InterviewSolution
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InterviewSolution
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A stream-lined body falls through air from a height `h` on the surface of a liquid . Let `d` and `D` denote the densities of the materials of the body and the liquid respectively, if `D gt d`, then the time after which the body will be intantaneously at rest, is:A. `sqrt((2h)/(g))`B. `sqrt((2h)/(g)(D)/(d))`C. `sqrt((2h)/(g)(d)/(D))`D. `sqrt((2h)/(g)) ((d)/(D-d))` |
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Answer» Correct Answer - D Velocity `u` of the body when it enters the liquid is given by `mgh=1/2mu^(2) or u = sqrt(2gh)` Let Volume of the body = `V` Mass of the body = `Vd` weight of the body = `V dg` Mass of liquid displaceed = `VD` weight of liquid displaced = `VDg` Net upward force = `VDg-VDg` `=V_(g)(D-d)` Retardation = `(n et weight)/(mass)` `=(V(D-d)g)/(Vd) = ((D-d)/(d))g` Acceleration `a = -((D-d)/(d))g` Final velocity , `v` in the liquid when the body is instantaneously at rest is zero. let the time taken be `t`. `v = u+at` `0=sqrt(2gh)-((D-d)/(d))g t * ((D-d)/(d))g t =sqrt(2gh)` `:. t= [(d)/(D-d)] sqrt((2h)/(g))`. |
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