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
Diffcult
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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 $x_{2}=8 \mathrm{cm}$
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