Question

A cubical block of edge 10 cm and mass 0.92kg floats on a tank of water with oil of relative density 0.6. The thickness of oil is 4cm above water. When the block attains equilibrium with four of its edge vertical:
(This question has multiple correct options)
A. 1cm of it will be above the free surface of oil.
B. 5cm of it will be underwater.
C. 2cm of it will be above the common surface of oil and water.
D. 8cm will be underwater.

Answer 1

The block will attain equilibrium and be at rest when the two buoyant forces exerted by oil and water will balance the gravitational force acting on the block. Also use the fact that the edges of the block are vertical.

Formula used:
${{F}_{g}}=mg$
where ${{F}_{g}}$ is gravitational force on mass m and g is acceleration due to gravity.
${{F}_{B}}=\rho Vg$
where ${{F}_{B}}$ is buoyant force on a body in a liquid of density $\rho$ with volume V submerged in the liquid.

Complete step by step answer:
The block will experience a gravitational pull (in the downward direction) exerted by the earth. And the two liquids will also exert a force of buoyancy (in the upward direction) on the block. When the net force on the block is zero, it will attain an equilibrium.
In other words, the block will be at rest when the two buoyant forces will balance the gravitational force acting on the block.
i.e. ${{F}_{g}}={{F}_{B,oil}}+{{F}_{B,water}}$ ….. (i)
In this case, $m=0.92kg=920gm$
$\Rightarrow {{F}_{g}}=mg=920g$
Let ${{x}_{1}}$ height of the block be submerged in oil and ${{x}_{2}}$ height of the block be submerged in water.
Here, ${{x}_{1}}+{{x}_{2}}=10cm$ ….. (ii)
Therefore, the volume of the block submerged in oil is ${{V}_{1}}=10\times 10\times {{x}_{1}}=100{{x}_{1}}$.
And that of in water is ${{V}_{2}}=10\times 10\times {{x}_{2}}=100{{x}_{2}}$.
Density of water is ${{\rho }_{2}}=1gc{{m}^{-3}}$.
It is given that the relative density of the oil is 0.6. Then this means that the density of the oil is ${{\rho }_{1}}=1\times 0.6gc{{m}^{-3}}=0.6gc{{m}^{-3}}$.
Now, we know that the force of buoyancy applied the oil is ${{F}_{B,oil}}={{\rho }_{1}}{{V}_{1}}g$ ….. (iii).
Substitute the known values in (iii).
$\Rightarrow {{F}_{B,oil}}=(0.6)100{{x}_{1}}g=60{{x}_{1}}g$
Similarly,
${{F}_{B,water}}={{\rho }_{2}}{{V}_{2}}g$
And we found that ${{V}_{2}}=10\times 10\times {{x}_{2}}=100{{x}_{2}}$ and ${{\rho }_{2}}=1gc{{m}^{-3}}$.
$\Rightarrow {{F}_{B,water}}={{\rho }_{2}}{{V}_{2}}g=100{{x}_{2}}g$.
Now, substitute the known values in (i).
$\Rightarrow 920g=60{{x}_{1}}g+100{{x}_{2}}g$
$\Rightarrow 92=6{{x}_{1}}+10{{x}_{2}}$ ….. (iv)
And from (ii) we get that ${{x}_{1}}=10-{{x}_{2}}$.
Substitute this value in (iv).
$\Rightarrow 92=6(10-{{x}_{2}})+10{{x}_{2}}$
$\Rightarrow 92=60-6{{x}_{2}}+10{{x}_{2}}$
$\Rightarrow {{x}_{2}}=8cm$
Then this means that ${{x}_{1}}=10-{{x}_{2}}=10-8=2cm$.
Therefore, 2cm of the block is submerged in oil, that is above the common surface of oil and water and 8cm of the block is submerged in water.

Hence, options C and D are correct.

Note: Some students may not know what is meant by relative density of a liquid.
Relative density of a liquid is its density with respect to the density of water. It is equal to the ratio of the absolute density of the liquid to that of water. Others point to note that we could write the volume like above only because of the reason that the edges of the block are vertical.