Question

An object is put one by one in three liquids having different densities. The object floats with $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$, $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$11$}\right.$, and $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.$ parts of their volumes outside the liquid surface in liquids of densities $d_1$, $d_2$ and $d_3$ respectively. Which of the following statement is correct?

• A. $d_1>d_2>d_3$
• B. $d_1>d_2<d_3$
• C. $d_1<d_2>d_3$
• D. $d_1<d_2<d_3$

Answer 1

The correct option is D

$d_1<d_2<d_3$

Explanation for Correct Option:

1. As there are three different liquids, their densities ($d_1,d_2,d_3$) are different.
2. When the body is immersed in these liquids, it will experience an upward force, which depends upon the densities of the liquids.
3. Buoyant Force may be defined as when a rigid object is submerged in a fluid (completely or partially), there exists an upward force on the object that is equal to the weight of the fluid that is displaced by the object.
4. In this case, there are 3 liquids with varying densities.
5. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.=0.11$,$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$11$}\right.=0.18$ and $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$7$}\right.=0.43$. These are the volume outside of liquids of densities $d_1,d_2$and $d_3$​ respectively.
6. This means that the buoyant force is maximum for $d_3$​ and minimum for $d_1$​.
7. As the buoyant force is proportional to density, the densities of the liquids are in the order $d_1<d_2<d_3$​.

Thus, option D is the correct option.