If two liquids are mixed in equal volumes, their resultant density is 5 gr/cc and mixed in equal masses their resultant density is 4.8 g/cc . The ratio of their densities is $\frac{2}{k}$ .Then K is

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Solution

The correct option is A 3
Let the densities be $D_{1}$ and $D_{2}$

$D e n s i t y = \frac{M a s s}{V o l u m e}$

Case 1 : Equal Volume

$⟹ M_{1} = D_{1} \times V$

and

$M_{2} = D_{2} \times V$

Density combined $= \frac{(M_{1} + M_{2})}{2 V} = \frac{(D_{1} + D_{2})}{2} = 5$

$⟹ D_{1} + D_{2} = 10$

Case 2 : Equal Masses

$⟹ V_{1} = \frac{M}{D_{1}}$

$⟹ V_{2} = \frac{M}{D_{2}}$

Density combined $= \frac{2 M}{(M / D_{1} + M / D_{2})} = \frac{2 D_{1} D_{2}}{(D + 1 + D_{2})} = 4.8$

$\frac{D_{1} \times D_{2}}{10} = 2.4$

$⟹ D_{1} \times D_{2} = 24$

$i m p l i e s D_{1} (10 - D_{1}) = 24$

$⟹ - D_{1}^{2} + 10 D_{1} = 24$

$⟹ D_{1}^{2} - 10 D_{1} + 24 = 0$

$⟹ D_{1}^{2} - 6 D_{1} - 4 D_{1} + 24 = 0$

$⟹ (D_{1} - 6) (D_{1} - 4) = 0$

Then,

$D_{1} = 6$ or $4$

$D_{2}$ are $4 & 6$

$\frac{D_{1}}{D_{2}} = \frac{6}{4} = \frac{3}{2}$ is not of form $\frac{2}{k}$

while $\frac{D_{1}}{D_{2}} = \frac{4}{6} = \frac{2}{3} = \frac{2}{k} ⟹ k = 3$

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