A) \[2R\]
B) \[4R\]
C) \[\frac{1}{4}R\]
D) \[\frac{1}{2}R\]
Correct Answer: D
Solution :
Acceleration due to gravity \[g=\frac{GM}{{{R}^{2}}}\] According to the question \[\frac{G{{M}_{p}}}{R_{p}^{2}}=\frac{G{{M}_{e}}}{R_{e}^{2}}\] \[\Rightarrow \] \[\frac{G\times \frac{4}{3}\pi R_{p}^{3}{{\rho }_{p}}}{R_{p}^{2}}=\frac{G\times \frac{4}{3}\pi R_{e}^{2}{{\rho }_{e}}}{R_{e}^{2}}\] \[{{R}_{p}}{{\rho }_{p}}={{R}_{e}}{{\rho }_{e}}\] \[{{R}_{p}}\times 2{{\rho }_{e}}={{R}_{e}}{{\rho }_{e}}\] \[{{R}_{p}}=\frac{{{R}_{e}}}{2}=\frac{R}{2}\]
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