A metallic sphere of mass 1 kg and volume 2 × 10⁻⁴ m³ is completely immersed in water. The buoyant force exerted by water on the sphere is: (Given: density of water = 1000 kg/m³, g = 10 m/s²)

Correct — C, 2 N. Archimedes' principle says the buoyant force equals the weight of the fluid displaced, so the only quantities that matter are the displaced volume, the density of the fluid and g. Because the sphere is completely immersed, the volume it displaces is its own volume, 2 times 10 to the power minus 4 cubic metres. The mass of that displaced water is density times volume, 1000 times 2 times 10 to the power minus 4, which is 0.2 kilograms, and its weight is 0.2 times 10, which is 2 newtons. That is the upthrust. Notice what the sphere's own mass does and does not do. It plays no part in the upthrust at all; it only tells you that the sphere weighs 10 newtons, and since 10 newtons is far more than the 2 newtons pushing up, the sphere sinks — its apparent weight in water being 10 minus 2, that is 8 newtons.

• (a)0·5 N — Does not follow from any correct combination of the data. It would need a displaced mass of 0.05 kg, a quarter of the water actually displaced.
• (b)1·5 N — A near miss designed to catch a slip in the power of ten or in the multiplication. Density times volume gives 0.2 kg, and 0.2 times 10 is exactly 2.
• (d)2·5 N — Another arithmetic near miss. Check the calculation by tracking units: kg per cubic metre times cubic metres gives kilograms, and kilograms times metres per second squared gives newtons.

The upthrust on an immersed body comes from the increase of fluid pressure with depth: the pressure on the lower surface exceeds that on the upper, and the difference, integrated over the body, is a net upward force equal to the weight of the displaced fluid. Written as a formula, buoyant force equals the density of the fluid times the volume displaced times g. For a fully immersed body the volume displaced is the body's own volume; for a floating body it is only the submerged part, and it adjusts itself until the upthrust equals the weight.

The instructive feature of this item is the extra datum. Supplying the sphere's mass tempts you to use it, but the mass of the immersed body never enters the buoyancy formula — only the fluid's density, the displaced volume and g. A quick consistency check is worth running once the answer is out: the sphere's density is mass over volume, 1 divided by 2 times 10 to the power minus 4, which is 5000 kilograms per cubic metre, five times that of water. A body five times as dense as water should experience an upthrust one-fifth of its weight, and indeed 2 newtons is one-fifth of 10 newtons. Consistency of that kind is the fastest way to catch a factor-of-ten error.

• Buoyant force equals the density of the fluid multiplied by the volume of fluid displaced multiplied by g.
• For a completely immersed body the volume displaced equals the volume of the body itself.
• Here the upthrust is 1000 x 2 x 10 to the power minus 4 x 10 = 2 N.
• The mass of the immersed body does not enter the buoyancy calculation; it determines the weight and hence whether the body sinks.
• The sphere's density is 5000 kg per cubic metre, five times that of water, so it sinks, and its apparent weight in water is 10 N minus 2 N, that is 8 N.

• Using the mass of the body in the buoyancy formula instead of the density of the fluid.
• Slipping a power of ten when handling a volume written in scientific notation.
• Forgetting that the displaced volume equals the body's volume only when the body is fully immersed.

As a direct numerical on upthrust with a redundant datum supplied, or as a follow-on asking for the apparent weight of the same body in water.

• practice — not a real PYQ

  A body of volume 5 × 10⁻⁴ m³ is completely immersed in water. The buoyant force on it is: (density of water = 1000 kg/m³, g = 10 m/s²)

  • (a)0·5 N
  • (b)5 N
  • (c)50 N
  • (d)500 N

  Answer(b) 5 N — the upthrust is 1000 × 5 × 10⁻⁴ × 10 = 5 N, and the body's own mass is irrelevant to it.

• practice — not a real PYQ

  A body weighing 10 N in air experiences an upthrust of 2 N when fully immersed in water. Its apparent weight in water is:

  • (a)12 N
  • (b)10 N
  • (c)8 N
  • (d)2 N

  Answer(c) 8 N — the apparent weight is the true weight less the upthrust, 10 N minus 2 N.