Sydney Shaffer

2023-03-13

Density of a mixture is $\rho $, when fluids of densities ${\rho}_{1}$ and ${\rho}_{2}$ are mixed in the same proportion. Then:

A)$\rho =\frac{{\rho}_{1}+{\rho}_{2}}{2}$;

B)$\rho =\frac{2{\rho}_{1}{\rho}_{2}}{{\rho}_{1}+{\rho}_{2}}$;

C)$\rho =\frac{{\rho}_{1}^{2}}{{\rho}_{2}}$;

D)$\rho =\frac{{\rho}_{2}^{2}}{{\rho}_{1}}$

A)$\rho =\frac{{\rho}_{1}+{\rho}_{2}}{2}$;

B)$\rho =\frac{2{\rho}_{1}{\rho}_{2}}{{\rho}_{1}+{\rho}_{2}}$;

C)$\rho =\frac{{\rho}_{1}^{2}}{{\rho}_{2}}$;

D)$\rho =\frac{{\rho}_{2}^{2}}{{\rho}_{1}}$

saltarenqa3

Beginner2023-03-14Added 1 answers

The right response is A $\rho =\frac{{\rho}_{1}+{\rho}_{2}}{2}$

As we know,

${\rho}_{mixture}=\frac{\text{Total mass}}{\text{Total volume}}\phantom{\rule{0ex}{0ex}}[\text{mass}=\text{volume}\times \text{density}]$

Let each liquid's volume be v. [equal proportion]

$\Rightarrow \rho =\frac{({\rho}_{1}\times v+{\rho}_{2}\times v)}{v}+v\phantom{\rule{0ex}{0ex}}\Rightarrow \rho =\frac{({\rho}_{1}+{\rho}_{2})v}{2v}\phantom{\rule{0ex}{0ex}}\Rightarrow \rho =(\frac{{\rho}_{1}+{\rho}_{2}}{2})$

As we know,

${\rho}_{mixture}=\frac{\text{Total mass}}{\text{Total volume}}\phantom{\rule{0ex}{0ex}}[\text{mass}=\text{volume}\times \text{density}]$

Let each liquid's volume be v. [equal proportion]

$\Rightarrow \rho =\frac{({\rho}_{1}\times v+{\rho}_{2}\times v)}{v}+v\phantom{\rule{0ex}{0ex}}\Rightarrow \rho =\frac{({\rho}_{1}+{\rho}_{2})v}{2v}\phantom{\rule{0ex}{0ex}}\Rightarrow \rho =(\frac{{\rho}_{1}+{\rho}_{2}}{2})$