A cubical block of wood of edge $10 c m$ and mass $0. 92 k g$ floats on a tank of water with oil of rel. density $0.6$ . The thickness of oil is $4 c m$ above water. When the block attains equilibrium with four of its sides edges vertical:

1 c m

of it, will be above the free surface of cal.

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5 c m

of it will be under water.

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2 c m

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

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8 c m

of it will be under water.

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Solution

The correct options are
C $2 c m$ of it will be above the common surface of oil and water.
D $8 c m$ of it will be under water.
at equilibrium weight of the block is equal to buoyancy force.
let $x_{1} i s s u b m e r g e d i n o i l a n d x_{2} i n w a t e r$
$0.92 \times 10 = 0.1 \times 0.1 \times x_{1} \times 600 \times 10 + 0.1 \times .01 \times x_{2} \times 1000 \times 10$
also $x_{1} + x_{2} = 10 / 100 m$
on solving we get $x_{1} = 2 c m a n d x_{2} = 8 c m$

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

A cubical block of wood of edge $10 c m$ and mass $0. 92 k g$ floats on a tank of water with oil of rel. density $0.6$ . The thickness of oil is $4 c m$ above water. When the block attains equilibrium with four of its sides edges vertical: