The ratio of masses of a cubical block of wood and a chunk of concrete is $\frac{3}{5}$ so that the combination just floats with entire volume submerged under water. If specific gravity of wood is $0.5$, then specific gravity of concrete will be
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$\mathrm{v}_{1}=\frac{\mathrm{m}_{1}}{\rho_{1}}$
$\mathrm{v}_{2}=\frac{\mathrm{m}_{2}}{\rho_{2}}$
$\frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\frac{3}{5} \mathrm{given}$
$\Rightarrow\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g}=\rho_{w}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right) \mathrm{g}$
$\mathrm{m}_{1}+\mathrm{m}_{2}=\rho_{\mathrm{w}}\left(\frac{\mathrm{m}_{1}}{\rho_{1}}+\frac{\mathrm{m}_{2}}{\rho_{2}}\right)$
divided by $\mathrm{m}_{2}$ both side
$1+\frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\left(\frac{\rho_{\mathrm{w}}}{\rho_{1}}\right) \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}+\frac{\rho_{\mathrm{w}}}{\rho_{2}}$
$1+\frac{3}{5}=\frac{3}{5} \times 0.5+\frac{1}{x}$
$\frac{8}{5}=\frac{6}{5}+\frac{1}{x} \Rightarrow \frac{1}{x}=\frac{2}{5}$
$x=2.5$
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