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  • A cubical block of density ρ is floating on the surface of water. Out of its height L, fraction x is submerged in water.

A cubical block of density \rho is floating on the surface of water. Out of its height L, fraction x is submerged in water. The vessel is in an elevator accelerating upwards with acceleration a. What is the fraction immersed?

Answers (1)

Through the principle of floatation when the block gets submerged in water

Vpg = V^{'} Pw g

Where V’ = volume of water displaced by block

V’ = base area of block x height of block inside the water

V^{'} = L^{2}x

Let V be the volume of the block and Pb by the density of the block

Hence,

V= L^{2}\times Pw

\frac{Pb}{Pw }=\frac{x}{L} ---------(1)

x =\frac{Pb}{Pw} L

as the block lifts upward, the acceleration = (g + a)

weight of L3 Pb block = m(g+a) = V \times Pb(g+a) = L^{3} Pb (g+a)

let x1 be the part of block which is submerged.

We know that the buoyant force will be equal and opposite to the weight of the block.

L^{3} Pb (g+a) = x1 L^{2} Pw (g+a)

So,

\frac{Pb}{Pw}=\frac{x1}{L}

So, x1=\frac{Pb}{Pw}L ------------(2)

From (1) and (2) we can infer that\frac{Pb}{Pw}L is the part of the block which is submerged in water. This is independent of the acceleration of the block (upwards, downwards or at rest)

Posted by

infoexpert24

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