KCSE 2025 Physics P1 Q18 — Archimedes' Principle & Upthrust

KCSE 2025 Form 4 Fluids

Published

The Question

“This is a Section B question on Archimedes' principle and floating. (a) State Archimedes' principle. (b) A metal block is lowered on a thread into a measuring cylinder of water and weighed on a spring balance. (b)(i) Describe how to find the mass of water displaced from the initial volume V₁ and final volume V₂. (b)(ii) Describe how to find the upthrust from its weight in air and its apparent weight in water, and state how this verifies Archimedes' principle. (c)(i) State two ways of reducing the surface tension of a liquid. (c)(ii) An object sinks fully when placed in water but floats partly in glycerine; explain this observation.”

1

(a) State Archimedes' principle

Archimedes' principle states that when a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces.

2

(b)(i) Find the mass of water displaced

First read the initial volume V₁ of water in the measuring cylinder. Then lower the metal block in on a thread until it is fully submerged and read the new volume V₂. The volume of water pushed aside is V₂ minus V₁, and the mass of that displaced water is the density of water multiplied by the displaced volume.

Vd=V2−V1V_d = V_2 - V_1
Mass=ρwater×Vd\text{Mass} = \rho_{\text{water}} \times V_d
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(b)(ii) Find the upthrust and verify the principle

Weigh the metal block in air with the spring balance to get its true weight, then weigh it again while it is fully submerged to get its smaller, apparent weight in water. The upthrust is the weight in air minus the weight in water. Archimedes' principle is verified when this upthrust equals the weight of the water displaced found in part (b)(i).

Upthrust U=Wair−Wwater\text{Upthrust } U = W_{\text{air}} - W_{\text{water}}
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(c)(i) Reduce the surface tension of a liquid

The surface tension of a liquid can be reduced in two ways: by raising its temperature, or by adding a detergent or soap to the liquid.

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(c)(ii) Explain floating in glycerine but sinking in water

Glycerine is denser than water, so for the same displaced volume it provides a greater upthrust. The object's density is greater than that of water, so in water its weight beats the upthrust and it sinks. But its density is less than that of glycerine, so glycerine can supply enough upthrust to balance the object's weight before it is fully immersed — hence it floats partly submerged in glycerine.

Final Result

The upthrust on the block is U = W(air) − W(water), and this equals the weight of the water displaced, ρ_water × V_d × g, which verifies Archimedes' principle. Surface tension is reduced by heating the liquid or adding a detergent; the object floats in glycerine because glycerine is denser than the object while water is less dense.

Why this method works

Archimedes' principle follows from pressure increasing with depth: the fluid pushes up harder on the bottom of a submerged body than it pushes down on the top, and that pressure difference is the upthrust. Working out the sum shows it always equals the weight of the fluid the body pushes out of the way, no more and no less. That is why two independent measurements must agree: the weight of water collected in the cylinder (ρ_water × V_d × g) and the loss of weight of the block when submerged (W_air − W_water) are two views of the same force. Whether a body floats or sinks is then a contest between its own weight and the largest upthrust the fluid can give — settled entirely by comparing densities, which is exactly why the object behaves differently in water and in the denser glycerine.

✓

A submerged object always weighs less on the balance than in air; the missing weight (W_air − W_water) is precisely the weight of the fluid whose place the object has taken.