Question

A sample of adulterated milk has a density of 1032 kg/${m^3}$ while pure milk has a density of 1080 kg/${m^3}$. Then the volume of pure milk in a sample of 10L of adulterated milk is:
A) 0.5L
B) 1L
C) 2L
D) 4L

Answer 1

Hint: This question can be easily solved using the relationship between mass, volume and density. First, we need to find the mass of both adulterated and pure milk in kilograms and then proceed to find the mass of water added. Then we can finally use these two quantities to find the volume of pure milk in the given sample of adulterated milk.

Formula Used:
Through the definition of density we know that:
$\rho = \dfrac{M}{V}$ Where M is the Mass and V is the velocity.

Complete step by step answer:
First we have to find the mass of $10L$ of adulterated milk $({M_A})$ in kilogram. In order to change density to mass we have to multiply it by ${10^{ - 3}}$ as $1L = {10^{ - 3}}{m^3}$. Thus:
${M_A} = 1032 \times \left( {10 \times {{10}^{ - 3}}} \right)$
Thus we get, Mass of adulterated milk, ${M_A} = 10.32kg$.
Let the Volume of pure milk be, ${V_P}$ . Then, mass of pure milk in kilogram can be expressed as:
${M_P} = 1080 \times {V_P}$

Since we know that $M = \rho \times V$, assuming ${\rho _W}$ to be the density of water and ${V_W}$ to be the volume of water we can write:
Mass of water added $ = {\rho _W}{V_W}$
Now this mass of water is added to change pure milk to adulterated milk. Thus we can write the expression:
${\rho _W}{V_W} = {M_A} - {M_P}$
In order to change the unit of density to litre we have to multiply it by ${10^{ - 3}}$. Then we have:
${10^{ - 3}} \times {V_P} = {M_A} - {M_P}$

Now, the volume of water added is equal to the volume of pure milk subtracted from the volume of adulterated milk. So,
${V_W} = ({10^{ - 3}} \times 10 - {V_P})$
Also, the Mass of pure milk can be written as, ${M_P} = 1080 \times {V_P}(\because M = \rho V)$
Substituting these values in the above equation and also putting the Mass of adulterated milk in the equation, we have:
${10^{ - 3}} \times \left( {10 \times {{10}^{ - 3}} - {V_P}} \right) = 10.32 - 1080{V_P}$
Now making ${V_P}$ the subject of the formula we have:
$
  80{V_P} = 0.32 \\
\Rightarrow {V_P} = \dfrac{{0.32}}{{80}} = 0.004{m^3} \\
$
In order to change the volume obtained into litres, we have to multiply it by ${10^3}$. Thus,
${V_P} = 4L$

Thus we can say that a sample of $10L$ adulterated milk contains $4L$ of pure milk. Thus the correct option is (D).

Note: Always remember to change all units accordingly. Students usually make a mistake in converting units given in ${m^3}$ to $L$.