Question

a body floats with 1upon3rd of its volume outside water and 3upon4th of its volume outside liquid then the density of liquid is​

Answer 1

SOLUTION

Let the total volume of the body be V.

Let the total volume of the body be V. It density be 'd' and the density of water be q = 1000kg/m^3

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

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)× 10^3 kg/m^3

hope it helps ✔️

Answer 2

Explanation:

firstly you have to know the correct formula if a body floats 1/3 of its volume it means 1/3 of x outside water density = mass/volume multiply both and put the value in he formula you will get the answer