Question

An object is put one by one in 3 liquids having different densities. The object floats with $\frac{1}{9}$, $\frac{2}{11}$ and $\frac{3}{7}$ parts of their volumes outside the liquid surface in liquids of densities $d_1$, $d_2$ and $d_3$ respectively. Arrange the densities in ascending order of their magnitude.

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

Answer 1

The correct option is A $d_1<d_2<d_3$

$\frac{1}{9}=0.11;\frac{2}{11}=0.18;\frac{3}{7}=0.43$
These are the volumes outside of liquid of densities $d_1$, $d_2$ and $d_3$ respectively. This means buoyant force is maximum for d3 since buoyant force is directly proportional to the density.
Hence, $d_1<d_2<d_3$.