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A solid floats with $1 / 4$ th of its volume above the surface of water, the density of the solid is
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$750 \mathrm{~kg} \mathrm{~m}^{-3}$
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
$\begin{aligned} \mathrm{V} \rho g & =\frac{3 \mathrm{~V}}{4} \times 10^3 \mathrm{~g} \\ \rho & =\frac{3}{4} \times 1000=750 \mathrm{kgm}^{-3}\end{aligned}$
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
$\begin{aligned} \mathrm{V} \rho g & =\frac{3 \mathrm{~V}}{4} \times 10^3 \mathrm{~g} \\ \rho & =\frac{3}{4} \times 1000=750 \mathrm{kgm}^{-3}\end{aligned}$
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