InterviewSolution
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InterviewSolution
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If `g` is acceleration due to gravity on the surface of the earth, having radius `R`, the height at which the acceleration due to gravity reduces to `g//2` isA. R/2B. `sqrt2R`C. `R//sqrt2`D. `(sqrt2-1)R` |
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Answer» Correct Answer - d `(g_(2))/(g_(1))=((R)/(R+h))^(2)` `(g)/(2g)=((R)/(R+h))^(2)` `(1)/(sqrt2)=(R)/(R+h)` `therefore" "R+h=sqrt2R` `h=sqrt2R-R=R(sqrt2-1).` |
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