When a concrete structure is built, and engineer often performs a slump test on the concrete to ensure that it is of suitable strength and consistency. The engineer fills a truncated cone with bottom diameter 200mm, top diameter 100mm and height 300mm with concrete. The engineer then places the cone on the grounf and lifts off the cone. After period of time, she measures the amount of that the top of the cone has slumped to decide if concrete is appropriate.
a) Original cone from which the slump test cone was derived had height of 600m. What was volume of original cone? Express answer to nearest cubic mm.
The formula for this is:
V = 1/3 * 3.14 * r2 * h
r2 means r squared, I can't make it small.
I went 100m * 200m and came up with the answer 20000. To get the radius I just divided 20000 by 2 and got 10000.
This is what I did:
1/3 * 3.14 * 10000 * 10000 * 600
I got 62800000000mm
Is that right?
To find the volume of the original cone, you need to use the formula for the volume of a cone, which is V = 1/3 * π * r^2 * h, where V is the volume, π is the mathematical constant pi (approximately 3.14), r is the radius, and h is the height.
In this case, you are given the dimensions of the truncated cone derived from the original cone: bottom diameter of 200mm, top diameter of 100mm, and height of 300mm.
To find the radius of the top and bottom of the truncated cone, you need to divide the diameter by 2. So, the radius of the bottom is 200mm / 2 = 100mm, and the radius of the top is 100mm / 2 = 50mm.
Now, to find the volume of the original cone, substitute the values into the formula:
V = 1/3 * π * ((100mm)^2 + (50mm)^2 + 100mm * 50mm) * 600mm
V = 1/3 * 3.14 * (10,000mm^2 + 2,500mm^2 + 5,000mm^2) * 600mm
V = 1/3 * 3.14 * 17,500mm^2 * 600mm
V ≈ 1/3 * 3.14 * 10,500,000mm^3
V ≈ 1/3 * 33,030,000mm^3
V ≈ 11,010,000mm^3
Therefore, the volume of the original cone is approximately 11,010,000 cubic mm.