A body weights 75 gm in air, 51 gm when
completely immersed in unknown liquid and 67
gm when completely immersed in water. Find
the density of the unknown liquid
(A) 6 gm / cm (B) 4 gm / cm
(C) 3 gm /cm (D) 8 gm/cm
Answers
Answer ⇒ The density of unknown liquid is 3 g/cm³.
Explanation ⇒ Weight of body in Air = 75 gm.
Weight of body in unknown liquid = 51 gm.
Loss in weight in unknown liquid = 75 - 51
= 24 gm.
Upthrust = loss in weight in unknown liquid
Vσg = 24 ------(i)
where V is the volume of body since body is completely immersed in unknown liquid, and σ is the density of unknown water.
Now, Weight of body in water = 67 gm.
Loss of weight in water = 75 - 67 = 8 gm.
∴ Upthrust = loss in weight.
∴ Vρg = 8 -----(ii)
where V is the volume of body since body is completely immersed in water, and ρ is the density of water.
(i) ÷ (ii)
σ/ρ = 24/8
∴ σ = 3 × ρ
∴ σ = 3 × 1
∴ σ = 3 g/cm³.
Hence, the density of unknown liquid is 3 g/cm³.
Hope it helps.
Weight of body in Air = 75 gm.
Weight of body in unknown liquid = 51 gm.
Loss in weight in unknown liquid = 75 - 51
= 24 gm.
Upthrust = loss in weight in unknown liquid
Vσg = 24 ------(i)
where V is the volume of body since body is completely immersed in unknown liquid, and σ is the density of unknown water.
Now, Weight of body in water = 67 gm.
Loss of weight in water = 75 - 67 = 8 gm.
∴ Upthrust = loss in weight.
∴ Vρg = 8 -----(ii)
where V is the volume of body since body is completely immersed in water, and ρ is the density of water.
(i) ÷ (ii)
σ/ρ = 24/8
∴ σ = 3 × ρ
∴ σ = 3 × 1
∴ σ = 3 g/cm³.
Hence, the density of unknown liquid is 3 g/cm³.