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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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