InterviewSolution
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InterviewSolution
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A wooden ball of density ` rho` is immersed in a liquid of density `sigma` to a depth H and then released. The height h above the surface of which the ball rises will beA. HB. `(sigma)/(rho)`C. `((sigma-rho)/(rho))H`D. `(rho)/(sigma)H` |
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Answer» Correct Answer - C Let the volume of wooden ball be V. Weight of the ball `=V rho g`. Upward thrust on the ball due to the liquid `=V sigma g`. Net upward force on the ball `=V sigma g- V rho g= V g(sigma-rho)` Upward ac celeration in the ball `a=("force")/(mass)=(V g(sigma-rho))/(V rho)=((sigma-rho)g)/(rho)` Velocity on reaching the surface `v=sqrt(2 a H)` , If h is the height to which the ball rises above the surface of liquid, then `v=sqrt(2 gh)` . Hence `sqrt(2 a H)=sqrt(2 gh)` or `2a H=2 gh` or `h=(a H)/(g)=((sigma-rho)/(rho))(g H)/(g)= ((sigma-rho)/(rho))H` |
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