Question

A liquid mixture of volume $V$, has two liquids as its ingredients with densities $\alpha$ and $\beta$. If density of the mixture is $\sigma$, then mass of the first liquid in mixture is

• A. $\frac{\alpha V[\sigma \beta +1]}{\beta [\alpha +\sigma ]}$
• B. $\frac{\alpha V[\sigma -\beta ]}{[\sigma +\beta ]}$
• C. $\frac{\alpha V(\beta -\sigma )}{\beta -\alpha }$
• D. $\frac{\alpha V[1-\sigma \alpha ]}{\beta [\alpha -\sigma ]}$

Answer 1

Answer: (C)

Let mass of liquid with density $\alpha =M_1$
Let mass of liquid with density $\beta =M_2$
Total volume $=V$
Net density of mixture $=\sigma$
Now,
Total mass $=M_1+M_2$
$\Rightarrow V\sigma =M_1+M_2$
$\Rightarrow M_2=V\sigma -M_1$ ...(1)
$\left[\because \frac{\text{ Total mass }}{V}=\sigma \right]$
Now,
$\sigma =\frac{\text{ Total mass }}{\text{ Total volume }}=\frac{\left(M_1+M_2\right)}{\left(\frac{M_1}{\alpha }\right)+\left(\frac{M_2}{\beta }\right)}$
Substituting value of $M_2$ from equation (1)
$\sigma =\frac{M_1+\left(V\sigma -M_1\right)}{\frac{M_1}{\alpha }+\frac{\left(V\sigma -M_1\right)}{\beta }}$
Solving this we get
$M_1=\frac{\alpha V(\beta -\sigma )}{\beta -\alpha }$