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# 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 the reason for each case.

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## Step 1: Given dataSide of the cube, $s=5cm$Side of the cube, $s=0.05m$Volume of the cube, $v=0.000125{m}^{3}$Reduced side of the cube, ${s}_{r}=4cm$Reduced side of the cube, ${s}_{r}=0.04cm$Volume of the reduced cube, ${v}_{r}=0.000064{m}^{3}$Step 2: AssumptionsDensity of water, ${\rho }_{w}=1000kg/{m}^{3}$Density of saturated salt solution$={\rho }_{s}$Buoyant force exerted on the cube of side $5cm$ immersed in water$={f}_{b{w}_{}}$Buoyant force exerted on the cube of side $5cm$ immersed in saturated salt solution$={f}_{bs}$Buoyant force exerted on the cube of side $4cm$ immersed in water$={f}_{4}$Acceleration due to gravity, $g=10m/{s}^{2}$Step 3: Predicting the relative magnitude of the buoyant force exerted on the cube of side $5cm$ immersed in water and saturated salt solutionBuoyant force exerted on the cube of side $5cm$ immersed in water, ${f}_{bw}={\rho }_{w}×g×V$ ………………….(a)Buoyant force exerted on the cube of side $5cm$ immersed in the saturated salt solution, ${f}_{bs}={\rho }_{s}×g×V$ …………………….(b)From equations (a) and (b), it is clear that"the buoyant force exerted on the cube is directly proportional to the density of the liquid in which the cube is immersed.Since acceleration due to gravity and the volume of the cube remains constant, therefore the “density” is the deciding parameter to depict whether the cube experiences more buoyant force in water or saturated salt solution.But the density of the salt solution is always greater than water i.e ${\rho }_{s}>{\rho }_{w}$Therefore, ${f}_{bs}>{f}_{bw}$Hence, the buoyant force exerted on the cube will be more when it is immersed in a saturated salt solution.Step 4: Estimating the buoyant force exerted on the cubes of the side $5cm$ and $4cm$Buoyant force exerted on the cube of the side $5cm$ immersed in water, ${f}_{bw}={\rho }_{w}×g×V$ Substituting the given values in the above equation, we get${f}_{bw}=1000kg/{m}^{3}×10m/{s}^{2}×0.000125{m}^{3}$${f}_{bw}=1.25N$Buoyant force exerted on the cube of the side $4cm$ immersed in water, ${f}_{4}={\rho }_{w}×g×{v}_{r}$Substituting the given values in the above equation, we get${f}_{4}=1000kg/{m}^{3}×10m/{s}^{2}×0.000064{m}^{3}$${f}_{4}=0.64N$Hence, the buoyant force will be more on the cube of the side $5cm$ as compared to the cube of the side $4cm$.  Suggest Corrections  0      Similar questions  Explore more