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(D) The gravitational field intensity $I$ due to a point mass $m$ at a distance $r$ is given by $I = \frac{Gm}{r^2}$.Since there are infinite masses,t

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(D) The gravitational field intensity $I$ due to a point mass $m$ at a distance $r$ is given by $I = \frac{Gm}{r^2}$.Since there are infinite masses,the total gravitational field intensity at point $O$ is the sum of the intensities due to each mass:$I_{total} = \sum \frac{Gm_i}{r_i^2} = G \sum \frac{3}{r_i^2} = 3G \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \dots ight)$.This is a geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.Substituting the values,$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$I_{total} = 3G 	imes \frac{4}{3} = 4G$.

Infinite number of masses each of $3 \ kg$ are placed along a straight line at the distances of $1 \ m, 2 \ m, 4 \ m, 8 \ m, \ldots$ from a point $O$ on the same line. If $G$ is the universal gravitational constant,then the magnitude of gravitational field intensity at $O$ is (in $G$)

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