question_answer

Four pairs (A, B, C, D) of bodies with their masses and separation are given in the following table.

Pair	Masses	Separation
A	20kg, 30kg	1m
B	40kg, 60kg	0.5m
C	30kg, 50kg	2m
D	10kg, 40kg	2.5m

Arrange the above pairs in the decreasing order of magnitude of gravitational force (a) D, C, B, A                       (b) A, D, B, C (c) B, A, C, D                       (d) C, A, D, B

Answer:

\[{{F}_{A}}=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\] \[=\frac{6.67\times {{10}^{-11}}\times 20\times 30}{{{1}^{2}}}=40\times {{10}^{-9}}=4\times {{10}^{-8}}N\] \[{{F}_{B}}=\frac{6.67\times {{10}^{-11}}\times 40\times 60}{{{(0.5)}^{2}}}\] \[=\frac{160}{0.25}\times {{10}^{-9}}=64\times {{10}^{-8}}N=64\times {{10}^{-8}}N\] \[{{F}_{C}}=\frac{6.67\times {{10}^{-11}}\times 30\times 50}{{{2}^{2}}}\] \[=25\times {{10}^{-9}}=2.5\times {{10}^{-8}}N=2.5\times {{10}^{-8}}N\] \[{{F}_{D}}=\frac{6.67\times {{10}^{-11}}\times 10\times 40}{{{(2.5)}^{2}}}\] \[=4.27\times {{10}^{-9}}N=0.427\times {{10}^{-8}}N\] \[{{F}_{A}}=40\times {{10}^{-9}}N;\,\,{{F}_{B}}=64\times {{10}^{-8}};\] \[{{F}_{C}}=2.5\times {{10}^{-8}}N;\,\,{{F}_{D}}=0.427\times {{10}^{-8}}N\] \[{{T}_{1}}=T\]Pairs in the decreasing order of magnitude of gravitational force is B, A, C, D.