A) \[1\]
B) \[2.25\]
C) \[4/9\]
D) \[0.12\]
Correct Answer: A
Solution :
The acceleration due to gravity (g) is given by \[g=\frac{GM}{{{R}^{2}}}\] where M is mass, G the gravitational constant and R the radius. Since, planets have a spherical shape \[V=\frac{4}{3}\pi {{r}^{3}}\] Also, mass (M) = volume (V) \[\times \] density \[(\rho )\] \[g=\frac{G\frac{4}{3}\pi {{R}^{3}}\rho }{{{R}^{2}}}\] \[\Rightarrow \] \[g=\frac{4G\pi \rho R}{3}\] Given, \[{{R}_{1}}:{{R}_{2}}=2:3\] \[{{\rho }_{1}}:{{\rho }_{2}}=\frac{3}{2}\] \[\therefore \] \[\frac{{{g}_{1}}}{{{g}_{2}}}=\frac{{{\rho }_{1}}{{R}_{1}}}{{{\rho }_{2}}{{R}_{2}}}=\frac{3}{2}\times \frac{2}{3}=1\]
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