How much does a solid aluminum cylinder weighing 165 pounds per cubic foot measure with a diameter of 4 feet and a height of 5 feet?

How much does a solid aluminum cylinder weighing 165 pounds per cubic foot measure with a diameter of 4 feet and a height of 5 feet?

• A. 8,000 lbs
• B. 10,362 lbs
• C. 12,000 lbs
• D. 14,600 lbs

Answer: (B)

To determine the weight of the solid aluminum cylinder, we need to find its volume first. The formula for the volume of a cylinder is: \[ V = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height of the cylinder. Given that the diameter of the cylinder is 4 feet, the radius \( r \) is half of that, which is 2 feet. The height \( h \) is given as 5 feet. Plugging these values into the volume formula: \[ V = \pi (2 \text{ ft})^2 (5 \text{ ft}) \] \[ V = \pi (4 \text{ ft}^2) (5 \text{ ft}) \] \[ V = 20\pi \text{ ft}^3 \] Now, to calculate the actual volume using the approximate value of \( \pi \) (3.14): \[ V \approx 20 \times 3.14 = 62.8 \text{ ft}^3 \] Next, we need to find the weight of the cylinder. The weight can be calculated by multiplying the volume by the density of aluminum, which

To determine the weight of the solid aluminum cylinder, we need to find its volume first. The formula for the volume of a cylinder is:

[ V = \pi r^2 h ]

where ( r ) is the radius and ( h ) is the height of the cylinder.

Given that the diameter of the cylinder is 4 feet, the radius ( r ) is half of that, which is 2 feet. The height ( h ) is given as 5 feet. Plugging these values into the volume formula:

[ V = \pi (2 \text{ ft})^2 (5 \text{ ft}) ]

[ V = \pi (4 \text{ ft}^2) (5 \text{ ft}) ]

[ V = 20\pi \text{ ft}^3 ]

Now, to calculate the actual volume using the approximate value of ( \pi ) (3.14):

[ V \approx 20 \times 3.14 = 62.8 \text{ ft}^3 ]

Next, we need to find the weight of the cylinder. The weight can be calculated by multiplying the volume by the density of aluminum, which