A) \[1:4\]
B) \[1:5\]
C) \[1:2\]
D) \[3:2\]
Correct Answer: C
Solution :
Acceleration due to gravity at earths surface is given by \[g=\frac{GM}{{{R}^{2}}}\] ?..(i) Since, earth is assumed to be spherical in shape, its mass is \[M=\]volume\[\times \]density \[=\frac{4}{3}\pi {{R}^{3}}\rho \] Given, \[{{\rho }_{e}}={{\rho }_{p}}=\rho ,\text{ }{{G}_{p}}=2{{G}_{e}}\] \[\therefore \] \[\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{{{G}_{e}}\left( \frac{4}{3}\pi R_{e}^{3} \right)\rho \times R_{p}^{2}}{R_{e}^{2}\times R_{p}^{3}\times 2{{G}_{e}}}\] \[1=\frac{{{G}_{e}}R_{e}^{3}\times R_{p}^{2}}{R_{e}^{2}\times R_{p}^{3}\times 2{{G}_{e}}}\] \[[\because {{G}_{p}}=2{{G}_{e}}]\] \[1=\frac{{{R}_{e}}}{2{{R}_{p}}}\] \[\Rightarrow \] \[\frac{{{R}_{p}}}{{{R}_{e}}}=\frac{1}{2}\]
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