Question

A body floats with one-third of ts volume outside water and three-fourth of its volume outside the another liquid.If density of water is 10³ kg m^-3 ,then the density of other liquid is _______

Answer 1

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 other liquid be ‘p’ and the volume of body inside that liquid is = V – 3V/4 = V/4

Again the buoyant force is = (V/4)pg

Since, in both case the body floats so,

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

=> p = 8q/3 = 8 × 1000/3 = (8/3) × 10³ kg/m³