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 :

\frac{W - W_{1}}{W - W_{2}}

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\frac{W - W_{2}}{W - W_{1}}

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\frac{W_{1} - W_{2}}{W - W_{1}}

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\frac{W_{1} - W_{2}}{W - W_{2}}

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Solution

The correct option is A $\frac{W - W_{1}}{W - W_{2}}$
Weight of the body in any liquid = Weight in air- Buoyancy force

$W_{1} = W - F_{1}$

Buoyancy force in water $F_{1} = V ρ_{w} g$

$W_{1} = W - V ρ_{w} g$ ...(1)

in another liquid

$W_{2} = W - F_{2}$

Buoyancy force in liquid $F_{1} = V ρ_{l} g$

$W_{2} = W - V ρ_{l} g$ ...(2)

solving (1) and (2)

$\frac{ρ_{l}}{ρ_{w}} = \frac{W - W_{1}}{W - W_{2}}$

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A piece of steel has a weight W in air. W1 when completely immersed in water and W2 when completely immersed in an unknown liquid. the relative density (specific gravity) of liquid is :

The correct option is A W−W1W−W2Weight of the body in any liquid = Weight in air- Buoyancy forceW1=W−F1Buoyancy force in water F1=VρwgW1=W−Vρwg ...(1)in another liquid W2=W−F2Buoyancy force in liquid F1=VρlgW2=W−Vρlg ...(2)solving (1) and (2)ρlρw=W−W1W−W2

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 :