A) \[g=3g\]
B) \[g=\frac{g}{9}\]
C) \[g=9g\]
D) \[g=27g\]
Correct Answer: A
Solution :
The acceleration due to gravity on the new planet can be found using the relation \[g=\frac{GM}{{{R}^{2}}}\] ?(i) but \[M=\frac{4}{3}\pi {{R}^{3}}\rho ,\,\rho \] being density. Thus, Eq. (i) becomes \[\therefore \] \[g=\frac{G\times \frac{4}{3}\pi {{R}^{3}}\rho }{{{R}^{2}}}\] \[=G\times \frac{4}{3}\pi R\rho \] \[\Rightarrow \] \[g\propto R\] \[\therefore \,\,\,\,\,\,\,\,\frac{g}{g}=\,\frac{R}{R}\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{g}{g}=\frac{3R}{R}=3\] \[\Rightarrow \,\,\,\,\,\,\,\,g=3g\]
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