An iron piece of mass 860g and a density of 8g/cm³ is suspended by a rope so that it is partially submerged in oil of density 0.9g/cm³. Find the tension in the string

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An iron piece of mass 860g and a density of 8g/cm³ is suspended by a rope so that it is partially submerged in oil of density 0.9g/cm³. Find the tension in the string.

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Solution

Mass = 0.860 kg
Density of iron = 8000 kg/m³
Density of oil = 900 kg/m³
Gravitational acceleration = 9.8 m/s²

Volume of the iron piece = mass / density = 0.860 / 8000 = 1.075 × 10⁻⁴ m³

Since the piece is half-submerged, the submerged volume = 0.5 × 1.075 × 10⁻⁴ = 5.375 × 10⁻⁵ m³

Buoyant force = density of oil × submerged volume × g
= 900 × 5.375 × 10⁻⁵ × 9.8 ≈ 0.474 N

Weight of the iron = mass × g = 0.860 × 9.8 = 8.428 N

Tension in the string = weight – buoyant force
= 8.428 – 0.474 = 7.954 N

Final answer: Tension in the String= 7.95 N

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