Question

What is the effective density ($\rho$) of the mixture of water and milk where both are having same masses of 1 $kg$.

• A. $\rho =$$\frac{{\rho }_{!}{\rho }_2}{{\rho }_1+{\rho }_2}$
• B. $\rho =$$\frac{{\rho }_{!}{\rho }_2}{2{\rho }_1+2{\rho }_2}$
• C. $\rho =$$\frac{4{\rho }_{!}{\rho }_2}{{\rho }_1+{\rho }_2}$
• D. $\rho =$$\frac{2{\rho }_{!}{\rho }_2}{{\rho }_1+{\rho }_2}$

Answer 1

The correct option is D $\rho =$$\frac{2{\rho }_{!}{\rho }_2}{{\rho }_1+{\rho }_2}$

Given:
Mass of water $=$ Mass of milk $m$ $=1kg$
Let's assume volume of water and milk is $v_1$ and $v_2$.
Mass of water,
$m=$${\rho }_1{v}_1$
$v_1=$$\frac{m}{{\rho }_1}$
Mass of milk,
$m=$${\rho }_2{v}_2$
$v_2=$$\frac{m}{{\rho }_2}$

total mass of water and milk is: $m+m=2m=2kg$
Total volume of water and milk is:$v_1+v_2$$=$$\frac{m}{{\rho }_1}+\frac{m}{{\rho }_2}$

Now, the effective density of the mixture is:
$\rho =$$\frac{Totalmass}{Totalvolume}$
$\rho =$$\frac{2m}{\frac{m}{{\rho }_1}+\frac{m}{{\rho }_2}}$
$\rho =$$\frac{2}{\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2}}$
$\rho =$$\frac{2{\rho }_{!}{\rho }_2}{{\rho }_1+{\rho }_2}$