A homeowner wishes to drain her swimming pool by siphoning the water, whose depth is h, into a nearby gully a distance H below it, where H is much larger than h (see figure). She uses a pipe of cross-sectional area a, and the pool water has surface area A.

Engineering

Physics

Equation of Continuity

Question

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Linked Question 1

How long does it take to empty the pool if h = 2m, H = 22 m, A = 50 m², a = 5 cm²?

2 × 10⁴ sec.

10⁴ sec.

2 × 10⁵ sec.

10⁵ sec.

$y = \sqrt{2 g (H - h)}$

$5 \times 10^{- 4} \times \sqrt{20 \times 120} = \frac{50 \times 2}{t}$

$t = \frac{20 \times 10^{4}}{20} = 10^{4} \sec$ .

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Linked Question 2

In the siphon arrangement of the last question, the pipe develops a leak at a point above the water surface of the pool. What happens to the water flow?

Rate of volume flow increases but velocity of water decreases.

Rate of volume flow as well as velocity of water remains same.

Rate of volume flow remains same but velocity of water increases.

Siphon stops working.

No Solution.

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