A) 1/2 R
B) 2 R
C) 4 R
D) ¼ R
Correct Answer: A
Solution :
[a] \[g=\frac{GM}{{{R}^{2}}}\text{ }also\text{ }M=d\times \frac{4}{3}\pi {{R}^{3}}\] \[\therefore g=\frac{4}{3}d\pi R\]at the surface of planet \[{{g}_{p}}=\frac{4}{3}\left( 2d \right)\pi R',{{g}_{e}}=\frac{4}{3}\left( d \right)\pi R\] \[{{g}_{e}}={{g}_{p}}\Rightarrow dR=2d\,R'\Rightarrow R'=R/2\]You need to login to perform this action.
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