Question

a cylinder made of Copper and Aluminium floats in Mercury of density 13.6 gram per centimetre cube such that 0.26 part of it is below Mercury find the density of the solid​

Answer 1

The density of the solid can be calculated as follows:

• As per the question, Density of Mercury = ρ(hg) = 13.6 gram per centimeter cube

• Consider V(solid)= Volume of solid cylinder.

• ∴ Volume of mercury displaced by the solid cylinder = V(hg) = 0.26 V(solid)

• Now, By Law of Flotation,

Weight of solid cylinder = Weight of mercury displaced by the solid cylinder.

⇒ V(solid) * ρ(solid) * g = V(hg) * ρ(hg) * g

⇒ V(solid) * ρ(solid) = 0.26 V(solid) * 13.6

⇒ ρ(solid) = 0.26 * 13.6

                 = 3.536  gram per centimeter cube

• Therefore, the density of solid is 3.536  gram per centimeter cube.

Answer 2

The density of the solid is 3.536 gcm^-3.

• From given, we have,
• density of mercury
• ρHg = 13.6 gcm^-3
• 0.26 part of the solid cylinder is below mercury
• Volume of mercury displaced by the immersed part of the solid cylinder
• VHg = 0.26 VSolid
• where,

• VSolid is the volume of solid cylinder.
• we have formula, by law of flotation,
• VSolid × ρsolid × g = VHg × ρHg × g
• weight of the solid cylinder = weight of mercury displaced by immersed part of solid cylinder
• ⇒
• VSolid × ρsolid = VHg × ρHg
• VSolid × ρsolid = 0.26 VSolid × 13.6
• ρsolid = 0.26 × 13.6
• ρsolid = 3.536
• Therefore, the density of the solid cylinder = 3.563 gcm^-3