Forces, Density and Pressure: Question 5

Syllabus 4.3

Multiple choice AS 1 mark

A small solid sphere of volume 2.0×10−4 m32.0\times10^{-4}\text{ m}^3 is fully submerged in water of density 1000 kg m−31000\text{ kg m}^{-3}. Take g=9.81 m s−2g = 9.81\text{ m s}^{-2}.

What is the upthrust acting on the sphere?

Choose an answer to check it, then compare with the worked solution below.

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Worked solution

Setting up the problem

Archimedes’ principle states that the upthrust on a submerged object equals the weight of fluid it displaces: F=ρgVF = \rho g V where ρ\rho is the density of the fluid, VV is the volume of fluid displaced (here, the full volume of the sphere, since it is fully submerged), and gg is the acceleration of free fall.

Calculating the upthrust

F=ρgV=(1000)(9.81)(2.0×10−4)F = \rho g V = (1000)(9.81)(2.0\times10^{-4}) F=9810×2.0×10−4F = 9810 \times 2.0\times10^{-4} F=1.962 N≈1.96 NF = 1.962\text{ N} \approx 1.96\text{ N}

Why the other options are wrong

  • A (0.20 N0.20\text{ N}): this omits gg entirely, i.e. calculates ρV\rho V instead of ρgV\rho g V.
  • C (2.00 N2.00\text{ N}): this uses g=10 m s−2g = 10\text{ m s}^{-2} instead of the given value g=9.81 m s−2g = 9.81\text{ m s}^{-2}.
  • D (19.6 N19.6\text{ N}): this is ten times too large, from a decimal/power-of-ten slip in the volume (e.g. using 2.0×10−3 m32.0\times10^{-3}\text{ m}^3 instead of 2.0×10−4 m32.0\times10^{-4}\text{ m}^3).

Final answer

  • Upthrust on the sphere =1.96 N= \boxed{1.96}\text{ N}, option B.