Question 1205006
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The volume of a frustum of a cone of height h is given by the formula
V = ⅓πh(R² + Rr + r²)
Where R and r are the radii of its circular ends.
Hollow metal hemispheres of internal and external diameters 12.5cm and 17.5cm respectively, 
are to be cast from molten contents of a ladle in the shape of a frustum of a cone 
of depth 1.2m and radii of ends 0.6m and 0.635m. Calculate the number of hemispheres that can be cast.
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<pre>
The volume of the hollow hemisphere is  {{{(1/2)*(4/3)*3.14159265*((17.5/2)^3-(12.5/2)^3)}}} = 891.754 cm^3 (rounded).


The volume of the frustum is  {{{(1/3)*3.14159265*120*(60^2 + 60*63.5 + 63.5^2)}}} = 1437875.54  cm^3 (rounded).


The number of hemispheres is an integer number closest from the bottom to the ratio

    {{{1437875.54/891.754}}} = 1612.413.


<U>ANSWER</U>.  1612 hemispheres.
</pre>

Solved.


Honestly, I don't know, for what reason there is a need to write more about these mechanical calculations.