Question

A cube is placed in a tub of water and floats with 2.50 m of the cube submerged. An identical cube is placed in a tub of an unknown liquid, and floats with 1.50 m of it submerged. What is the density of the unknown liquid in kg/m3?

Answer 1

Application of Archimedes principle for buoyant force.

Ans) lube of base area Let a cube А and hei of is 8 mass M, votume V, height Deusity of material of cube HIC weight of the cube (An Area of base) mg = 8vg W = Dersity = Mass volume This cube is immersed in water, subme in water to the T+ depth h= 2.5om Ale to Archimedes principle, h 1 the buoyant force will be equal to the M9 mg of the weight water displaced submerged the su cube by F = ptg Vw = Density of water V = Volume submerged

To float, W = In tg svg Sow (Ah) - 3(AH) co Sw h 3 3 H So, the fraction given by of volume submerged is u f . V V Р Su H W Giren that h=2.50 in water Sw = lovo regime je 1000X 2.50 H 2500 Н. 4 another liquid in of When it is submerged unknown density 9", ch' = 1.som Øn 5 Answer .. 8 Н. Equate ③ and ④ 1000X 2.50 H S'x150 H P= 1666.7 kylm?