Question

An adulterated sample of milk has a density $1032\, kg \,m^{-3}$ ,while pure milk has a density of $1080\, kg \,m^{-3}$. Then the volume of pure milk in a sample of $10$ litres of adulterated milk is

• A. $1$ litre
• B. $2$ litre
• C. $3$ litre
• D. $4$ litre

Answer 1

Answer: (D)

Mass of adulterated milk
$M_{A} = 1032 \times (10 \times 10^{-3}) \, kg$
or $M_{A} = 10.32 \, kg$ $(\because\, 1 \,litre=10^{-3} m^{3})$
Mass of pure milk $M_{p}=1080\times V_{p}$
$\therefore $ Mass of water added $= \rho_{w}V_{w}=M_{A}-M_{p}$
$\therefore 10^{3}\times$ (volume of water added) $= M_{A} - M_{p}$
$\therefore 10^{3}\times\left(10\times10^{-3}-V_{p}\right)=10.32-1080 V_{p}$
or $80 V_{p}=0.32$
or $V_{p}=\frac{0.32}{80}=\frac{1}{250} m^{3}=\frac{1000}{250}$
litre = $4$ litre.