Question

A cylinder made of copper and aluminium floats in mercury of density $13.6{\text{g/cm}}^3$, such that ${0.75}^{th}$ part of it is below mercury. What is the density of the cylinder?

Answer 1

Given:
Density of mercury $={\rho }_{Hg}=$ $13.6{\text{g/cm}}^3$
${0.75}^{th}$ part of cylinder is below mercury.
Let the volume pf the solid cylinder be $=V_s$
Volume of mercury displaced by immersed part of solid $=V_{Hg}$ $=0.75V_s$

By the law of floatation:
Weight of the solid cylinder $=$ Weight of the mercury displaced by the immersed part of solid cylinder.
$=V_s\times {\rho }_s\times g=V_{Hg}\times {\rho }_{Hg}\times g$
$=V_s\times {\rho }_s=0.75V_s\times 13.6$
${\rho }_s=0.75\times 13.6=10.2{\text{g/cm}}^3$
So, the density of the solid cylinder is $10.2{\text{g/cm}}^3$