Correct Answer - 2 and 6
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`