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
Answers
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: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