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 sampled of 10 litres of adulterated milk is:

1 litre

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2 litre

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3 litre

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4 litre

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Solution

The correct option is D 4 litre
Mass of adulterated milk
$M_{A} = 1032 \times (10 \times 10^{- 3}$ ) kg
or $M_{A} = 10.32 k g$ ( $∵ 1 l i t r e = 10^{- 3} m^{3}$ )
$∴$ Mass of pure milk $M_{p} = 1080 \times V_{p}$
$∴$ Mass of water added = $ρ_{w} V_{w} = M_{A} - M_{p}$
$∴$ 10 $^{3} \times$ (volume of water added)= $M_{A} - M_{p}$
$∴ 10^{3} \times (10 \times 10^{- 3} - V_{p}) = 10.32 - 1080 V_{p}$
or 80 $V_{p} = 0.32$
or $V_{p}$ = $\frac{0.32}{80}$
$= \frac{1}{250} m^{3} = \frac{1000}{250} l i t r e = 4 l i t r e$ .

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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 sampled of 10 litres of adulterated milk is:

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 sampled of 10 litres of adulterated milk is: