A test mass of 5g is placed at the mid-point on the line between two point masses 5kg and 6 kg separated by 10cm. The net gravitational field at that point where test mass is placed is

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12G towards 5kg

400G toward 6kg

40G towards 5kg

300G towards 6kg

answer is B.

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Detailed Solution

$m_{1} = 5 k g ⇒ m_{2} = 6 k g$

${\vec{E}}_{n e t} = {\vec{E}}_{1} + {\vec{E}}_{2} g r a v i t a t i o n f i e l d s t r e n g t h = E_{1} = \frac{G m_{1}}{r_{1}^{2}} t o w a r d s m_{1} h e r e G = u n i v e r s a l g r a v i t a t i o n c o n s \tan t; r_{1} = d i s \tan c e o f u n i t m a s s f r o m m a s s m_{1} E_{1} = \frac{G (5)}{{(5 \times 10^{- 2})}^{2}} = \frac{10000 G}{5} E_{1} = 2000 G t o w a r d s m_{1} E_{2} = \frac{G m_{2}}{r_{2}^{2}} t o w a r d s m_{2} h e r e r_{2} = d i s \tan c e o f u n i t m a s s f r o m m a s s m_{2} = \frac{G \times 6}{{(5 \times 10^{- 2})}^{2}} = \frac{60000}{25} G = \frac{12000}{5} G = 2400 G t o w a r d s m_{2} ∴ E_{n e t} = E_{2} - E_{1} t o w a r d s m_{2} (6 k g) n e t g r a v i t a t i o n f i e l d s t r e n g t h = E_{n e t} = 2400 G - 2000 G = 400 G t o w a r d s 6 k g$