Question

A cube of external dimension $20cm$ has an inner cubical portion of side $10cm$ whose density is twice that of the outer portion. If this cube is just floating in a liquid of density $1g/c{m}^3$, then the density of the inner portion is $\frac{x}{y}g/c{m}^3.$ Find $(x+y)$

Answer 1

Answer: 17

$mg=\left[2{\rho }_1\times 10^3+{\rho }_1\times \left(20^3-10^3\right)\right]g$
$=9{\rho }_1\times 10^{3}g={\rho }_l\times 8\times 10^{3}g$
$\Rightarrow {\rho }_1=\frac{8}{9}g/c{m}^3$