An object weighing 420 N in air is immersed in water after being tied to a string connected to a bal
An object weighing 420 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 380 N. Find the density of the object [density of water = 1.00 x 103 kgm-3, g = 9.8 ms-2]
An object weighing 420 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 380 N. Find the density of the object [density of water = 1.00 x 103 kgm-3, g = 9.8 ms-2]
The scale reads less in water because water pushes up on the object. Upthrust = weight in air − apparent weight = 420 − 380 = 40 N.
Upthrust equals the weight of water displaced, so 40 = ρw × V × g. V = 40 ÷ (1.00 × 10³ × 9.8) ≈ 4.08 × 10⁻³ m³. This is also the volume of the object.
Mass of object = 420 ÷ 9.8 ≈ 42.86 kg. Density = mass ÷ volume = 42.86 ÷ 4.08 × 10⁻³ ≈ 1.05 × 10⁴ kgm⁻³.
The 10³ values come from forgetting that the volume is a few thousandths of a cubic metre.
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