A) \[2.5\] R
B) \[4.5\] R
C) \[7.5\] R
D) \[1.5\] R
Correct Answer: C
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 (9R - x). By universal law of gravitation, \[F=\frac{GM\times 5M}{{{(12R-x)}^{2}}}\] Acceleration of smaller sphere, \[{{a}_{small}}=\frac{F}{M}=\frac{G\times 5M}{{{(12R-x)}^{2}}}\] Acceleration of bigger sphere, \[{{a}_{big}}=\frac{F}{5M}=\frac{GM}{{{(12R-x)}^{2}}}\] \[\Rightarrow \] \[x=\frac{1}{2}{{a}_{small}}{{t}^{2}}=\frac{1}{2}\frac{G\times 5M}{{{(12R-x)}^{2}}}{{t}^{2}}\] ?. (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 \[\frac{x}{9R-x}=5\] \[\Rightarrow \] \[x=45R-5x\] \[\Rightarrow \] \[6x=45R\,\,\,\,\Rightarrow \,\,\,\,x=7.5R\]You need to login to perform this action.
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