Question

A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density $0.8$ and with its vertex submerged. When another liquid of relative density $\rho$ is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is $0.10m$ and the radius of the circular base is $0.05m$. Find the specific gravity $\rho$ is given. (Round it to the closest integer).

Answer 1

Let the volume of cylinder be V and density to be ${\rho }_1$

$V{\rho }_{1}g=\frac{V\times 800\times g}{27}$..(1) ,Volume of cone with height is h/3 is V/27

$V{\rho }_{1}g+V\rho g/27=\frac{V\times 800g}{8}$ ..(2) ,Volume of cone with height is h/2 is V/8

from above equation we $\rho =\frac{800\times 19}{8}$

Specific gravity =$\frac{1900}{1000}=1.9or2$