Question

an object is put in turn , in thee liquids having different densities.the object floats with 3/5,2/9 and 8/11 parts of its volume inside the liquid surface in liquids of densities p1,p2 and p3 respectively.then the relation between p1,p2 and p3 is?

Answer 1

Remember this point: When an object is put in a liquid of density D, and if it floats with x/y portion of volume of the object is inside the liquid, then density of object is (x/y)*D

here, liquid density = p1
volume of object inside liquid = 3/5
density of object = 3/5 p1

liquid density = p2
volume of object inside liquid = 2/9
density of object = 2/9 p2

liquid density = p3
volume of object inside liquid = 8/11
density of object = 8/11 p3

So the relation is 3P1/5 = 2P2/9 = 8P3/11

Answer 2

let the volume of the object be V units
let the density of the object be  p units.

As per Archimedes law:
        V * p =  3/5 * V * p1
      mass of the object = mass of the liquid displaced
         =>  p = 0.60 p1    =>  p1 = 5/3 * p

  Similarly,  p = 2/9 * p2 =>    p2 = 9/2 * p    or  p2 = 2.7 * p1
                 p = 8/11 p3  =>   p3 =  11/8 * p  or  p3 = 0.825 * p1

 =>  p1 : p2 : p3 =  5/3 : 9/2 :  11/8
                          = 40 : 108 :  33