A) 1.5R
B) 2.5 R
C) 4.5 R
D) 7.5 R
Correct Answer: D
Solution :
Let at O there will be a collision. If smaller sphere moves x distance to reach at O, then bigger sphere will move a distance of\[\left( 9R-x \right)\]. \[F=\frac{GM\times 5M}{{{(12R-x)}^{2}}}\] \[{{a}_{small}}=\frac{F}{M}=\frac{G\times 5M}{{{(12R-x)}^{2}}}\] \[{{a}_{big}}=\frac{F}{5M}=\frac{GM}{{{(12R-x)}^{2}}}\] \[x=\frac{1}{2}{{a}_{small}}\,{{t}^{2}}=\frac{1}{2}\frac{G\times 5M}{{{(12R-x)}^{2}}}{{t}^{2}}\] ?(i) \[(9R-x)=\frac{1}{2}{{a}_{big}}{{t}^{2}}=\frac{1}{2}\frac{GM}{{{(12R-x)}^{2}}}{{t}^{2}}\] ?(ii) Thus, dividing Eq. (i) by Eq. (ii), we get \[\frac{x}{9R-x}=5\] \[\Rightarrow \] \[x=45R-5x\] \[\Rightarrow \] \[6x=45R\] \[\therefore \] \[x=7.5R\]You need to login to perform this action.
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