Gravitation Question 315
Three particles P, Q and R are placed as per . 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:
A) $ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in ST direction only
B) $ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in SQ direction and $ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in SU direction
C) $ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in SQ direction only
D)$ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in SQ direction and $ [\frac{\sqrt{3}GM^{2}}{2d^{2}}] $ in ST direction
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Answer:
Correct Answer: C
Solution:
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In horizontal direction $ [Net,force=\frac{G\sqrt{3}mm}{12d^{2}}\cos {{30}^{o}}-\frac{Gm^{2}}{4d}\cos {{60}^{o}}] $
$ [=\frac{Gm^{2}}{8d^{2}}-\frac{Gm^{2}}{8d^{2}}=0] $
In vertical direction, Net force $ [=\frac{G\sqrt{3}m^{2}}{12d^{2}}\cos {{60}^{o}}+\frac{G\sqrt{3}m^{2}}{3d^{2}}+\frac{Gm^{2}}{4d^{2}}\cos {{30}^{o}}] $
$ \[=\frac{\sqrt{3}Gm^{2}}{24d^{2}}+\frac{\sqrt{3}Gm^{2}}{3d^{2}}+\frac{\sqrt{3}Gm^{2}}{8d}\] $
$ \[=\frac{\sqrt{3}Gm^{2}}{d^{2}}\left[ \frac{1+8+3}{24} \right]=\frac{\sqrt{3}Gm^{2}}{2d^{2}}\] $ along SQ-
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