# A pencil in a beaker of water

You have a beaker containing 100ml of (slightly salty) water, standing on a digital scale. The scale has been calibrated to the beaker so that it shows only the weight of the water, which is exactly 100.0 grams. The beaker is perfectly cylindrical, and the depth of the water in it is exactly 5 cm.

You have a pencil which is half the density of the water. The rear end of the pencil is flat and has a surface area of 0.5 square centimetres. You put the rear end of the pencil into the water until it barely touches the bottom, and hold it here.

What will the scale say? (Hand holding the pencil not pictured)

• sorry, but this seems not on-topic, according to the scope defined in the help center. such off-topic posts may get deleted or closed. please check the help center or faqs on the meta site to see what questions you should/ can ask here on P.SE. happy puzzling! ;) – Omega Krypton Oct 16 at 13:42
• I'm not sure this is completely on topic, but there are a couple of interesting ways to get the wrong answer. – Jaap Scherphuis Oct 16 at 13:42
• So far this has been borne out. None of the answers are correct.... – Jaap Scherphuis Oct 16 at 14:57
• @OmegaKrypton You don't see any irony in explaining the concept of "on-topic" and recommending the help centre to someone who's been active on this site for 2 years more than you? – Rand al'Thor Oct 16 at 15:14
• @Brandon_J I wouldn't be so quick to make the conclusion that because it's a phsyics problem that it must be off-topic. The meta consensus mentions that one aspect of a math/physics puzzle is a counterintuitive result, which this question certainly has. The solution also requires clever thinking to avoid wrong pathways, and I would argue that the situation presented in the puzzle is fairly unexpected/interesting, certainly not a standard textbook problem. Thus, I think this question should be left open – PiIsNot3 Oct 16 at 23:25

Well it seems to me that

The delta weight will be equal to the force of buoyancy against the pencil. This is Archimedes Principle in that the buoyant force is equal to the weight of the water displaced.

Also, as @Daniel Mathias points out

the water level goes up. So we have initially $$5 \pi r^2 = 100$$. Afterwards our new height is $$x$$ so $$x \pi r^2 - 0.5 x = 100$$. Solving this gives $$x = 5.128205$$.

Therefore

The weight of the water displaced is .5 sq. cm x 5.13 cm x 1000 kg/ cubic meter (the density of water) x 1/1000000 cubic meter/ cubic cm which comes out to be .002564 Kg or 2.564 grams

So the scale would read very close to

102.564 grams

• I think you need to account for the fact that the pencil is half the density of water - you're displacing ~2.5g of water, but replacing it with 1.25g of pencil. If you insert a "pencil" made of vacuum, the buoyant force should be greater, but if you insert a "pencil" made of water, the buoyant force should be 0. This answer suggests the pencil material is irrelevant. – Nuclear Wang Oct 16 at 14:49
• i do think the pencil material is irrelevant, only the volume affects the buoyant force. Of course I may be wrong, too. :) – SteveV Oct 16 at 14:51
• the difference there is the force applied by gravity. that has been removed from this puzzle because he's holding it in place (either keeping it from sinking if it's made of lead or pushing it down if it's made of Styrofoam) – SteveV Oct 16 at 14:56

The forces on the beaker+water system are

gravity, acting downward (100g)

and

the force you exert to counterbalance the buoyancy of the pencil (a volume of 2.5cm^3, a density of 0.5g/cm^3, so effective weight of 1.25g)

and

balancing those, the upward reaction force from the weighing scale, which determines what weight the scale reads.

So I think the scale will measure

101.25g.

(I am adopting the convenient approximation, already present in the question itself, that the density of water is exactly 1g/cm^3, which isn't exactly right but is very close.)

But that's not quite right because

the water level will rise a bit when the pencil goes in, so the volume of water displaced by the pencil is slightly larger than the figure above. (Thanks to Daniel Mathias in comments for noticing this; I missed it.) Let's be more precise. The beaker contained 100cm^3 of water at a height of 5cm so its base area is 20cm^2; now a column of base-area 0.5cm^2 contains pencil rather than water, so the area has decreased by a ratio of 39/40 and the height has therefore increased by 40/39. So the depth is now 200/39cm or about 5.13cm, so instead of 1.25g as above we need 40/39*1.25g=50/39g ~= 1.28g. So the scale, if accurate enough, will measure 101.28g instead of 101.25g.

• What, exactly, is the depth of the water with the pencil in place? I think this is an important detail. – Daniel Mathias Oct 16 at 14:44
• Oh, yeah, that's an important little correction term to include. – Gareth McCaughan Oct 16 at 15:30
• Fixed now. Thanks! – Gareth McCaughan Oct 16 at 15:34
• ... Though on reflection I think I agree with SteveV's different answer :-). – Gareth McCaughan Oct 16 at 16:14

I'll post my own answer to the question, mainly so that I can highlight and explain some of the pitfalls of the puzzle.

It takes force to hold the pencil in place. The pencil is buoyant, to keep it just touching the bottom of the beaker means you are pushing down. Therefore the weight shown by the scale must increase.
If the pencil were glued to the bottom of the beaker, then there would be no net force on the system - the beaker pulls the pencil down, the pencil pulls the beaker up by the same amount - and the measured weight would not change. But that is not the situation we have here.

The density (or weight) of the pencil is irrelevant, and was provided in the question as a red herring. As SteveV pointed out, only the volume of the displaced water matters. Whether the pencil were made of lead or styrofoam, you are holding it in place so it displaces the same amount of water either way. Your hand pushes down and combines with the gravity on the pencil, and together they counteract the upwards force from the Archimedes principle of the displaced water. The latter is the same regardless, and it is the reaction force against that water displacement that gets added to the scale's measurement. Changing the weight of the pencil only changes the force that your hand exerts, but not the water displacement, and not the reading on the scale.

When you push the pencil into the water, the water level rises. Daniel Matthias was the first to point this out. The rising level in turn increases the amount of displaced water. You need to take the final water level into account when calculating the total volume of displaced water. The really eagle-eyed among you might notice that the drawing in the question does indeed have a water level rise of a few pixels.

The quickest way to calculate the answer is as follows.
The original cylinder of water has a base or surface area of $$\frac{100\text{ ml}}{5\text{ cm}} = 20\text{ cm}^2$$.
After the pencil is inserted, that has effectively been reduced to $$19.5\text{ cm}^2$$
The height of the water has then effectively become $$\frac{20}{19.5}$$ times higher. The scales will show the weight you would have if the all displaced volume, i.e. the volume of the submerged pencil, were replaced by water. So it is as if the beaker contained only a cylinder of water but at this new water level. It therefore shows $$\frac{20}{19.5}*100 = 102.564$$, or $$102.6\text{ g}$$ when rounded on a 1-decimal display.