Question

Two liquids of equal masses filled in a vessel, their densities are P1= 13.6 x 10³ kg/m³ and
P2 = 0.806 x 10³ kg/m³ respectively.
Then the density of mixure of fluid will be
a) 0.7×10³ kg/m³
b) 0.6×10³ kg/m³
c) 0.75×10³ kg/m³
d) 1.52×10³ kg/m³
​

Answer 1

Answer:

The density of mixure of fluid will be 1.52×10³ kg/m³

Explanation:

Step 1: Assume the mass of the two liquids. Then we can write the mass in terms of the density of the liquid mixture. Next, calculate the density of the mixture as the total mass divided by the total volume. Here, mass is the same as the volume should be converted in terms of mass and density.

Step 2: When two liquids of two different densities are mixed, they separate when we stop mixing them. The heavier liquid goes to the bottom as the density is high and the lighter liquid will deposit at the top layer as its density is less. The densest liquid is molasses and the least dense is the alcohol. The density of the final mixture after mixing different liquids with different densities, will be the ration between the total mass of the liquids and the total volume occupied by the liquids.

Step 3: Calculation:

$\begin{aligned}& \rho_1=13.6 \times 10^3 \mathrm{~kg} / \mathrm{m}^3 \\& \rho_2=0.806 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\end{aligned}$

Density of mixtures of equal masses

$\begin{aligned}& \rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}=\frac{2\left(13.6 \times 10^3 \times 0.806 \times 10^3\right)}{(13.6+0.806) 10^3} \\& =\frac{21.9232 \times 10^6}{14.406 \times 10^3}=1.52 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\end{aligned}$

Hence, The density of mixure of fluid will be 1.52×10³ kg/m³.

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