A piece of steel has a weight W in air,W_1 wh | vedclass

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(B) Let $V$ be the volume of the steel piece,$\rho_w$ be the density of water,and $\rho_L$ be the density of the unknown liquid.According to Archimede

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$\frac{W - W_2}{W - W_1}$

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(B) Let $V$ be the volume of the steel piece,$\rho_w$ be the density of water,and $\rho_L$ be the density of the unknown liquid.According to Archimedes' principle,the apparent weight of an object immersed in a fluid is $W_{apparent} = W - F_B$,where $F_B$ is the buoyant force.For water: $W_1 = W - V \rho_w g \implies W - W_1 = V \rho_w g$ (Equation $1$).For the unknown liquid: $W_2 = W - V \rho_L g \implies W - W_2 = V \rho_L g$ (Equation $2$).The relative density $(RD)$ of the liquid is defined as the ratio of the density of the liquid to the density of water: $RD = \frac{\rho_L}{\rho_w}$.Dividing Equation $2$ by Equation $1$: $\frac{W - W_2}{W - W_1} = \frac{V \rho_L g}{V \rho_w g} = \frac{\rho_L}{\rho_w}$.Therefore,$RD = \frac{W - W_2}{W - W_1}$.

$A$ piece of steel has a weight $W$ in air,$W_1$ when completely immersed in water,and $W_2$ when completely immersed in an unknown liquid. The relative density (specific gravity) of the liquid is:

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