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Difficulty: EasyFluids at Rest, Archimedes' Principle and Viscosity

A solid object of volume 0.002 m30.002\text{ m}^3 is completely immersed in water of density 1000 kg/m31000\text{ kg/m}^3. What is the magnitude of the upthrust exerted on the object by the water? [Take g=10 m/s2g = 10\text{ m/s}^2]

Answer: 20 N

Answer

The magnitude of the upthrust exerted on the object is 20 N20\text{ N}.
According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using U=V⋅ρ⋅gU = V \cdot \rho \cdot g. Substituting V=0.002 m3V = 0.002\text{ m}^3, ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3, and g=10 m/s2g = 10\text{ m/s}^2 yields U=0.002×1000×10=20 NU = 0.002 \times 1000 \times 10 = 20\text{ N}.

Step-by-Step Solution

1
Identify the given quantities from the problem statement.
V=0.002 m3V = 0.002\text{ m}^3, ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3, and g=10 m/s2g = 10\text{ m/s}^2.
These are the essential inputs required to calculate the weight of the displaced liquid.
2
Apply Archimedes' principle to find upthrust force.
U=Vρg=0.002×1000×10=20 NU = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}.
Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Key Concept

Archimedes' Principle and Upthrust
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Topics

Fluids at Rest, Archimedes' Principle and Viscosity

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