Two liquids when they are mixed in equal volumes resultant density is 9, when they are mixed in equal masses resultant density is 8. Find the densities of liquids
Answers
Answer:
Here the resultant density is,
\begin{gathered}\rho=\dfrac {m_1+m_2}{V_1+V_2}\quad\implies\quad (1)\\\\\\\rho=\dfrac {\rho_1V_1+\rho_2V_2}{V_1+V_2}\end{gathered}
ρ=
V
1
+V
2
m
1
+m
2
⟹(1)
ρ=
V
1
+V
2
ρ
1
V
1
+ρ
2
V
2
If V_1=V_2=V,V
1
=V
2
=V,
\begin{gathered}\rho=\dfrac {\rho_1V+\rho_2V}{V+V}\\\\\\\rho=\dfrac {V(\rho_1+\rho_2)}{2V}\\\\\\\rho=\dfrac {\rho_1+\rho_2}{2}\end{gathered}
ρ=
V+V
ρ
1
V+ρ
2
V
ρ=
2V
V(ρ
1
+ρ
2
)
ρ=
2
ρ
1
+ρ
2
So the resultant density is the arithmetic mean of the two individual densities at same volume.
From (1),
\rho=\dfrac {m_1+m_2}{\left(\dfrac {m_1}{\rho_1}+\dfrac {m_2}{\rho_2}\right)}ρ=
(
ρ
1
m
1
+
ρ
2
m
2
)
m
1
+m
2
If m_1=m_2=m,m
1
=m
2
=m,
\begin{gathered}\rho=\dfrac {m+m}{\left(\dfrac {m}{\rho_1}+\dfrac {m}{\rho_2}\right)}\\\\\\\rho=\dfrac {2m}{m\left(\dfrac {1}{\rho_1}+\dfrac {1}{\rho_2}\right)}\\\\\\\rho=\dfrac {2}{\left(\dfrac {\rho_1+\rho_2}{\rho_1\rho_2}\right)}\\\\\\\rho=\dfrac {2\rho_1\rho_2}{\rho_1+\rho_2}\end{gathered}
ρ=
(
ρ
1
m
+
ρ
2
m
)
m+m
ρ=
m(
ρ
1
1
+
ρ
2
1
)
2m
ρ=
(
ρ
1
ρ
2
ρ
1
+ρ
2
)
2
ρ=
ρ
1
+ρ
2
2ρ
1
ρ
2
Thus the resultant density is the harmonic mean of the two individual densities at same mass.
=================================
Given,
\begin{gathered}\dfrac {\rho_1+\rho_2}{2}=9\\\\\\\rho_1+\rho_2=18\quad\implies\quad (2)\end{gathered}
2
ρ
1
+ρ
2
=9
ρ
1
+ρ
2
=18⟹(2)
And,
\dfrac {2\rho_1\rho_2}{\rho_1+\rho_2}=8
ρ
1
+ρ
2
2ρ
1
ρ
2
=8
From (2),
\begin{gathered}\dfrac {2\rho_1\rho_2}{18}=8\\\\2\rho_1\rho_2=144\\\\\\4\rho_1\rho_2=288\end{gathered}
18
2ρ
1
ρ
2
=8
2ρ
1
ρ
2
=144
4ρ
1
ρ
2
=288
Again from (2),
\begin{gathered}(\rho_1+\rho_2)^2=18^2\\\\\\(\rho_1)^2+(\rho_2)^2+2\rho_1\rho_2=324\\\\\\(\rho_1)^2+(\rho_2)^2+2\rho_1\rho_2-4\rho_1\rho_2=324-288\\\\\\(\rho_1)^2+(\rho_2)^2-2\rho_1\rho_2=36\\\\\\(\rho_1-\rho_2)^2=6^2\\\\\\\rho_1-\rho_2=6\quad\implies\quad (3)\end{gathered}
(ρ
1
+ρ
2
)
2
=18
2
(ρ
1
)
2
+(ρ
2
)
2
+2ρ
1
ρ
2
=324
(ρ
1
)
2
+(ρ
2
)
2
+2ρ
1
ρ
2
−4ρ
1
ρ
2
=324−288
(ρ
1
)
2
+(ρ
2
)
2
−2ρ
1
ρ
2
=36
(ρ
1
−ρ
2
)
2
=6
2
ρ
1
−ρ
2
=6⟹(3)
Adding (2) and (3) implies,
\large\boxed {\rho_1=\mathbf {12}\ \mathbf{kg\ m^{-3}}}
ρ
1
=12 kg m
−3
And subtracting (3) from (2) implies,
\large\boxed {\rho_2=\mathbf {6}\ \mathbf{kg\ m^{-3}}}
ρ
2
=6 kg m
−3
Explanation:
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