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A ball of mass m and density rho is immersed in a liquid of density 3rho at a depth and released. To what height will the ball jump up above the surface of liquid? (neglected the resistance of water and air). |
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Answer» SOLUTION :Volume of ball `V=m/(rho)` Acceleration of ball INSIDE the liquid `a=(F_("net"))/m=("upthrust "-"WEIGHT")/m` `a=((m/(rho))(3rho)(g)-mg)/m=2G` (upwards) `:.` velocity of ball while reaching at surface `V=SQRT(2ah)=sqrt(4gh)` `:.` The ball will jump to a height `H=(v^(2))/(2g)=(4gh)/(2g)=2h` |
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