Gravitation

JEE Main level, single correct, medium

101 of 223

Question 101 of 223

Net gravitational force at the center of a square is found to be F1F_{1} when four particles having mass M,2M,3MM, 2 M, 3 M and 4M4 M are placed at the four corners of the square as shown in figure and it is F2F_{2} when the positions of 3M3 M and 4M4 M are interchanged. The ratio F1F2\frac{F_{1}}{F_{2}} is α5\frac{\alpha}{\sqrt{5}}. The value of α\alpha is ____\_\_\_\_.

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Answer: (D) 2

Let k=Gr22k = \frac{G}{r^2\sqrt{2}} where r=a/2r = a/\sqrt{2}.
Net field at center has components along xx and yy.
Config 1 (original):
gx=k(−M+2M+3M−4M)=0g_x = k(-M+2M+3M-4M) = 0,
gy=k(−M−2M+3M+4M)=4kMg_y = k(-M-2M+3M+4M) = 4kM.
F1=4kMF_1 = 4kM.
Config 2 (3M and 4M swapped):
gx=k(−M+2M+4M−3M)=2kMg_x = k(-M+2M+4M-3M) = 2kM,
gy=k(−M−2M+4M+3M)=4kMg_y = k(-M-2M+4M+3M) = 4kM.
F2=kM4+16=25 kMF_2 = kM\sqrt{4+16} = 2\sqrt{5}\,kM.
F1F2=425=25⇒α=2\frac{F_1}{F_2} = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}} \Rightarrow \alpha = 2.

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