How many 6 inch water lines are required to carry the same amount of water as one 12 inch water line, neglecting friction?

How many 6 inch water lines are required to carry the same amount of water as one 12 inch water line, neglecting friction?

• A. 2
• B. 4
• C. 6
• D. 8

Answer: (B)

To determine the number of 6-inch water lines needed to carry the same amount of water as one 12-inch water line, we can use the principle that the flow capacity of a pipe depends on its cross-sectional area. The area of a cylindrical pipe can be calculated using the formula for the area of a circle, which is A = πr², where r is the radius of the pipe. For a 12-inch diameter pipe: - The radius is 6 inches, so the area is A = π(6)² = π(36) = 36π square inches. For a 6-inch diameter pipe: - The radius is 3 inches, so the area is A = π(3)² = π(9) = 9π square inches. To find out how many 6-inch pipes are needed to equal the area of one 12-inch pipe, we can set up the following equation based on their areas: - Total area of the required 6-inch pipes = Area of the 12-inch pipe. - If we let n be the number of 6-inch pipes needed, we can express this as: n × 9π = 36π. To solve for n,

To determine the number of 6-inch water lines needed to carry the same amount of water as one 12-inch water line, we can use the principle that the flow capacity of a pipe depends on its cross-sectional area.

The area of a cylindrical pipe can be calculated using the formula for the area of a circle, which is A = πr², where r is the radius of the pipe.

For a 12-inch diameter pipe:

• The radius is 6 inches, so the area is A = π(6)² = π(36) = 36π square inches.

For a 6-inch diameter pipe:

• The radius is 3 inches, so the area is A = π(3)² = π(9) = 9π square inches.

To find out how many 6-inch pipes are needed to equal the area of one 12-inch pipe, we can set up the following equation based on their areas:

• Total area of the required 6-inch pipes = Area of the 12-inch pipe.

• If we let n be the number of 6-inch pipes needed, we can express this as:

n × 9π = 36π.

To solve for n,