A) 2.5R
B) 4.5R
C) 7.5R
D) 1.5R
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 \[F=\frac{GM\times 5M}{{{(12R-x)}^{2}}}\] \[{{a}_{small}}=\frac{F}{M}=\frac{G\times 5M}{{{(12R-x)}^{2}}}\] \[{{a}_{big}}=\frac{F}{5M}=\frac{GM}{{{(12R-x)}^{2}}}\] \[x=\frac{1}{2}{{a}_{small}}{{t}^{2}}\] \[=\frac{1}{2}\frac{G\times 5M}{{{(12R-x)}^{2}}}{{t}^{2}}\] ? (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 \[\therefore \] \[\frac{x}{9R-x}=5\] \[\Rightarrow \] \[x=45R-5x\] \[\Rightarrow \] \[6x=45R\] \[\Rightarrow \] \[x=7.5R\]You need to login to perform this action.
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