and 
are shown. The value of\[\frac{{{M}_{1}}}{{{M}_{2}}}\] is [JEE MAIN Held On 08-01-2020 Morning] 
A) \[\frac{1}{6}\]
B) \[\frac{1}{2}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
[a] From the diagram Gravitation field at the surface \[E=\frac{Gm}{{{r}^{2}}}\] \[\therefore {{E}_{1}}=\frac{G{{m}_{1}}}{{{r}^{2}}_{1}}\] and \[{{E}_{2}}=\frac{G{{m}_{2}}}{{{r}^{2}}_{2}}\] \[\therefore \frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)\] \[\therefore \frac{2}{3}={{\left( \frac{2}{1} \right)}^{2}}\left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)\] \[\Rightarrow \left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)=\frac{1}{6}\]
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