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As we know that pressure becomes greater with decrease of height in a liquid column. If an object having lower density than that of the liquid cut in a shape of a pyramid or a right circular cone is immersed upright in the liquid then is this possible at all for the object to sink? Because due to the shape the force on the surfaces apart from the base of the object will act downwards.

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By Archimedes principle, the upward buoyancy force the water exerts on an object is equal to the weight of the fluid displaced by that object.

Regardless of how you cut an object, in order for the object to sink, the weight of that object must exceed the upward buoyancy force.

This therefore means that the weight of the displaced water must be less than the weight of the object. This is only possible if the object is denser than water.

Therefore an object that is less dense than water will not sink regardless of the object's shape.

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As Mews says, Archimedes principle applies to any shape so you can't make an object sink just by changing its shape. If you're interested Floris gives a simple proof of Archimedes principle here.

But let me suggest an alternative sort of proof by pointing out that if your idea worked you could use it to make a perpetual motion machine. Suppose you take some flexible object and put some scaffolding inside it to force it into a shape that will sink. Put the object in the water, let it sink and extract work from it as it sinks (e.g. you could get it to turn a generator).

Now have the scaffolding automatically collapse at some convenient depth, so the water pressure will force the object into a sphere. We know a sphere of lower density than water floats, so our object will float up to the surface again.

Finally, re-erect the scaffolding to make the object sink again, extract more work, and repeat indefinitely to keep generating energy from nothing. Voila! A perpetual motion machine.

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  • $\begingroup$ So ultimately its impossible. $\endgroup$ Commented Jun 20, 2015 at 11:39
  • $\begingroup$ @Angelika: Yes. $\endgroup$ Commented Jun 20, 2015 at 14:00
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While in general I agree with other answers, there is some subtlety here. If you put your pyramid on the bottom of the liquid, and both the bottom of the liquid and the base of the pyramid are smooth enough, so the water (for example) cannot get under the pyramid (hydrophobic surfaces can help), the pyramid will stay at the bottom. The same would be true for a cube or other shapes though (http://www.aps.org/units/fed/newsletters/spring2006/mungan.html).

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  • $\begingroup$ Aren't you just describing a suction cup? $\endgroup$ Commented Jun 20, 2015 at 13:08
  • $\begingroup$ @MikeDunlavey: Not necessarily, but yes, this is an example of what I described. $\endgroup$
    – akhmeteli
    Commented Jun 20, 2015 at 13:15
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Mew's answer really explains the principle in general, however here is a little bit of further elaboration concerning shapes.

It is true that that the pin tipped shape of the object will direct all the weight at one point and apparently make it sink-friendly. BUT think about the big ugly wide upper part of the object. This wider part of the object is working against the tipped part by offering more resistance to sinking. So the sink-friendliness of the narrower part is being constantly undone by the sink-unfriendliness of the upper part. The real "sinkiness" of the object is finally determined by its overall density.

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