Question

# A solid of density $5000Kg/{m}^{3}$ weight $\left(0.5Kg\right)×g$ in air. It is completely immersed in water of density $1000Kg/{m}^{3}$. Calculate the apparent weight of the solid in water.

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Solution

## Step1: Given dataThe density of the solid is ${\rho }_{s}=5000Kg/{m}^{3}.$The weight of the solid is ${W}_{s}=\left(0.5Kg\right)×g$The density of water is ${\rho }_{w}=1000Kg/{m}^{3}$Step2: The apparent weight of a body in a solidVolume is the ratio of mass and density, i.e, $V=\frac{m}{\rho }$, where, m is the mass of the body and $\rho$ is the density of the body.We know the apparent weight of a body in a liquid is, ${W}_{a}=Actualweightofthebody-buoyancyforceonthebody$.The buoyancy force exerted on a body in a liquid is equal to the weight of the displaced water due to the body. Buoyancy is defined by the form, ${F}_{b}=\left(V×{\rho }_{w}\right)g$, where, V is the volume of the body and ${\rho }_{w}$ is the density of the liquid, and g is the acceleration due to gravity.Step3: DiagramStep4: Finding the buoyancy forceThe volume of the solid is $V=\frac{m}{\rho }=\frac{0.5}{5000}=10000\phantom{\rule{0ex}{0ex}}orV={10}^{-4}{m}^{3}................\left(1\right)$The buoyancy force exerted on the body when it is in the water is ${F}_{b}=\left(V×{\rho }_{w}\right)g=\left({10}^{-4}\right)×\left(1000Kg\right)g\phantom{\rule{0ex}{0ex}}or{F}_{b}=\left(0.1Kg\right)g.............\left(2\right)$Step5: Finding the apparent weightAs we know, ${W}_{a}=Actualweightofthebody-buoyancyforceonthebody$From equations 1 and 2 we get,The apparent weight of the body is ${W}_{a}=\left(0.5Kg\right)g-\left(0.1Kg\right)g\phantom{\rule{0ex}{0ex}}or{W}_{a}=\left(0.4Kg\right)g$Therefore, the apparent weight of the solid in water $\left(0.4Kg\right)g$.

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