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 g/"cc" , 8/3 g/"cc"`
B. 4 g/cc
C. `3/8` g /cc
D. 1

Answer

B

Explanation

Let the volume and density of the body be V and `rho` respectively and the densities of water and the liquid be `sigma_(w)`, and `sigma_(i)` respectively. As the body floats In the water
`therefore` Weight of the body = Weight of water displaced
i.e., `Vrhog =V_("in")sigma_(w)g` or `Vrhog = (V-V/3) sigma_(w)g =2/3 Vsigma_(w)g`
or `rho =2/3 sigma_(w)`...........(i)
In the liquid
Weight of the body = Weight of liquid displaced,
i.e. `Vrhog = V_(in)sigma_(i)g` or `Vrhog = (V-3/4V)sigma_(l)g =V/4 sigma_(l)g`
or `rho =1/4 sigma_(l)`.............(ii)
From eqns. (i) and (ii)
`2/3 sigma_(w) =1/4 sigma_(l)` or `sigma_(l) = 8/3 sigma_(w) = 8/3 xx 1 g//cc` (as `sigma_(w) =1 g//cc`)
Thus, the density of the liquid is `8/3` g/cc.