Question

# A body of weight W, experiences an upthrust R in water. What will be the apparent weight of the body and apparent density of the body when W>R?

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Solution

## Step 1: Given dataThe weight of the body is $W$.The upthrust experienced by the body is $R$.Step 2: Concept and formulaeWe know from the concept of Archimedes's principle that, when a body is fully or partially in a fluid, the upward thrust force (buoyancy force) experienced by a body equals the weight of the displaced fluid by the body.In this case, $theapparentweightofthebody\left(W\text{'}\right)=weightofthebody\left(Vg\rho \right)-upthrust\left(Vg\rho \text{'}\right).$If $\rho$ and $\rho \text{'}$ are the densities of water and the body, then the apparent weight of the body due to upthrust is $W\text{'}=Vg\rho -Vg\rho \text{'}$ and the apparent density is ${\rho }_{a}=\left(\rho -\rho \text{'}\right)$, where, V is the volume of the body, g is the acceleration due to gravity.Step 3: Finding the apparent weight and apparent densityNow, according to the question, $W>R$.We know, $apparentweight=bodyweight-upthrust$So, the apparent weight of the body is,$orW\text{'}=W-R\phantom{\rule{0ex}{0ex}}orW\text{'}=Vg\rho -Vg\rho \text{'}$And the apparent density is,${\rho }_{a}=\left(\rho -\rho \text{'}\right)$Therefore, the apparent weight is $\mathbit{W}\mathbf{-}\mathbit{R}$ and apparent density of the body is $\left(\rho -\rho \text{'}\right)$.

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