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A body of mass `m` is placed over a smooth inclined plane of inclination theta. Which iis placed over a lift which is moving up with an acceleration `a_(0)` . Base length o f the inclined plane is `L` . Calculate the velocity of the block with respect to lift at the bottom, if it is allowed to slide down from the top o fthe plane from rest.A. `sqrt(2(a_(0)+g)Lsintheta)`B. `sqrt(2(a_(0)+g)Lcostheta)`C. `sqrt(2(a_(0)+g)Ltantheta)`D. `sqrt(2(a_(0)+g)Lcottheta)` |
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Answer» Correct Answer - C Acceleration along the plane with respect to lift is, `a=(a_(0)+g)sintheta` Initial velocity=0 `v^(2)=u^(2)+2as` `v=sqrt(2(a_(0)+g)sintheta(l)/(costheta))` . |
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