Question 1112243
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a sphere of {{{highlight(cross(radious))}}} <U>radius</U> 2 cm is dropped into water contained in a cylindrical vessel of {{{highlight(cross(radious))}}} <U>radius</U> 4 cm. 
if the sphere is completely immersed then what will be the rise in level of water
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<pre>
The general formula/equation to solve the problem is THIS:


{{{pi*R^2*H}}} = {{{pi*R^2*h}}} + {{{(4/3)*pi*r^3}}},


where R is the radius of the cylindrical vessel, r is the radius of the immersed sphere, H is the water level in the vessel after immersing, 

h is the water level in the vessel before immersing.


The formula says that the volume below the water surface after submersing is equal to that before submersing PLUS the volume of the immersed body.


So the rise of the water level in the vessel is equal to 

H - h = {{{((4/3)*pi*r^3)/(pi*R^2)}}} = {{{(4/3)*r^3/R^2}}}.


Having the formula, you can calculate the final number on your own.
</pre>

<U>Once again</U>:  the water level rise in a cylindrical vessel is equal to the volume of the immersed body divided by the base area of the vessel.


The simple and basic idea behind this solution is that the volume of water in the vessel remains the same before and after submerging.



See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Water-level-rise-in-a-tank-after-submerging-bodies.lesson>Water level rise in a tank after submerging bodies</A> 

in this site, where you can find the solutions to other similar problems.


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Happy {{{pi}}}-day !



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And learn how to write the word "radius" correctly !