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InterviewSolution
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A small ball of density `rho` is immersed in a liquid of density `sigma(gtrho)` to a depth `h` and released. The height above the surface of water up to which the ball will jump is |
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Answer» Volume of ball `V=(m)/(rho)` Acceleration of ball moving in upward direction inside the liquid. `a=(F_("net"))/(m)=("upthrust-weight")/(m)=(V_("total")rho_(1)g-mg)/(m)` `a=(((m)/(rho))(3rho)(g)-mg)/(m)=2g` (upwards) `:.` velocity of ball while crossing the surface `v=sqrt(2ah)=sqrt(4gh)` `:.` the ball will jump to a height `H=(v^(2))/(2g)=(4gh)/(2g)=2h` |
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