Question

A solid float with $\dfrac{2}{3}$ of its volume immersed in liquid and with $\dfrac{3}{4}$ of its value immersed in another liquid. What fraction of its volume will be immersed if it floats in a homogeneous mixture formed of equal volumes of the liquid:
A) $\dfrac{6}{7}$.
B) $\dfrac{8}{{11}}$.
C) $\dfrac{{11}}{{16}}$.
D) $\dfrac{{12}}{{17}}$.

Answer 1

Hint: Density is defined as the ratio of mass and volume. The buoyancy force is defined as the upthrust which is applied by the liquid on the body which is floating on the liquid. The upthrust will act on the bodies which are floating on the liquid or is partly or completely immersed in the liquid.

Formula used:The formula of the density is given by,
$ \Rightarrow density = \dfrac{{mass}}{{volume}}$

Complete step by step solution:
It is given in the problem that a solid float with $\dfrac{2}{3}$ of its volume immersed in liquid and with $\dfrac{3}{4}$ of its value immersed in another liquid and we need to find the fraction of its volume which is immersed if it floats in a homogeneous mixture formed of equal volumes of the liquid.
Density of the mixture of liquid is${\rho _m}$ and density of liquid 1 is ${\rho _1}$ and density of liquid 2 is ${\rho _2}$.
The solid is $\dfrac{2}{3}V$ in the first liquid and $\dfrac{3}{4}V$ in the second liquid.
The formula of the density is given by,

$ \Rightarrow density = \dfrac{{mass}}{{volume}}$
$ \Rightarrow {\rho _1} = \dfrac{3}{2}\rho $………eq. (1)

And the density of the liquid 2 is equal to,

$ \Rightarrow {\rho _2} = \dfrac{4}{3}\rho $………eq. (2)
$ \Rightarrow 2V{\rho _m} = V{\rho _1} + V{\rho _2}$
$ \Rightarrow 2{\rho _m} = {\rho _1} + {\rho _2}$
$ \Rightarrow {\rho _m} = \dfrac{{{\rho _1} + {\rho _2}}}{2}$………eq. (3)

Since the weight is equal to the buoyancy force as the body is at equilibrium so we get,

$ \Rightarrow V'g{\rho _m} = V\rho g$(since the density of solid is $\rho $).
$ \Rightarrow V'{\rho _m} = V\rho $………eq. (4)

Replacing the value of equation (3) into equation (4) we get,

$ \Rightarrow V'{\rho _m} = V\rho $
$ \Rightarrow V'\left( {\dfrac{{{\rho _1} + {\rho _2}}}{2}} \right) = V\rho $
$ \Rightarrow \dfrac{{V'}}{2}\left( {{\rho _1} + {\rho _2}} \right) = V\rho $
$ \Rightarrow \dfrac{{V'}}{2}\left( {{\rho _1} + {\rho _2}} \right) = V\rho $

Replacing the value of densities in the above relation from equation (1) and equation (2) we get,

$ \Rightarrow \dfrac{{V'}}{2}\left( {{\rho _1} + {\rho _2}} \right) = V\rho $
$ \Rightarrow \dfrac{{V'}}{2}\left( {\dfrac{3}{2}\rho + \dfrac{4}{3}\rho } \right) = V\rho $
$ \Rightarrow \dfrac{{17\rho V'}}{{12}} = V\rho $
$ \Rightarrow \dfrac{{17V'}}{{12}} = V$
$ \Rightarrow V' = V\left( {\dfrac{{12}}{{17}}} \right)$.
The fraction of volume immersed of the total volume is equal to$V' = V\left( {\dfrac{{12}}{{17}}} \right)$.

The correct answer is option D.

Note: The body floats on the liquid because the weight of the body is less than the buoyancy force applied on the body of the liquid. The body is in equilibrium in the different liquids this means that the buoyancy force applied by the mixture of the liquid is equal to the weight of the body.