A solid floats with 1 / 4 th of its volume above the surface of...

Let V and $\rho$ be volume and density of solid respectively and $\rho^{\prime}$ be the density of water i.e., $\mathrm{p}^{\prime}=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ Weight of body $=\mathrm{V} \rho g$ Volume of solid body outside water $=\mathrm{V} / 4$ $\therefore$ Volume of solid body inside water $=\mathrm{V}-\mathrm{V} / 4=3 \mathrm{~V} / 4$ Weight of water displaced by solid $=\frac{3 \mathrm{~V}}{4} \times 10^3 \times \mathrm{g}$ As solid body is floating, then Weight of body = Weight of water displaced by it $$\begin{aligned} \mathrm{V} \rho g & =\frac{3 \mathrm{~V}}{4} \times 10^3 \mathrm{~g} \\ \rho & =\frac{3}{4} \times 1000=750 \mathrm{kgm}^{-3}\end{aligned}$$