$(a)$ $A$ cube of side $5\, cm$ is immersed in water and then in a saturated salt solution. In which case will it experience a greater buoyant force? If each side of the cube is reduced to $4\, cm$ and then immersed in water,what will be the effect on the buoyant force experienced by the cube as compared to the first case for water? Give reasons for each case.
$(b)$ $A$ ball weighing $4\, kg$ with a density of $4000\, kg\, m^{-3}$ is completely immersed in water of density $10^3\, kg\, m^{-3}$. Find the force of buoyancy on it. (Given $g = 10\, m\, s^{-2}$)

(N/A) $(i)$ The cube will experience a greater buoyant force in the saturated salt solution because the buoyant force is directly proportional to the density of the fluid $(F_b = \rho V g)$. Since the density of the salt solution is greater than that of water,the buoyant force in the salt solution is higher.
$(ii)$ The smaller cube ($4\, cm$ side) will experience a lesser buoyant force because the buoyant force is directly proportional to the volume of the displaced liquid. Since the volume of the smaller cube is less than the initial cube,the volume of water displaced is less,resulting in a lower buoyant force.
$(b)$ The volume of the ball $V = \frac{\text{mass}}{\text{density}} = \frac{4\, kg}{4000\, kg\, m^{-3}} = 10^{-3}\, m^3$.
Buoyant force $(F_b)$ = weight of the liquid displaced = $\text{density of water} \times \text{volume of ball} \times g$.
$F_b = 1000\, kg\, m^{-3} \times 10^{-3}\, m^3 \times 10\, m\, s^{-2} = 10\, N$.