When equal volumes of two metals are mixed together the specific gravity of alloy is 4. When equal masses of the same two metals are mixed together the specific gravity of the alloy becomes 3. find specific gravity of each metal?
(specific gravity `=("density of substance")/("density of water"))`

Answer

2 and 6

Explanation

Let `m_(1),m_(2)` = masses of two metals,
`V_(1), V_(2)` = volumes of two metals.
If `rho` is the density of mixture of two metals, then
`rho =(m_(1) + m_(2))//(V_(1) + V_(2))`
When equal volume of two metals of density `rho_(1)` and `rho_(2)` are mixed, then `V_(1) = V_(2) = V` and `m_(1) =rho_(1) V` and `m_(2) = rho_(2) V` . If `rho` is the density of mixture of two metals, then
`rho=((rho_(1)V+rho_(2)V))/(V+V) = (rho_(1) + rho_(2))/(2)`
or `rho_(1) + rho_(2) =2rho =2xx4=8` ....(i)
When equal masses are mixed, then
`m_(1) =m_(2)=m` and `V_(1) =m//rho_(1)` and `V_(2) =m//rho_(2)`
So, `rho=(m+m)/((m//rho_(1))+(m//rho_(2))) = (2rho_(1)rho_(2))/(rho_(1)+rho_(2)) =3`
or `rho_(1)rho_(2) =(3)/(2) (rho_(1) + rho_(2)) =(3)/(2)xx8 =12` ...(ii)
Solving (i) and (ii), we get,
`rho_(1) =2` and `rho_(2) =6`