How is the volume of a pipe calculated?

How is the volume of a pipe calculated?

• A. π x Diameter x Length
• B. 0.785 x Diameter x Diameter x Length
• C. Diameter x Diameter x Height
• D. π x Diameter x Height

Answer: (B)

The volume of a pipe is calculated using the formula that represents the volume of a cylinder. A pipe essentially functions as a cylindrical shape, where the key parameters are its diameter and length. The correct formula, which takes into account the radius of the cylinder, is expressed as: \[ \text{Volume} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height (or length) of the cylinder. Since the radius is half of the diameter, and the area of the circle (cross-section of the pipe) is \( \pi r^2 \), you can also express it using the diameter. By substituting the radius in terms of diameter (where \( r = \frac{Diameter}{2} \)), you arrive at: \[ \text{Volume} = \pi \left(\frac{Diameter}{2}\right)^2 Length = \pi \frac{Diameter^2}{4} Length \] However, this can also be conveniently represented as: \[ \text{Volume} = 0.785 \times Diameter^2 \times Length \] Thus, the correct choice takes the necessary geometric relationships into account

The volume of a pipe is calculated using the formula that represents the volume of a cylinder. A pipe essentially functions as a cylindrical shape, where the key parameters are its diameter and length.

The correct formula, which takes into account the radius of the cylinder, is expressed as:

[

\text{Volume} = \pi r^2 h

]

where ( r ) is the radius and ( h ) is the height (or length) of the cylinder. Since the radius is half of the diameter, and the area of the circle (cross-section of the pipe) is ( \pi r^2 ), you can also express it using the diameter.

By substituting the radius in terms of diameter (where ( r = \frac{Diameter}{2} )), you arrive at:

[

\text{Volume} = \pi \left(\frac{Diameter}{2}\right)^2 Length = \pi \frac{Diameter^2}{4} Length

]

However, this can also be conveniently represented as:

[

\text{Volume} = 0.785 \times Diameter^2 \times Length

]

Thus, the correct choice takes the necessary geometric relationships into account