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This was the answer to a question:

When a metal sphere is dipped in a liquid of density $\rho$, with the aid of a thread, find the pressure at bottom of the vessel.

Now I personally feel as the ball displaces liquid, thus increasing height , the pressure would increase (Be greater than the original pressure). however, my book and many other sources disagree, why is it that the pressure doesn't increase?

edit:- someone pointed put the size may matter The dimensions of the vessel aren't given, so I assumed it to be of any reasonable size, bigger than the ball, , not not so big that the ball would be insignificant

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    $\begingroup$ Please state which book(s) you are referring to. In the context of the question what does the word dipped mean. Suspended from a string, dropped in $\dots$ ? $\endgroup$
    – Farcher
    Dec 10, 2023 at 10:38
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    $\begingroup$ Not all answers in books are correct. $\endgroup$
    – Farcher
    Dec 10, 2023 at 11:34
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    $\begingroup$ Depends on the shape of the container, no? If you dip the ball in the ocean, then the height won't change (not by any measurable amount). So the pressure won't change either. If you dip it in a vessel that's only slightly wider than the cross section of the sphere, then the height, (and the pressure below the ball), will rise significantly. $\endgroup$ Dec 10, 2023 at 16:02
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    $\begingroup$ What matters is actually the volume of the ball v.s. the cross-sectional area of the top surface of the liquid. The ball displaces a weight of water equal to $\rho gV$, but the pressure increase is $\rho g\Delta h$ where $\Delta h=V/A$ where, in turn, $A$ is the cross-sectional area of the top surface of the liquid. Note that the string is holding up $(\rho_{\text{metal}}-\rho)gV$ of the weight of the metal sphere. Everything needs to be accounted for. Nothing is weird. $\endgroup$ Dec 11, 2023 at 5:20
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    $\begingroup$ This question is a duplicate of "If an object suspended by a string is immersed in a liquid, will its weight contribute to the pressure at the bottom of the container?" asked on 25 November. $\endgroup$ Dec 13, 2023 at 14:17

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i don't know which question you are talking about, i just googled it and it says the pressure increases but since i am writing an answer, i need to explain why enter image description here

as you can see in the figure, increased height is 'h' and for the part where solid is present , you can replace it with the fluid having weight as that of the buoyant force (Weight of water displaced = W metal - Tension) and the result would be the same because buoyant force (in fact any force) exists in pair. So the ball would also be applying equal and opposite to the force experienced by it , on the liquid, which would also be equivalent to weight of displaced fluid. Hence, you can simply replace that ball with the weight equal to the buoyant force.

Pressure at the bottom = Atmospheric Pressure + pgh

Edit: More intuitive way to look at this is: There was liquid in the container now you poured more liquid on it which weighs equal to the buoyant force experienced by the body when it is actually immersed (in this case, Weight of displaced fluid = Weight of body- Tension), and now the height increases to h. So pressure would definitely increase.

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  • $\begingroup$ The use of formula is flawed! $$P = \rho g h$$ applies for homogeneous liquid column of height h and density $$\rho$$, this is not a homogeneous liquid! You have a metal sphere in it, it will increase density of stuff(liquid+sphere) system. $\endgroup$
    – Qwerty
    Dec 13, 2023 at 10:16
  • $\begingroup$ please read my answer. Also you are actually calculating pressure at the bottom of the vessel $\endgroup$
    – PinkAura
    Dec 13, 2023 at 12:51
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    $\begingroup$ @Qwerty i'm unable to add comment to your answer as i don't have that much reputation points so i am writing this here, "Before immersing the sphere π‘Š=π‘Šπ‘“π‘™π‘’π‘–π‘‘ After immersing π‘Š=π‘Šπ‘“π‘™π‘’π‘–π‘‘+π‘Šπ‘ π‘β„Žπ‘’π‘Ÿπ‘’ π‘Šπ‘ π‘β„Žπ‘’π‘Ÿπ‘’=0" Firstly, we see the net force acting on the "fluid" for calculation of pressure that's, Pressure = (F net)/Area, Secondly, there's weight of the initial water + opposite reaction of buoyant force acting on the fluid so ,Pressure at the bottom=( F buoyant + Weight of liquid)/ Area; which is indeed greater than initial pressure. $\endgroup$
    – PinkAura
    Dec 13, 2023 at 13:15
  • $\begingroup$ buoyant force acts on body not on fluid, let's keep it simple, just see the weight of system, it's not increasing, actually all net force on sphere is borne by string! $\endgroup$
    – Qwerty
    Dec 13, 2023 at 13:26
  • $\begingroup$ @Qwerty Buoyant force acts on the body but by THE THIRD LAW OF NEWTON, we have body applying the same force on the fluid. Also, the weight of the body is acted on the body and not the fluid so it would not affect the pressure anyway. We see the net force acting on the bottom surface by the fluid for pressure calculation.(This point would be more clear if you draw FBD) $\endgroup$
    – PinkAura
    Dec 13, 2023 at 13:42
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$$P_{fluid} = P + \rho gh$$

where,

  • $P_{fluid}$ is the pressure at a point in the fluid
  • $P$ is the pressure at a reference point
  • $\rho$ is the density of the fluid
  • $g$ is the acceleration due to gravity (usually 9.8 m/s$^2$)
  • $h$ is the height of the fluid column above or below the point

This formula can be derived by considering a small volume element of fluid with area $A$ and height $h$. The weight of this volume element is:

$$W = \rho V g$$

where,

  • $W$ is the weight of the volume element
  • $\rho$ is the density of the fluid
  • $V$ is the volume of the volume element
  • $g$ is the acceleration due to gravity

The pressure exerted by this weight on any surface in contact with it is:

$$P = \frac{W}{A} = \frac{\ m g}{A} =\frac{\rho V g}{A} = \rho V g$$

Let's keep it simple and use

$$P = \frac{\ W }{A}$$

W = weight of stuff above the area upon which pressure has to be calculated.

Before immersing the sphere $$W = W_{fluid}$$ After immersing $$W = W_{fluid} + W_{sphere}$$

$$W_{sphere} = 0$$

Since whatever may be the net force, tension in string is always gonna balance it!

Hence $$W = W_{fluid} + W_{sphere}$$ $$W = W_{fluid} + 0$$ $$W = W_{fluid} $$

$$P = \frac{\ W}{A}$$

W is unchanged A is unchanged

Hence P remains same and thus the total pressure at bottom.

Concerning the height of fluid it's going to increase obviously but it won't affect the net pressure on bottom, proof for this can be made by assuming a two thick disks one upon another in place of sphere and then doing calculation for pressure at bottom, this will show that whatever number of disks are introduced between two disks result will be same.

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    $\begingroup$ The string will only support a portion of the sphere's weight, not all of it. The rest of the weight would be supported by the fluid and thence to the bottom of the chamber. $\endgroup$
    – BowlOfRed
    Dec 13, 2023 at 8:51
  • $\begingroup$ @BowlOfRed correct πŸ’―, and hence I have written "Since whatever may be the net force, tension in string is always gonna balance it!" Net force is combination of buoyant force, force due to liquid above sphere and force of gravity, tension in string will always be equal to resultant of all these forces and hence the sphere will be weightless. The proof is very simple, hold something in your hand and place it just above weighing machine, machine will weigh nothing until you let the object free! (Air is fluid and your hand is acting as string) $\endgroup$
    – Qwerty
    Dec 13, 2023 at 10:06
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    $\begingroup$ Think about what would happen if the sphere was a balloon full of water being lowered into the container with initial water height $h_i$ and initial bottom pressure $P_i=\rho \mathrm{g} h_i$. (The balloon material has negligible mass.) Once lowered into the water, the sphere would be neutrally buoyant and there would be no tension in the string, so it could be removed. We now effectively have vessel full of water of height $h_f>h_i$, so the pressure at the bottom is $P_f=\rho \mathrm{g} h_f>P_i$. $\endgroup$ Dec 13, 2023 at 15:26

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