A cubical block of wood of edge 10 cm and mass 0.92 kg floats on a tank of water with oil of rel. de

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9-1.Fluid Mechanics

hard

A cubical block of wood of edge $10$ $cm$ and mass $0.92$ $kg$ floats on a tank of water with oil of rel. density $0.6$ to a depth of $4$ $cm$ above water. When the block attains equilibrium with four of its sides edges vertical

A

$1$ $cm$ of it will be above the free surface of oil.

B

$8 $ $cm$ of it will be under water.

C

$2$ $cm$ of it will be above the common surface of oil and water.

D

$B$ and $C$ both

Solution

at equilibrium weight of the block is equal to buoyancy force.

let $x_{1}$ is submerged in oil and $x_{2}$ in water

$0.92 \times 10=0.1 \times 0.1 \times x_{1} \times 600 \times 10+0.1 \times .01 \times x_{2} \times 1000 \times 10$

also $x_{1}+x_{2}=10 / 100 \mathrm{m}$

on solving we get $x_{1}=2 \mathrm{cm}$ and $x_{2}=8 \mathrm{cm}$

Standard 11

Physics

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(KVPY-2013)