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 liquid is

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$\frac{w - w_{2}}{w - w_{1}}$

$\frac{w_{1} - w_{2}}{w - w_{1}}$

$\frac{w_{1} - w_{2}}{w - w_{2}}$

$\frac{w - w_{1}}{w - w_{2}}$

answer is B.

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Detailed Solution

$(w - w_{1}) = V ρ_{w} g$ _________(i)

$(w - w_{2}) = V ρ_{L} g$ _________(ii)

Dividing eq. (2) by (1), we have

$\frac{ρ_{L}}{ρ_{w}} = \frac{w - w_{2}}{w - w_{1}} =$ relative density of known liquid

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