A) 8R
B) 9R
C) 10R
D) 20R
Correct Answer: B
Solution :
[b] \[g=\frac{GM}{{{R}^{2}}}...(1)\] \[g''=\frac{g}{100}\] \[g''=\frac{GM}{{{(R+H)}^{2}}}\] \[\frac{g}{100}=\frac{GM}{{{(R+H)}^{2}}}...(2)\] by (1) & (2) \[100={{\left[ \frac{R+H}{R} \right]}^{2}}\] \[\frac{R+H}{R}=10\] \[R+H=10R\] \[H=9R\]You need to login to perform this action.
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