Question

A body float with (1/3) of its volume outside water and (3/4) of its volume outside another liquid the density of another liquid

Answer 1

Answer:

Volume = V

density= d

density of water =q = 1000 kg/m3

weight in air = w= Vgd

Buoyant force= (2V/3)qg

density of other liquid =p

volume of body inside water= V- 3V/4= V/4

buoyant force = (V/4)pg

as in both situation the body will float so,

(2V/3)qg = (V/4)pg

p = 8q/ 3

p = 8 x 1000/3

p= 8 /3 x 10³

Answer 2

Answer:8/3g/cc

Explanation:   v(sub)/v = density of solid/density of fluid

     for  floating body = v/3/4=density of solid/ density of liquid                                      = 3v/4= d of solid/d of fluid                                                                                                   =3v/4 (1000)= d of solid                 (d of water = 1000kg/m)                                                          fb of body outside the other liquid = v/1/3=density of solid / density of liquid                    =v/3=d of solid / d of liquid                                                                                               v-3v/4=v/4   ; v-v/3=2v/3                                                                                                      in this case fb1 = fb2                    (fb = dgv)                                                                                v/4dg=2v/3dg             (g gets cancelled)           (by shifting v/4 to rhs) we get             8v/3v (1000) = d of liquid              (v gets cancelled)                                                        =8/3  g/cc