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1. |
The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is : |
Answer» Given `g/9=g(R/(R+h))^(2)rArr R/(R+h)=1/3` `3R=R+hrArr 2R=h` |
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