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Difficulty: EasyFluids at Rest, Archimedes' Principle and Viscosity

A solid object has a mass of 0.50 kg0.50\text{ kg} and a volume of 2.0×10−4 m32.0 \times 10^{-4}\text{ m}^3. If it is completely immersed in a liquid of density 800 kg/m3800\text{ kg/m}^3, what is the magnitude of the upthrust exerted on the object? [Take g=10 m/s2g = 10\text{ m/s}^2]

  1. 1.6 N1.6\text{ N}Answer
  2. B
    5.0 N5.0\text{ N}
  3. C
    3.4 N3.4\text{ N}
  4. D
    0.16 N0.16\text{ N}

Answer

The magnitude of the upthrust exerted on the object is 1.6 N1.6\text{ N}.
By Archimedes' Principle, upthrust is equal to the weight of the liquid displaced: U=ρliquidVgU = \rho_{\text{liquid}} V g. Substituting ρ=800 kg/m3\rho = 800\text{ kg/m}^3, V=2.0×10−4 m3V = 2.0 \times 10^{-4}\text{ m}^3, and g=10 m/s2g = 10\text{ m/s}^2 yields 1.6 N1.6\text{ N}.

Step-by-Step Solution

1
Identify the given values and state Archimedes' Principle
Volume of displaced liquid V=2.0×10−4 m3V = 2.0 \times 10^{-4}\text{ m}^3, density of liquid ρ=800 kg/m3\rho = 800\text{ kg/m}^3, acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2. Upthrust formula is U=ρVgU = \rho V g.
According to Archimedes' Principle, the upward buoyant force (upthrust) equals the weight of the liquid displaced by the submerged object.
2
Calculate the upthrust
U=800 kg/m3×(2.0×10−4 m3)×10 m/s2=1.6 NU = 800\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 1.6\text{ N}.
Multiplying fluid density by submerged volume and gravitational acceleration gives the force in newtons.

Key Concept

Archimedes' Principle and Upthrust
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Topics

Fluids at Rest, Archimedes' Principle and Viscosity

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