Question

question
a cast iron pipe used as a water main for a home has an outer diameter of 11
in. if the pipe has a thickness of 0.35 in, what volume of iron does 23 in of
pipe contain? answer in cubic inches. round your answer to the nearest
hundredth if necessary. (note: diagram is not drawn to scale.)
answer
attempt 1 out of 2
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Explanation:

Step1: Find the outer radius

The outer diameter is \(d_{outer}=11\) inches, so the outer radius \(r_{outer}=\frac{d_{outer}}{2}=\frac{11}{2} = 5.5\) inches.

Step2: Find the inner radius

The thickness of the pipe is \(t = 0.35\) inches. The inner radius \(r_{inner}=r_{outer}-t=5.5 - 0.35=5.15\) inches.

Step3: Use the formula for the volume of a hollow - cylinder

The volume of a hollow - cylinder (the volume of the iron in the pipe) is \(V=\pi h(r_{outer}^{2}-r_{inner}^{2})\), where \(h = 23\) inches.
Substitute the values:

$$
LATEXBLOCK0
$$

$$V = 23\times3.14\times3.7275$$

$$V=23\times11.70435$$

$$V = 269.19905\approx269.20$$

Answer:

\(269.20\) cubic inches