Question

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

Answer 1

$m g =\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 / cm ^{3} $