Pipes and Cisterns – Three taps with flow proportional to square of diameter: Three inlet taps have diameters 1 cm, 4/3 cm, and 2 cm. The volume flow through each is proportional to the square of its diameter. The largest (2 cm) tap alone fills the tank in 61 minutes. If all three taps are opened together, how long will the tank take to fill?

Introduction / Context:
When flow is proportional to the square of diameter (v ∝ d^2), relative rates can be set in those ratios. One tap’s absolute time fixes the unit scaling, allowing us to compute the combined rate and time.

Given Data / Assumptions:

• Diameters: 1 cm, 4/3 cm, 2 cm ⇒ squares 1, 16/9, 4.
• Largest (2 cm) alone fills in 61 min.
• Tank capacity normalized to W; proportionality is linear in rates.

Concept / Approach:
Let “1 unit” of rate correspond to the 1 cm tap. Since 4 units correspond to the 2 cm tap and that equals W/61 per min, we get the unit rate and can sum all three taps’ rates.

Step-by-Step Solution:
Let 4 units = W/61 per min ⇒ 1 unit = W/244 per min.Total units = 1 + 16/9 + 4 = (9 + 16 + 36)/9 = 61/9.Combined rate = (61/9) * (W/244) = W * 61 / 2196 per min.Time = W / (W * 61 / 2196) = 2196 / 61 = 36 min.

Verification / Alternative check:
Because 61 × 36 = 2196, the arithmetic cancels exactly, giving an integer number of minutes.

Why Other Options Are Wrong:
44, 45, 155/18, and 40 min do not align with the squared-diameter proportionality anchored by the 61-minute reference.

Common Pitfalls:
Using diameters (d) instead of d^2 for proportionality, or mixing units.

Final Answer:
36 min