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 plan can be found using the relation \[g=\frac{GM}{{{R}^{2}}}\] 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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