Gravitation Question 174

Question: 3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is [DCE 2005]

Options:

A) Zero

B) $ [\frac{3GM}{L^{2}}] $

C) $ [\frac{9GM}{L^{2}}] $

D) $ [\frac{12}{\sqrt{3}},\frac{GM}{L^{2}}] $

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Answer:

Correct Answer: A

Solution:

Due to three particles net intensity at the centre $ [I={\vec I_{A}}+{\vec I_{B}}+{\vec I_{C}}=0] $.

Because out of these three intensities one equal in magnitude and the angle between each other is$ [120{}^\circ ] $ .



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