Question

A wooden block floats in glycerinein such a way that it's 2/5 thvolume remains above the surface. If relative density wood is 0.78 calculate relative density of glycerine

Answer 1

$$\begin{gathered} \mathbf{From}\mathbf{}\mathbf{laws}\mathbf{}\mathbf{of}\mathbf{}\mathbf{floatation}\mathbf{,} \\ \mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}=\mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{glycerine}\mathrm{displaced}\mathrm{by}\mathrm{wood} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{wooden}\mathrm{block}=\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{glycerine}\mathrm{displaced}\mathrm{by}\mathrm{wood} \\ {\mathrm{V}}_{\mathrm{wood}}\times {\mathrm{RD}}_{\mathrm{wood}}={\mathrm{V}}_{\mathrm{gly}}\times {\mathrm{RD}}_{\mathrm{gly}} \\ {\mathrm{V}}_{\mathrm{wood}}\times {\mathrm{RD}}_{\mathrm{wood}}=\left(1-\frac{2}{5}\right){\mathrm{V}}_{\mathrm{wood}}\times {\mathrm{RD}}_{\mathrm{gly}}[\mathrm{since}2/5\mathrm{volume}\mathrm{of}\mathrm{wood}\mathrm{remains}\mathrm{above}\mathrm{the}\mathrm{surface}] \\ {\mathrm{V}}_{\mathrm{wood}}\times {\mathrm{RD}}_{\mathrm{wood}}=\left(\frac{3}{5}\right){\mathrm{V}}_{\mathrm{wood}}\times {\mathrm{RD}}_{\mathrm{gly}} \\ {\mathrm{RD}}_{\mathrm{wood}}=\frac{3}{5}\times {\mathrm{RD}}_{\mathrm{gly}} \\ {\mathrm{RD}}_{\mathrm{gly}}=\frac{5}{3}\times 0.78 \\ =1.3 \end{gathered}$$