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(B) The magnitude of the gravitational field intensity at the origin is the sum of the gravitational field intensities due to each mass $m = 1 \ kg$.T

Vedclass · Accepted Answer

$\frac{4G}{3}$

Vedclass · Answer

(B) The magnitude of the gravitational field intensity at the origin is the sum of the gravitational field intensities due to each mass $m = 1 \ kg$.The formula for gravitational field intensity is $I = \frac{Gm}{r^2}$.Summing the intensities due to all masses,we get:$I = G 	imes 1 \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \dots ight)$This is a geometric progression $(GP)$ with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite $GP$ is given by $S = \frac{a}{1 - r}$.Substituting the values:$I = G \left( \frac{1}{1 - 1/4} ight) = G \left( \frac{1}{3/4} ight) = \frac{4G}{3}$.

An infinite number of masses,each of $1 \ kg$,are placed on the $+ve \ X$-axis at $1 \ m, 2 \ m, 4 \ m, \dots$ from the origin. The magnitude of the gravitational field at the origin due to this distribution of masses is:

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