A cubical block of wood of specific gravity 0 | vedclass

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(C) Let the mass of wood be $M$ and the mass of concrete be $m$. The density of water is $\rho_w$. The specific gravity is the ratio of the density of

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$3/5$

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(C) Let the mass of wood be $M$ and the mass of concrete be $m$. The density of water is $\rho_w$. The specific gravity is the ratio of the density of the substance to the density of water.Volume of wood $V_w = \frac{M}{0.5 \rho_w} = \frac{2M}{\rho_w}$.Volume of concrete $V_c = \frac{m}{2.5 \rho_w} = \frac{m}{2.5 \rho_w} = \frac{0.4m}{\rho_w}$.For the combination to float with its entire volume submerged,the total weight must equal the buoyant force:$(M + m)g = (V_w + V_c) \rho_w g$$M + m = \left(\frac{2M}{\rho_w} + \frac{0.4m}{\rho_w}\right) \rho_w$$M + m = 2M + 0.4m$$m - 0.4m = 2M - M$$0.6m = M$$\frac{M}{m} = \frac{0.6}{1} = \frac{6}{10} = \frac{3}{5}$.

$A$ 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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