Multiple choice

Two substances of densities ${ \rho }{ 1 }$ and ${ \rho }{ 2 }$ 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 ${ \rho }{ 1 }$ and ${ \rho }{ 2 }$ are :

  1. ${ \rho }_{ 1 }=6$ and ${ \rho }_{ 2 }=2$
  2. ${ \rho }_{ 1 }=3$ and ${ \rho }_{ 2 }=5$
  3. ${ \rho }_{ 1 }=12$ and ${ \rho }_{ 2 }=4$

  4. None of these
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A Correct answer
Explanation

Equal volume: (rho1 + rho2)/2 = 4, so rho1 + rho2 = 8. Equal mass: 2/(1/rho1 + 1/rho2) = 3, so 2*rho1*rho2 / (rho1 + rho2) = 3. 2*rho1*rho2 / 8 = 3, so rho1*rho2 = 12. Solving x^2 - 8x + 12 = 0 gives x = 6 and 2.

AI explanation

When two liquids are mixed in equal volumes, the density of the mixture is the simple average: (rho1 + rho2)/2 = 4, meaning rho1 + rho2 = 8. When mixed in equal masses, the density formula is 2(rho1)(rho2)/(rho1 + rho2) = 3. Substitute rho2 = 8 - rho1 into the second equation to get 2(rho1)(8 - rho1)/8 = 3. Solving this quadratic equation yields the values rho1 = 6 and rho2 = 2 (or vice versa).

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