Two point masses 3kg each separated by a distance of 8cm. The net gravitational field intensity at a point 3cm away on the perpendicular bisector from the line joining the two masses is

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$4 \times 10^{- 7} N / k g$

$1.6 \times 10^{- 7} N / k g$

$3 \times 10^{- 7} N / k g$

$6 \times 10^{- 7} N / k g$

answer is B.

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Detailed Solution

$G r a v i t a t i o n a l f i e l d s t r e n g t h = E = \frac{G m}{r^{2}} h e r e u n i v e r s a l g r a v i t a t i o n c o n s \tan t = G; m a s s = m; d i s \tan c e = r; r e s o l v e E i n t o c o m p o n e n t s$

at ‘P’ field due to each of the masses have two components
$E_{1} \cos 37^{0}$ will cancel $E_{2} \cos 37^{0}$
$∴$ Net gravitational field at point P will be
$E_{P} = E_{1} \sin 37^{0} + E_{2} \sin 37^{0}$

convert distance 5cm into meter as $r = 5 \times 10^{- 2} m$

$E_{1} = \frac{G \times 3}{{(5 \times 10^{- 2})}^{2}}; E_{2} = \frac{G \times 3}{{(5 \times 10^{- 2})}^{2}} w e o b s e r v e E_{1} = E_{2} E_{P} = 2 \times E_{1} \sin 37 [∵ E_{1} = E_{2}] E_{P} = 2 \times \frac{G \times 3}{{(5 \times 10^{- 2})}^{2}} E_{P} = \frac{6 \times 6.67 \times 10^{- 11}}{25 \times 10^{- 4} \times 5} E_{P} = 1.6 \times 10^{- 7} N / k g$