A cubical block of wood of specific gravity $0.5$ and a chunk of concrete of specific gravity $2.5$ are fastened together. The ratio of the mass of wood to the mass of concrete which makes the combination to float with its entire volume submerged under water is
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Let the mass of wood and concrete be $M$ and
$m$ respectively. Then
$\frac{M}{0.5}+\frac{m}{2.5}=(M+m)$
$\mathrm{Or}$ $M \left(\frac{1}{0.5}-1\right)=m\left(1-\frac{1}{2.5}\right)$
$\mathbf{o r}$ $M=m\left(\frac{1.5}{2.5}\right)=m\left(\frac{3}{5}\right)$
$\therefore$ $\frac{M}{\operatorname{m}}=\frac{3}{5}$
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