A body floats in water with one third of its volume above the surface of water if it is placed in oil it floats with half of its volume above the surface of the oil what will be the specific gravity of the Oil.

Let the total volume of the body be V. Its density be ‘d’ and the density of water be q = 1000 kg/m3

Weight of the body in air, W = Vdg

Volume of body inside water = V – V/3 = 2V/3

So, buoyant force is = (2V/3)qg

Let the density of the oil be ‘p’ and the volume of body inside that liquid is = V – V/2 = V/2

Again the buoyant force is = (V/2)pg

Since, in both case the body floats so,

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

=> p = 4q/3 

Hence specific gravity = 4/3

Regards

Arun (askIITians forum expert)