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}}] $
Show Answer
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 ] $ .
BETA
