Question

A piece of paraffin wax of density $ 0.9gc{c^{ - 1}} $ floats on water. A layer of turpentine of density $ 0.87gc{c^{ - 1}} $ is added on top of water until the wax is entirely submerged. The ratio of the volume of wax immersed in water to that in turpentine is

Answer 1

Hint :In order to solve this question, we are going to use the phenomenon of the buoyancy such that the buoyant force is equal to the weight of the liquid displaced , as we are given the densities of the two liquids so, the volume multiplied by the densities give us the required quantities in the equation.
the buoyant force $ {F_b} $ equals the weight of the liquid displaced $ {W_L} $ , hence the equation that can be written is
 $ {F_b} = {W_L} $
Weight of the liquid is: $ specific gravity \times volume $

Complete Step By Step Answer:
In the given question, we are given the density of the paraffin wax as $ 0.9gc{c^{ - 1}} $ and the density of the turpentine is $ 0.87gc{c^{ - 1}} $
Now, as we know that the buoyant force $ {F_b} $ equals the weight of the liquid displaced $ {W_L} $ , hence the equation that can be written is
 $ {F_b} = {W_L} $
Putting the values in this equation, we get,
 $ 0.9V = 1 \times {V_1} + 0.87 \times {V_2} $
We know that
 $ V = {V_1} + {V_2} $
Using this in the equation given above, we get
 $ 0.9\left( {{V_1} + {V_2}} \right) = {V_1} + 0.87{V_2} \\
   \Rightarrow 0.9{V_1} + 0.9{V_2} = {V_1} + 0.87{V_2} \\
   \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{3}{{10}} \\ $
Hence, the ratio of the volume of wax immersed in water to that in turpentine is $ \dfrac{3}{{10}} $ .

Note :
Buoyancy is the phenomenon which causes the objects to float over the surface of the liquids. The buoyant force is equal to the weight of the liquid displaced by the floating objects. Note that the densities given in this question are gram per centimeter cube that signifies the specific gravity of liquid.