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iitjee10
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Homework Statement
There is a cylindrical container that has a small hole of radius ##a ( < R)## at the bottom. A sphere of radius ##R## and density ##\rho_{s} > \rho_{w}## is placed in the cylinder such that it completely covers the hole (a part of it sticks out as in attached figure). The container is now filled with water (##\rho_{w}##) to a height ##h##.
The objective is to determine the force exerted by the water on the sphere. (specifically on its dependence with the height of water)
Homework Equations
I have two methods in mind, but not sure which is correct.
- Using Archimedes Principle
- Using pressure integration
The Attempt at a Solution
Method 1:
If we assume that the hole along with the portion that remains outside the container did not exist .. and the resulting contact surface between the flat portion of the sphere and the bottom of the cylinder was not perfectly smooth, then the force by the liquid would be equal to the weight of the liquid displaced (##V \rho_{w}g##).
This force would be comprised of the force on the bottom surface and the force on the remaining cylindrical surface. Now, the force on the bottom surface is ##\rho_{w}hg \pi a^{2}##.
So the force on the cylindrical section is ## V \rho_{w} g - \rho_{w} h g \pi a^{2}##.
Since this is the force we want for the original question, this should be the answer after calculating ## V ## appropriately.
Method 2.
This involves integration.
Consider the diagram.
The pressure at a circular (shell) element that makes an angle ##\theta## is given by ##P = (h - R \cos \alpha - R \cos \theta) \rho_{w}g ##. The elemental area ## A = 2 \pi R \sin \theta R d \theta##.
So the net vertical force on the element (downward) is ## F_{v} = PA \cos \theta##
Integrating this from ## \theta = 0 ## to ## \theta = \pi - \alpha ## should give the answer.
However, evaluating this integral does not seem to match the answer for the above method.
Is one of the two methods conceptually wrong?
Any help would be highly appreciated.