The ratio of masses of a cubical block of woo | vedclass

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(A) Let $m_1$ and $v_1$ be the mass and volume of the wood,and $m_2$ and $v_2$ be the mass and volume of the concrete.Given the ratio of masses: $\fra

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) Let $m_1$ and $v_1$ be the mass and volume of the wood,and $m_2$ and $v_2$ be the mass and volume of the concrete.Given the ratio of masses: $\frac{m_1}{m_2} = \frac{3}{5}$.Let $\rho_1$ be the density of wood and $\rho_2$ be the density of concrete. The specific gravity of wood is $S_1 = \frac{\rho_1}{\rho_w} = 0.5$,so $\rho_1 = 0.5 \rho_w$.For the combination to just float with its entire volume submerged,the total weight must equal the buoyant force:$(m_1 + m_2)g = \rho_w (v_1 + v_2)g$$m_1 + m_2 = \rho_w (\frac{m_1}{\rho_1} + \frac{m_2}{\rho_2})$Dividing both sides by $m_2$:$\frac{m_1}{m_2} + 1 = \frac{\rho_w}{\rho_1} \cdot \frac{m_1}{m_2} + \frac{\rho_w}{\rho_2}$Since $\frac{\rho_w}{\rho_1} = \frac{1}{S_1} = \frac{1}{0.5} = 2$ and $\frac{\rho_w}{\rho_2} = S_2$ (specific gravity of concrete):$\frac{3}{5} + 1 = 2 \cdot \frac{3}{5} + S_2$$\frac{8}{5} = \frac{6}{5} + S_2$$S_2 = \frac{8}{5} - \frac{6}{5} = \frac{2}{5} = 0.4$Wait,re-evaluating the calculation: $\frac{1}{S_2} = \frac{8}{5} - \frac{6}{5} = \frac{2}{5} \Rightarrow S_2 = 2.5$.Thus,the specific gravity of concrete is $2.5$.

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 its entire volume submerged under water. If the specific gravity of wood is $0.5$,then the specific gravity of concrete will be:

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