First slide

Fluid statics

Question

Two substances of densities $ρ_{1} and ρ_{2}$ are mixed in equal volume and the relative density of mixture is 4. When they are mixed in equal masses, the relative density of the mixture is 3. The values of $ρ_{1} and ρ_{2}$ are:

Moderate

A

$ρ_{1} = 6 and ρ_{2} = 2$
B

$ρ_{1} = 3 and ρ_{2} = 5$
C

$ρ_{1} = 12 and ρ_{2} = 4$
D

none of these

Solution

When the substance are mixed in equal volumes, then
${Vρ}_{1} + {Vρ}_{2} = 2 V \times 4 - - - - - - (i)$
When the two substances are mixed in equal masses, then
$\frac{m}{ρ_{1}} + \frac{m}{ρ_{2}} = \frac{2 m}{3} - - - - (ii)$
From eq.(i): $ρ_{1} + ρ_{2} = 8$
From eq. (ii): $\frac{1}{ρ_{1}} + \frac{1}{ρ_{2}} = \frac{2}{3}$
or $\frac{ρ_{1} + ρ_{2}}{ρ_{1} ρ_{2}} = \frac{2}{3}$
or $\frac{9}{ρ_{1} ρ_{2}} = \frac{2}{3} or ρ_{1} ρ_{2} = 12$ ------(iv)
Now, $ρ_{1} - ρ_{2} = {[{(ρ_{1} + ρ_{2})}^{2} - 4 ρ_{1} ρ_{2}]}^{\frac{1}{2}}$ ------(v)
Solving eqs. (iii) and (v), we get $ρ_{1} = 6 and ρ_{2} = 2$

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