Question

An object is put in three liquids having different densities one by one. The object floats with 1/9, 2/11 and 3/7 parts of its volume outside the surface of liquids of densities d1, d2 and d3 respectively. Which of the followimg is the correct order of the densities of the three liquids?
a) d1>d2>d3.
b) d2>d3>d1.
c) d1 d) d3>d2>d1.

{AMU +2 ENTRANCE TEST 2016-17}

Answer 1

HELLO DEAR,

AS BUOYANT FORCE IS DIRECTLY

PROPORTIONAL TO DENSITY OF THE LIQUID

AND MORE BUOYANT FORCE MORE THE

FLOATING.

----------V1 < V2 < V3-----------

So,

---------d1 < d2 < d3-----------
OR,

 Here, volumes of object 
outside the liquids of 
densities d1, d2, d3
 respectively, are given as: 


-------1/9= 0.11, ----------

------2/11 = 0.18,--------

---------3/7=0.43-------



 This means buoyant face


 is maximum in liquid with 

density d3 and minimum in

 liquid withdensity d2.

As buoyant force exerted 


by a liquid is directly


 proportional to its density,

Therefore, correct order of

 the densities of three


 liquids is: d1 < d2 <d3.




I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer is [d] d3>d2>d1
for simply doing this.
take the volume of that object 693 cm³.
(LCM of denominators of fractions 9,11 and 7 as it will be more easy than others in solving this question because time plays an important role in exams).

In d1, volume of that object above the surface of water is 693 x 1/9.
that will be 77 cm³ above the surface of water.

in d2, volume of that object above the surface of water is 693 x 2 / 11.
that will be 126 cm³ above the surface of water.

in d3, volume of that object above the surface of water is 693 x 3/ 7.
that will be 297 cm³ above the surface of water.

As we saw that D3 exerts more force on that object than d2 while d2 exerts more force than d1 on that object.

we know that more force is exerted by the liquid of high densities.

so D3>D2>D1

PLEASE MARK IT AS A BRAINLIEST ANSWER.