Question

A metal cube is found to float in a liquid of density $2gc{m}^{-3}$ with $\frac{1}{2}cm$ of its vertical side above the liquid. On placing a weight of $144g$ over its top, it just submerges in the liquid. Find the specific gravity of the metal cube?

Answer 1

Let the side of metal be L.

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

Now the weight of 144 g weight will be balanced by the buoyancy force due to $1/2cm$ length of metal cube $(\rho =2gc{m}^{-3})$

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

This force is equal to weight of 144gm.

So $L^2\times \frac{1}{2}\times 2\times 10=144\times 10$

$L=12cm$

When no weight was placed then $1/2$ cm was above the liquid.

So total length under the water was $L-\frac{1}{2}=\frac{23}{2}cm$

Weight of block=buoyancy force

${\rho }_{metal}\times L^3\times g=\rho \times L^2\times \frac{23}{2}\times g$

${\rho }_{metal}=\frac{23}{12}gc{m}^{-3}$