Question

A body weighs 75 g in air ,70.5 g in water and 66 g in liquid.Calculate relative density of liquid

Answer 1

Given W₁ = 75g;W₂ = 70.5g;W₃ = 66g

W₁-W₂→Weight of water
W₁-W₃→Weight of  liquid

R.D → Weight of  liquid/Weight of water
= (75g - 66g)/(75g - 70.5g)
= 9g/4.5g
= 2

Therefore R.D of liquid is 2

Answer 2

Answer:

The relative density of the liquid is 2

Explanation:

Given that,

The weight of the body in air is 75gm

$= > W_{ba}=75gm$

Weight of body in water is 70.5gm

$= > W_{bw}=70.5gm$

Therefore weight of the water is calculated as

$W_{w}=W_{ba} - W_{bw} \\ = 75 - 70.5 = 4.5gm$

Also given that the weight of the body in liquid is 66gm

$= > W_{bl}=66gm$

Therefore,the weight of the liquid is calculated as

$W_{l}=W_{ba} - W_{bl} \\ = 75 - 66 = 9gm$

The relative density or specific gravity of a liquid is the ratio of weight of liquid to the weight of water.Therefore,

$R.D=\frac{W_{l}}{W_{w}}$

Substituting the known values we get

$R.D=\frac{9}{4.5} = 2$

Therefore,the relative density of the liquid is 2

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