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

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 ms-²) (A) 3.24 m(B) 2.24 m(C) 3.25 m(D) 4.24 m
Answer» he 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 mHope it helps :)