Question

A wooden cube is found to float 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

Let the side of wood be L.

So the of base is $A=L^2$

$\rho =1.2gc{m}^{-3}$

Now the weight of 67.5 g weight will be balanced by the buoyancy force due to $1/4cm$ length of metal cube $\rho =2gc{m}^{-3}$

So buoyancy force is $F=A\times L_1\rho g=L^2\times \frac{1}{4}\times 2\times 10$

This force is equal to weight of 67.5gm.

So $L^2\times \frac{1}{4}\times 1.2\times 10=67.5\times 10$

$L=15cm$

When no weight was placed then $1/4$ cm was above the liquid.

So total length under the water was $L-\frac{1}{4}=\frac{59}{4}cm$

Weight of block=buoyancy force

${\rho }_{wood}\times L^3\times g=\rho \times L^2\times \frac{59}{4}\times g$

${\rho }_{wood}=1.18gc{m}^{-3}$