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The height at which the acceleration due to gravity becomes `g//9` in terms of R the radius of the earth isA. 2RB. `(R )/sqrtr(3)`C. `(R )/(2)`D. `sqrt(2)R` |
Answer» Correct Answer - A `g=(GM)/(r+h)^(2)` acceleration due to gravity at height h ? `(g)/(g)=(GM)/(R^(2)).(R^(2))/(R+h)^(2)=g(r )/(R+h)^(2)` `rarr 1/9=(R )/(R+n)^(2) rarr (R )/(R+h)=1/3` `rarr 3R =R +h rarr 2R =h` |
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