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))`

Explanation

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 potenrial energy decreases by an amount equal to mgh = `10^(-1)xx10xx1=1J` The work done against the buoyant force is, W = buoyant force x displacement
`vpgxxh=4/3pi^3xxpgxxh`
Where p is the density ot water. Substituting thbe values,
`W=4/3xx22/7(5xx10^(-2))^3xx1000xx10`
=5.24J (app)
Total energy E below the water surface is given by subtracting equation (1) from (2)
`therefore` E=5.24-1
=4.24J
This .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.