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A body of mass m is on the top point of a smooth hemisphere of radius 5 m It is released to slide down the surface when its velocity is `5 m//s` . At this instant the angle made by the radius vector of the body with verticle is `( g=10 m//s^(2))`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - C `mv^(2)/(r)=mg cos theta ` `v^(2)=r g cos theta ` `therefore cos theta =(v^(2))/(rg)=(25)/(5xx10)=(1)/(2)` ` theta =60^(@)` |
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