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Question 1178404: in a laboratory, a chemist's measuring glass is conical in shape. if it is 8cm deep and 3 cm across the mouth, find the distance on the slant edge between the markings for 1 cc and 2 cc.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**1. Visualize the Cone**
Imagine a cone representing the measuring glass.
* **Height (h):** 8 cm
* **Diameter (d):** 3 cm
* **Radius (r):** d/2 = 3/2 = 1.5 cm
**2. Volume of a Cone**
The volume (V) of a cone is given by:
* V = (1/3)πr²h
**3. Similar Cones**
When we fill the cone to different levels, we create smaller cones that are similar to the larger cone. This means their dimensions are proportional.
**4. Finding the Radii and Heights for 1 cc and 2 cc**
Let's denote:
* r1, h1: radius and height for 1 cc
* r2, h2: radius and height for 2 cc
We know:
* 1 = (1/3)πr1²h1
* 2 = (1/3)πr2²h2
Also, by similarity:
* r1/h1 = r/h = 1.5/8
* r2/h2 = r/h = 1.5/8
From these proportions, we have:
* r1 = (1.5/8)h1
* r2 = (1.5/8)h2
Substitute these into the volume equations:
* 1 = (1/3)π((1.5/8)h1)²h1 = (1/3)π(2.25/64)h1³
* 2 = (1/3)π((1.5/8)h2)²h2 = (1/3)π(2.25/64)h2³
Solve for h1 and h2:
* h1³ = (64 * 3) / (2.25π) ≈ 27.147
* h1 ≈ ∛27.147 ≈ 3.003 cm
* h2³ = (64 * 6) / (2.25π) ≈ 54.294
* h2 ≈ ∛54.294 ≈ 3.784 cm
Now find r1 and r2:
* r1 = (1.5/8) * 3.003 ≈ 0.563 cm
* r2 = (1.5/8) * 3.784 ≈ 0.710 cm
**5. Finding the Slant Edge Distances**
Let:
* s1: slant edge distance for 1 cc
* s2: slant edge distance for 2 cc
Using the Pythagorean theorem (in 3D):
* s1 = √(r1² + h1²) = √(0.563² + 3.003²) ≈ √(0.317 + 9.018) ≈ √9.335 ≈ 3.055 cm
* s2 = √(r2² + h2²) = √(0.710² + 3.784²) ≈ √(0.504 + 14.319) ≈ √14.823 ≈ 3.850 cm
**6. Finding the Distance Between Markings**
The distance on the slant edge between the 1 cc and 2 cc markings is:
* s2 - s1 ≈ 3.850 - 3.055 ≈ 0.795 cm
**Therefore, the distance on the slant edge between the markings for 1 cc and 2 cc is approximately 0.795 cm.**
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