A body floats with one-third of its volume outside water and `3//4` of its volume outside another liquid. The density of another liquid is :

Options

A. `((9)/(4))xx10^3(kg)/(m^3)`
B. `((4)/(9))xx10^3(kg)/(m^3)`
C. `((8)/(3))xx10^3(kg)/(m^3)`
D. `((3)/(9))xx10^3(kg)/(m^3)`

Answer

C

Explanation

`f_("in")=1-(1)/(3)=(2)/(3)`,`f_("in")'=1-(3)/(4)=(1)/(4)`
As `f_("in")=(rho)/(sigma)` and `f_("in")'=(rho)/(sigma')`,`(f_("in"))/(f_("in")')=((rho)/(sigma))/((rho)/(sigma')=(sigma')/(sigma)`
(`rho` is the density of the body, `sigma` that of water and `sigma'` that of the liquid)
Thus, `(sigma')/(sigma)=(f_(in))/(f_("in")')=((2)/(3))/((1)/(4))=(8)/(3)`
or `sigma'=((8)/(3))sigma=((8)/(3))xx10^3(kg)/(m^3)`