Apiece 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 liquid is
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$w_{1}=w-v \cdot \rho_{w} \cdot g$
$\omega_{2}=w -V \cdot \rho_{L} \cdot g$
$w -w _{1}=V \rho_ w g$
$w -w _{2}=v \rho_{L} g$
$\frac{V \rho _L g }{V \cdot \rho_{w} g}$ $=\frac{w-w_{2}}{w-w_{1}}$
$R \cdot D=\frac{\rho_{L}}{\rho_{w}}$
$R \cdot D=\frac{w-w_{2}}{w-w_{1}}$
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