When a ball is suspended in a fluid, why doesn't the pressure at the bottom of the container, and thus the hight of the fluid increase?

Question

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

Answer 1

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

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.

Answer 2

$$P_{fluid} = P + \rho gh$$

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.