A single circular pipe of diameter x centimetres is to be replaced by six smaller circular pipes, each of diameter 12 cm, to convey a liquid in a laboratory. If the flow speed of the liquid is kept the same in all pipes, what is the value of x?

Introduction / Context:
This question explores how multiple smaller pipes can replace a single larger pipe while maintaining the same flow rate. When the liquid speed is the same in each pipe, the volume flow rate is proportional to the total cross sectional area of the pipes. Therefore, the area of the single large pipe must equal the combined area of the six smaller pipes.

Given Data / Assumptions:

• Single pipe has diameter x cm and radius x / 2 cm.
• Each of the six smaller pipes has diameter 12 cm and radius 6 cm.
• Speed of the liquid is the same in all pipes.
• Flow rate is directly proportional to cross sectional area.

Concept / Approach:
The cross sectional area A of a circular pipe is:
A = π * r^2. For equivalent flow, the area of the single large pipe must equal the sum of the areas of the six smaller pipes:
Area of large pipe = 6 * area of one small pipe. Expressing both sides in terms of π and radii, we can solve for x, the diameter of the large pipe.

Step-by-Step Solution:
Step 1: Area of one small pipe. Radius of small pipe r_s = 12 / 2 = 6 cm. Area_s = π * 6^2 = 36π cm^2. Step 2: Total area of six small pipes. Total area_small = 6 * 36π = 216π cm^2. Step 3: Area of large pipe. Radius of large pipe r_L = x / 2. Area_L = π * (x / 2)^2 = π * x^2 / 4. Step 4: Equate total areas. π * x^2 / 4 = 216π. Step 5: Cancel π from both sides and solve for x^2. x^2 / 4 = 216 ⇒ x^2 = 864. Step 6: Take square root. x = sqrt(864) = sqrt(144 * 6) = 12 * sqrt(6). Using sqrt(6) ≈ 2.449, x ≈ 12 * 2.449 ≈ 29.39 cm.

Verification / Alternative check:
We can verify by substituting x ≈ 29.39 cm back. Radius r_L ≈ 14.695 cm. Area_L ≈ π * 14.695^2 ≈ π * 216, which matches the total area of six small pipes, 216π cm^2. Therefore the diameter of about 29.39 cm is consistent with the requirement for equal flow.

Why Other Options Are Wrong:
14.69 cm is approximately the radius, not the diameter, and would give an area only about one quarter of the required total area.
18.65 cm and 22.21 cm are too small and would yield total flow capacity less than that of six smaller pipes combined.
25.00 cm is closer but still short; its area is π * (12.5)^2, which is significantly less than 216π cm^2.

Common Pitfalls:
A common mistake is to directly equate diameters or to sum diameters instead of equating areas. Another pitfall is forgetting that area depends on the square of the radius, so combining six pipes does not mean just multiplying the diameter by six. Careful use of the area formula and attention to radius versus diameter are essential here.

Final Answer:
The required diameter of the single equivalent pipe is approximately 29.39 cm.