A liquid mixture of volume V has two liquids | vedclass

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(C) Let the mass of the liquid with density $\alpha$ be $M_1$ and the mass of the liquid with density $\beta$ be $M_2$.Total volume $= V$.Total densit

Vedclass · Accepted Answer

$\frac{\alpha V(\beta-\sigma)}{\beta-\alpha}$

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(C) Let the mass of the liquid with density $\alpha$ be $M_1$ and the mass of the liquid with density $\beta$ be $M_2$.Total volume $= V$.Total density of the mixture $= \sigma$.Total mass $= M_1 + M_2 = V\sigma$.Therefore,$M_2 = V\sigma - M_1 \dots (1)$.The density of the mixture is given by $\sigma = \frac{	ext{Total Mass}}{	ext{Total Volume}} = \frac{M_1 + M_2}{\frac{M_1}{\alpha} + \frac{M_2}{\beta}} \dots (2)$.Substituting equation $(1)$ into equation $(2)$:$\sigma = \frac{V\sigma}{\frac{M_1}{\alpha} + \frac{V\sigma - M_1}{\beta}}$.$\frac{M_1}{\alpha} + \frac{V\sigma - M_1}{\beta} = \frac{V\sigma}{\sigma} = V$.$\frac{M_1}{\alpha} - \frac{M_1}{\beta} = V - \frac{V\sigma}{\beta}$.$M_1 \left( \frac{\beta - \alpha}{\alpha\beta} ight) = V \left( \frac{\beta - \sigma}{\beta} ight)$.$M_1 = \frac{V(\beta - \sigma)}{\beta} 	imes \frac{\alpha\beta}{\beta - \alpha} = \frac{\alpha V(\beta - \sigma)}{\beta - \alpha}$.

$A$ liquid mixture of volume $V$ has two liquids as its ingredients with densities $\alpha$ and $\beta$. If the density of the mixture is $\sigma$,then the mass of the first liquid in the mixture is ............

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