1.

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).

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`


Discussion

No Comment Found