Question

A body floats with 2/5th of its volume above the surface of water. Calculate the density of material in g/m

Answer 1

volume immersed/total volume=density of material/density of water

volume immersed = V-(2/5)×V
=3V/5

(3V/5)/V = d/1

{ density of water = 1gm/cm cube}

d= 3/5 = 0.6 g/cm^3

Answer 2

Given :

A body floats with 2/5th of its volume above the surface of water.

To Find :

The density of material in $g/cm^3$ .

Solution :

We know ,

$\dfrac{volume \ immersed}{total\ volume}=\dfrac{density\ of\ material}{density\ of \ water}$

Let , total volume be V and density be d .

So , volume immersed , $V-\dfrac{2}{5}V=\dfrac{3}{5}V$ .

Also ,

$\dfrac{\dfrac{3}{5}V}{V}=\dfrac{d}{1}\\\\d=\dfrac{3}{5}=0.6\ g/cm^3$ ( Density of water is $1\ g/cm^3$ )

Therefore , The density of material is  $0.6\ g/cm^3$ .

Hence , this is the required solution .