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
B
ρ1=3 and ρ2=5
C
ρ1=12 and ρ2=4
D
None of these

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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