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
$M_{1}=\frac{\alpha V(\beta-\sigma)}{\beta-\alpha}$