# 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?

2
by sippi
wht is the answer in terms of ?
I m getting 495M³/72

2015-11-01T17:52:48+05:30
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
thanks sis
U r welcme
how to chat with u???
2015-11-01T19:05:40+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

click on thanks button above; select best answer
tq sir pls ans other questions of sippi