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At what height the acceleration due to gravity decreasing by 51 % of its value on the surface of th earth ? |
Answer» As, `g_(h)=g((R)/(R+h))^(2) and g_(h)=g-51%` of `g=g-(51)/(100)g` `:. g_(h)=(49)/(100)g` `:. (49)/(100)g=g((R)/(R+h))^(2)` `:.(7)/(10)=(R)/(R+h)` `:. 7R+7h=10R` `:. 7h=3RrArrh=(3R)/(7)`. |
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