In making an alloy,a substance of specific gr | vedclass

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(C) The specific gravity of a substance is defined as the ratio of its density to the density of water $(d_w)$.$s = \frac{\rho}{\rho_w} \implies \rho

Vedclass · Accepted Answer

$\left[ \frac{m_1 + m_2}{(m_1/s_1 + m_2/s_2)} \right]$

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(C) The specific gravity of a substance is defined as the ratio of its density to the density of water $(d_w)$.$s = \frac{ho}{ho_w} \implies ho = s \cdot ho_w$.Total mass of the alloy = $m_1 + m_2$.Total volume of the alloy = $V_1 + V_2 = \frac{m_1}{ho_1} + \frac{m_2}{ho_2} = \frac{m_1}{s_1 ho_w} + \frac{m_2}{s_2 ho_w} = \frac{1}{ho_w} \left( \frac{m_1}{s_1} + \frac{m_2}{s_2} ight)$.The specific gravity of the alloy $(s_{alloy})$ is the ratio of the density of the alloy to the density of water:$s_{alloy} = \frac{ho_{alloy}}{ho_w} = \frac{	ext{Total Mass}}{	ext{Total Volume} 	imes ho_w}$.Substituting the values:$s_{alloy} = \frac{m_1 + m_2}{\left[ \frac{1}{ho_w} \left( \frac{m_1}{s_1} + \frac{m_2}{s_2} ight) ight] 	imes ho_w} = \frac{m_1 + m_2}{\frac{m_1}{s_1} + \frac{m_2}{s_2}}$.

In making an alloy,a substance of specific gravity $s_1$ and mass $m_1$ is mixed with another substance of specific gravity $s_2$ and mass $m_2$. Then,the specific gravity of the alloy is:

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