A) 3:2
B) 4:3
C) 5:3
D) 5:6
Correct Answer: C
Solution :
[c] \[{{\operatorname{M}}_{A}}=\sigma 4\pi {{R}^{2}}_{A},{{M}_{B}}=\sigma 4\pi R\], where a is surface density \[{{\operatorname{V}}_{A}}=\frac{-G{{M}_{A}}}{{{R}_{A}}},\,{{V}_{B}}=\frac{-G{{M}_{B}}}{{{R}_{B}}}\] \[\frac{{{\operatorname{V}}_{A}}}{{{\operatorname{V}}_{\operatorname{B}}}}=\frac{{{M}_{A}}}{{{M}_{\operatorname{B}}}},\,\frac{{{R}_{B}}}{{{R}_{\operatorname{A}}}}=\frac{\sigma 4\pi {{R}^{2}}_{A}}{\sigma 4\pi {{R}^{2}}_{\operatorname{B}}},\frac{{{R}_{B}}}{{{R}_{\operatorname{A}}}}=\frac{{{R}_{\operatorname{A}}}}{{{R}_{\operatorname{B}}}}\] Given \[\frac{{{\operatorname{V}}_{A}}}{{{\operatorname{V}}_{\operatorname{B}}}}=\,\frac{{{R}_{A}}}{{{R}_{B}}}=\frac{3}{4}\]then \[{{R}_{\operatorname{B}}}=\frac{4}{3}{{R}_{\operatorname{A}}}\] for new shell of mass M and radius R \[M={{\operatorname{M}}_{A}}+{{\operatorname{M}}_{\operatorname{B}}}=\sigma 4\pi {{R}^{2}}_{A}+\sigma 4\pi {{R}^{2}}_{\operatorname{B}}\] \[\sigma 4\pi {{R}^{2}}=\sigma 4\pi \left( {{R}^{2}}_{A}+{{\operatorname{R}}^{2}}_{B} \right)\] then \[\frac{\operatorname{V}}{{{\operatorname{V}}_{\operatorname{A}}}}=\frac{M}{M},\,\frac{{{R}_{\operatorname{A}}}}{{{R}_{\operatorname{B}}}}=\frac{\sigma 4\pi \left( {{R}^{2}}_{A}+{{R}^{2}}_{\operatorname{B}} \right)}{{{\left( {{R}^{2}}_{A}+{{R}^{2}}_{\operatorname{B}} \right)}^{1/2}}},=\frac{{{R}_{\operatorname{A}}}}{\sigma 4\pi {{R}^{2}}_{\operatorname{A}}}\]\[\frac{\sqrt{{{R}^{2}}_{A}+{{R}^{2}}_{\operatorname{B}}}}{{{R}_{A}}}=\frac{5}{3}\]
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