Question

A wooden cube is found to be floating in a liquid of density $1.2gc{m}^{-3}$ with $\frac{1}{4}cm$ of its vertical side above the liquid. On keeping a weight of 67.5g over its top it just submerges in the liquid. Find the specific gravity of wooden cube.

Answer 1

$d_l=1.2gc{m}^{-3}$
Let S be the side of cube.
Height of the vertical side
Inside the liquid $=(S-\frac{1}{4})cm$
Upthrust $=$ weight of the body
$S^2(S-\frac{1}{4})\times d_l\times g=S^3\times d_{cube}\times g$
$\therefore d_{cube}=\frac{(S-\frac{1}{4})d_l}{S}gc{m}^{-3}$

When a weight of 67.5 g is placed on the top;
$(S^3\times d_{cube}\times g)+67.5=S^3\times d_l\times g$
$S^3\times d_{cube}\times g+67.5=S^3\times 1.2\times g$
$\frac{S^3\times (S-\frac{1}{4})1.2}{S}\times g+67.5=S^3\times 1.2\times g$
$1.2S^{3}g-3S^{2}g+67.5S+1.2S^{3}g=0$
$S^2=+\frac{\left(67.5\right)S}{0.3g}$
$S=\frac{225}{9.8}=22.9cm$
$\therefore d_{cube}=\frac{(22.9-0.25)1.2}{22.9}$
$=1.19gc{m}^{-3}$
Specific gravity of wooden cube $=1.19$