Question

When equal volumes of two liquids are mixed, the specific gravity of mixture is
$\dfrac{7}{2}$. When equal masses of the same liquid are mixed, the specific gravity is $\dfrac{12}{7}$. The specific gravity of two liquids is:
(A) $1$ and $6$
(B) $2$and $6$
(C) $2$and $4$
(D) $3$and $6$

Answer 1

Density of water is always$1$. To find the specific gravity, we should know about the density of the substance. If we are finding specific gravity of the mixture with respect to water then the specific gravity will always be equal to the density of the mixture. Specific gravity is an absolute quantity.

Complete step-by-step answer:
Specific gravity is the ratio between the density of the substance and density of water. It is also known as relative density .Which is given by the below formula –
Specific gravity$=\dfrac{{{\rho }_{object}}}{{{\rho }_{water}}}$…………Eq(1)
Case 1: when equal volumes of the two liquids are mixed.
Volume of first liquid $={{V}_{a}}$
Volume of second liquid $={{V}_{b}}$
But it is given that volume of two liquids is equal
So,${{V}_{a}}={{V}_{b}}$, which will be equal to $V$.
We know that, density of substance $=\dfrac{mass}{volume}$
Density of first substance,${{d}_{a}}=\dfrac{{{m}_{a}}}{{{V}_{a}}}$
$\Rightarrow {{m}_{a}}={{V}_{a}}{{d}_{a}}$………. Eq (2)
Density of second substance,${{d}_{b}}=\dfrac{{{m}_{b}}}{{{V}_{b}}}$
$\Rightarrow {{m}_{b}}={{V}_{b}}{{d}_{b}}$………..Eq (3)
Now specific gravity of mixture $=\dfrac{{{\rho }_{mixture}}}{1}$(as density of water is always one)
So, specific gravity $=\dfrac{{{m}_{a}}+{{m}_{b}}}{{{V}_{a}}+{{V}_{b}}}$…….. Eq(4)
Putting the values of Eq (2) and Eq (3) in Eq (4)
We get,
specific gravity$=\dfrac{{{V}_{a}}{{d}_{a}}+{{V}_{b}}{{d}_{b}}}{{{V}_{a}}+{{V}_{b}}}$.
As volumes are equal, so it will be, specific gravity$=\dfrac{V({{d}_{a}}+{{d}_{b}})}{2V}$
It is given that specific gravity is $\dfrac{7}{2}$ when volumes are equal .
So, $\dfrac{7}{2}=\dfrac{{{d}_{a}}+{{d}_{b}}}{2}$
$\Rightarrow {{d}_{a}}+{{d}_{b}}=7$……..Eq (5)
Case 2: when equal masses of same liquid are mixed.
Mass of first liquid $={{m}_{a}}$
Mass of second liquid $={{m}_{b}}$
But it is given that masses are equal
So, ${{m}_{a}}={{m}_{b}}$ , which will be equal to $m$
Volume of first liquid $={{V}_{a}}$
Volume of second liquid $={{V}_{b}}$
Density of first substance, ${{d}_{a}}=\dfrac{{{m}_{a}}}{{{V}_{b}}}$
$\Rightarrow {{V}_{a}}=\dfrac{{{m}_{a}}}{{{d}_{a}}}$………..Eq (6)
Density of second substance,
$\Rightarrow {{V}_{b}}=\dfrac{{{m}_{b}}}{{{d}_{b}}}$……….Eq (7)
Specific gravity $=\dfrac{{{m}_{a}}+{{m}_{b}}}{{{V}_{a}}+{{V}_{b}}}$
$\Rightarrow$Specific gravity $=\dfrac{m+m}{m\left( \dfrac{1}{{{d}_{a}}}+\dfrac{1}{{{d}_{b}}} \right)}$ (as mass are equal)
Specific gravity $=\dfrac{2{{d}_{a}}{{d}_{b}}}{{{d}_{a}}+{{d}_{b}}}$
It is given that specific gravity is equal to $\dfrac{12}{7}$ when masses are equal
So, $\dfrac{12}{7}=\dfrac{2{{d}_{a}}{{d}_{b}}}{{{d}_{a}}+{{d}_{b}}}$
Putting Eq (5) in above equation , we get
${{d}_{a}}{{d}_{b}}=6$, this will be possible if ${{d}_{a}}=1$ and ${{d}_{b}}=6$
So the specific gravity of the first liquid is $1$ and the second liquid is $6$.

So, the correct answer is “Option A”.

Note: Specific gravity is unit less because the unit of numerator and denominator cancel out each other. Liquids that have light weight than water have specific gravity less than $1$ and the liquid which have heavier weight than water have specific gravity greater than $1$. Specific gravity of oil is less than water therefore it floats when water when oil is mixed in water.