A rubber ball of mass 100 g and radius 5 cm is submerged in water to a depth of 1 m and released. To what height will the ball jump up above the surface of water ? (Take g= 10 m s^(-2))

A rubber ball of mass 100 g and radius 5 cm is submerged in water to a

1.	A rubber ball of mass 100 g and radius 5 cm is submerged in water to a depth of 1 m and released. To what height will the ball jump up above the surface of water ? (Take g= 10 m s^(-2))
Answer» Solution :LET the rubber ball reach to a height 'h' above the surface of WATER. When the ball is taken to a depth of 1 m below the water surface, the GRAVITATIONAL potential energy decreases by an amount equal to MGH `=10^(-1)xx10xx1= 1 J` The work DONE against the buoyant force is, W =buoyant force x displacement `V rho g xx h =(4)/(3)pi r^(3) rho g xx h` Where `rho` is the density of water. Substituting the values, `W=(4)/(3)xx(22)/(7)(5xx10^(-2))^(3)xx1000xx10` 5.24 J (app) Total energy E below the water surface is given by subtracting equation (1) from (2) `therefore E=5.24-1` =4.24 J The 'E' is the potential energy of the ball above the water surface `therefore E=mgh =0.1xx10xxh` Equating the above with equation (3), we get `=0.1xx10xxh=4.24` h=4.24 m `therefore` The height reached by the ball above the water surface is 4.24 m