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Question

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 :


A
ρ1=6 and ρ2=2
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B
ρ1=3 and ρ2=5
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C
ρ1=12 and ρ2=4
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D
None of these
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Solution

The correct option is A $${ \rho }_{ 1 }=6$$ and $${ \rho }_{ 2 }=2$$
When substances are mixed in equal volume then density $$=\cfrac { { \rho  }_{ 1 }+{ \rho  }_{ 2 } }{ 2 } =4\quad $$

$$\Rightarrow { \rho  }_{ 1 }+{ \rho  }_{ 2 }=8...(i)$$

when substances are mixed in equal masses then density

$$\cfrac { 2{ \rho  }_{ 1 }{ \rho  }_{ 2 } }{ { \rho  }_{ 1 }+{ \rho  }_{ 2 } } =3$$

$$2{ \rho  }_{ 1 }{ \rho  }_{ 2 }=3({ \rho  }_{ 1 }+{ \rho  }_{ 2 })...(ii)$$

On solving (i) and (ii) we get

$$\quad { \rho  }_{ 1 }=6;\quad { \rho  }_{ 2 }=2$$

Physics

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