Question

Two liquids when they are mixed in equal volumes resultant density is 9,when they are mixed equal masses resultant density is 8.find the densities of liquids.

Answer 1

Here the resultant density is,

$\rho=\dfrac {m_1+m_2}{V_1+V_2}\quad\implies\quad (1)\\\\\\\rho=\dfrac {\rho_1V_1+\rho_2V_2}{V_1+V_2}$

If $V_1=V_2=V,$

$\rho=\dfrac {\rho_1V+\rho_2V}{V+V}\\\\\\\rho=\dfrac {V(\rho_1+\rho_2)}{2V}\\\\\\\rho=\dfrac {\rho_1+\rho_2}{2}$

So the resultant density is the arithmetic mean of the two individual densities at same volume.

From (1),

$\rho=\dfrac {m_1+m_2}{\left(\dfrac {m_1}{\rho_1}+\dfrac {m_2}{\rho_2}\right)}$

If $m_1=m_2=m,$

$\rho=\dfrac {m+m}{\left(\dfrac {m}{\rho_1}+\dfrac {m}{\rho_2}\right)}\\\\\\\rho=\dfrac {2m}{m\left(\dfrac {1}{\rho_1}+\dfrac {1}{\rho_2}\right)}\\\\\\\rho=\dfrac {2}{\left(\dfrac {\rho_1+\rho_2}{\rho_1\rho_2}\right)}\\\\\\\rho=\dfrac {2\rho_1\rho_2}{\rho_1+\rho_2}$

Thus the resultant density is the harmonic mean of the two individual densities at same mass.

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Given,

$\dfrac {\rho_1+\rho_2}{2}=9\\\\\\\rho_1+\rho_2=18\quad\implies\quad (2)$

And,

$\dfrac {2\rho_1\rho_2}{\rho_1+\rho_2}=8$

From (2),

$\dfrac {2\rho_1\rho_2}{18}=8\\\\2\rho_1\rho_2=144\\\\\\4\rho_1\rho_2=288$

Again from (2),

$(\rho_1+\rho_2)^2=18^2\\\\\\(\rho_1)^2+(\rho_2)^2+2\rho_1\rho_2=324\\\\\\(\rho_1)^2+(\rho_2)^2+2\rho_1\rho_2-4\rho_1\rho_2=324-288\\\\\\(\rho_1)^2+(\rho_2)^2-2\rho_1\rho_2=36\\\\\\(\rho_1-\rho_2)^2=6^2\\\\\\\rho_1-\rho_2=6\quad\implies\quad (3)$

Adding (2) and (3) implies,

$\large\boxed {\rho_1=\mathbf {12}\ \mathbf{kg\ m^{-3}}}$

And subtracting (3) from (2) implies,

$\large\boxed {\rho_2=\mathbf {6}\ \mathbf{kg\ m^{-3}}}$