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

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Solution

Let say Their Densities are $D_{1}$ & $D_{2}$
Density = $\frac{M a s s}{V o l u m e}$
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 = $\frac{2 M}{\frac{M}{D_{1}} + \frac{M}{D_{2}}}$ = $\frac{2 D_{1} \times D_{2}}{D_{1} + D_{2}}$ = 4.8
=> $\frac{D_{1} \times D_{2}}{10} = 2.4$
=> $D_{1} \times D_{2} = 24$
=> $D_{1} (10 - D_{1}) = 24$
=> $- D_{1} ² + 10 D_{1} = 24$
=> $D_{1} ² - 10 D_{1} + 24 = 0$
=> $D_{1} ² - 6 D_{1} - 4 D_{1} + 24 = 0$
=> $(D_{1} - 6) (D_{1} - 4) = 0$
=> $D_{1} = 6$ $o r$ $4$
$D_{2}$ are 4 and 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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