An object of volume V_0 and density d_0 is pl | vedclass

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(C) According to the principle of flotation, the weight of the object is equal to the weight of the liquid displaced by the submerged part of the obje

Vedclass · Accepted Answer

$\frac{d - d_0}{d}$

Vedclass · Answer

(C) According to the principle of flotation, the weight of the object is equal to the weight of the liquid displaced by the submerged part of the object.Let $V_{in}$ be the volume of the object submerged in the liquid.Weight of the object = $V_0 d_0 g$.Weight of the displaced liquid = $V_{in} d g$.Equating the two: $V_0 d_0 g = V_{in} d g$.This gives $V_{in} = V_0 \frac{d_0}{d}$.The volume of the object outside the liquid is $V_{out} = V_0 - V_{in}$.Substituting $V_{in}$: $V_{out} = V_0 - V_0 \frac{d_0}{d} = V_0 \left( 1 - \frac{d_0}{d} \right) = V_0 \left( \frac{d - d_0}{d} \right)$.The fraction of the volume outside the liquid is $\frac{V_{out}}{V_0} = \frac{d - d_0}{d}$.

An object of volume $V_0$ and density $d_0$ is placed in a liquid of density $d$. What fraction of its volume remains outside the liquid?

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