Question

A cubical iron block of side $ 10cm $ is floating on mercury in a vessel. What is the height of the block above the mercury level?
Given that the density of iron is $ 7.2gm/c.c. $ and density of mercury is $ 13.6gm/c.c. $

Answer 1

Hint : To solve this question, we need to use the Archimedes’ principle. We have to determine the volume of the block submerged inside the mercury in the form of the required height, whose weight has to be balanced with the total weight of the block.

Complete step by step answer
Let us assume the area of the cross section of the block be $ A $ .
According to the question, we have the side of the block $ d = 10cm $ . So the total volume of the block can be written as
 $\Rightarrow V = Ad $ ...............................(1)
 $ \Rightarrow V = 10A $ ……………………………...(2)
Now, we know that the mass of an object is given by
 $\Rightarrow m = \rho V $
According to the question, the density of the block is $ \rho = 7.2gm/c.c. $ Also from (2) we have $ V = 10A $ . Substituting these above, we have the total mass of the block as
 $\Rightarrow m = 72A $
Now, the weight of the block can be written as
 $\Rightarrow W = mg $
 $ \Rightarrow W = 72Ag $ ..................................(3)
Now, let the height of the block above the mercury level be $ h $ . As the total height of the block is equal to $ 10cm $ , so the height of the block submerged under the mercury is given by
 $\Rightarrow h' = 10 - h $
So from (1) we have the volume of the block submerged inside the mercury is
 $\Rightarrow {V_S} = A\left( {10 - h} \right) $
As the block occupy the space inside the mercury by displacing the mercury molecules, so the volume of the mercury displaced by the block will be equal to the volume of the block submerged, that is,
 $\Rightarrow V' = {V_S} $
 $ \Rightarrow V' = A\left( {10 - h} \right) $
The density of the mercury is $ 13.6gm/c.c. $ . So the mass of water displaced is given as
 $\Rightarrow m' = 13.6A\left( {10 - h} \right) $
And the weight of the water displaced is
 $\Rightarrow W' = m'g $
 $ \Rightarrow W' = 13.6Ag\left( {10 - h} \right) $ ...........................(4)
Now, according to the Archimedes’ principle, an object submerged inside a fluid experiences an upward force equal to the weight of the fluid displaced. So the upward force on the block is
 $\Rightarrow U = W' $
From (4)
 $ \Rightarrow U = 13.6Ag\left( {10 - h} \right) $ ..................................(5)
For the equilibrium of the block, the downward weight of the block will be equal to the upward force on the block. So we have
 $\Rightarrow U = W $
From (3) and (5)
 $\Rightarrow 13.6Ag\left( {10 - h} \right) = 72Ag $
Cancelling $ Ag $ from both the sides, we have
 $\Rightarrow 13.6\left( {10 - h} \right) = 72 $
Dividing both sides by $ 13.6 $
 $\Rightarrow 10 - h = \dfrac{{72}}{{13.6}} $
 $ \Rightarrow 10 - h = 5.29 $
On simplifying, we finally get
 $\Rightarrow h = 4.7cm $
Hence the height of the block above the mercury level is equal to $ 4.7cm $ .

Note
There is no need to convert the values of the height and the densities into SI units. All the units have been given in the CGS units, so we preceded our calculations in this system of units only and easily obtained the final answer.