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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. `rho_(1) = 6 and rho_(2)=2`B. `rho_(1)=3 and rho_(2) = 5`C. `rho_(1) = 12` and `rho_(2) = 4`D. noneof these

Answer» Correct Answer - A
When the substances are mixed in equal volumes then
`V rho_(1)+Vrho_(2) = 2V xx 4` …(1)
When the two substances are mixed in equal masses. Then,
`(m)/(rho_1) + (m)/(rho_2) = (2m)/(2)` ..(2)
From eq. (1) , `rho_(1)+rho_(2) = 8`
from eq. (2) , `1/(rho_1)+1/(rhp_2) = 2/3`
or `(rho_(1)+rho_(2))/(rho_(1)rho_(2)) = 2/3` ..(3)
or `9/(rho_(1)rho_(2)) = 2/3 or rho_(1)rho_(2) = 12` ..(4)
Now, `rho_(1) - rho_(2) = [(rho_(1)+rho_(2))^(2)-4 rho_(1)rho_(2)]^(1//2)` ..(5)
solving eqs, (3) and (5) , we get `rho_(1) = 6 and rho_(2)=2`.


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