Question

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 ms-²)

(A) 3.24 m

(B) 2.24 m

(C) 3.25 m

(D) 4.24 m​

Answer 1

:

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^−1×10×1 = 1J

The work done against the buoyant force is, W = buoyant force x displacement

Vρg×h=4/3πr^3*ρg×h

Where ρ is the density of water. Substituting the values,

W=43×227(5×10^−2)3×1000×10

5.24 J (approx)

Total energy E below the water surface is given by subtracting equation (1) from (2)

∴E=5.24−1

=4.24 J

The 'E' is the potential energy of the ball above the water surface

∴E=mgh=0.1×10×h

Equating the above with equation (3), we get

=0.1×10×h=4.24

h=4.24 m

∴ The height reached by the ball above the water surface is 4.24 m

Hope it helps :)