. The gravitational force on a fourth particle $S$ of mass $m$ is equal to

Options:

Answer:

Solution:

Along the direction parallel to TU

Net force $= \frac{G \sqrt{3} mm}{12 d^{2}} cos 30^{\circ} - \frac{{Gm}^{2}}{4 d^{2}} cos 60^{\circ}$

$= \frac{{Gm}^{2}}{8 d^{2}} - \frac{{Gm}^{2}}{8 d^{2}} = 0$

Along the direction perpendicular to TU
Net force $= \frac{G \sqrt{3} m^{2}}{12 d^{2}} cos 60^{\circ} + \frac{G \sqrt{3} m^{2}}{3 d^{2}} + \frac{{Gm}^{2}}{4 d^{2}} cos 30^{\circ}$

$= \frac{\sqrt{3} {Gm}^{2}}{24 d^{2}} + \frac{\sqrt{3} {Gm}^{2}}{3 d^{2}} + \frac{\sqrt{3} {Gm}^{2}}{8 d^{2}}$

$= \frac{\sqrt{3} {Gm}^{2}}{d^{2}} [\frac{1 + 8 + 3}{24}] = \frac{\sqrt{3} {Gm}^{2}}{2 d^{2}} along SQ$

Subject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...