# Fluid Mechanics Questions and Answers – Time Required to Empty a Reservoir with Weirs or Notches

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This set of Fluid Mechanics Questions and Answers for Aptitude test focuses on “Time Required to Empty a Reservoir Or a Tank with a Rectangular, Triangular & Cipolletti Weir or Notch”.

1. What is the reduction in crest length due to each end contraction?
a) 0.1H
b) 0.2H
c) 0.15H
d) 0.25H

Explanation: Francis through his experiment had derived this empirical relation.

2. In Francis formula, the effective length is –
a) L-0.2H
b) L-0.4H
c) L-0.3H
d) L-0.1H

Explanation: In Francis formula, the effective length is L-0.2H.

3. In Francis empirical expression for discharge, the relation between head of water and discharge is
a) Q is directly proportional to H
b) Q is directly proportional to H1.5
c) Q is directly proportional to H2.5
d) Q is directly proportional to H0.5

Explanation: In Francis empirical expression for discharge, the relation between head of water and discharge is Q is directly proportional to H1.5.

4. In Bazin’s formula, the discharge is inversely proportional to the length of weir.
a) True
b) False

Explanation: In Bazin’s formula, the discharge is directly proportional to the length of weir.

5. The head of water over a rectangular weir is 38 cm. The length of the crest of the weir end contraction suppressed is 1.3 m. Find the discharge using the Francis formula.
a) 0.56 m3/s
b) 0.75 m3/s
c) 0.85 m3/s
d) 0.69 m3/s

Explanation: Q = 1.84*L*H1.5 = 0.56 m3/s.

6. The head of water over a rectangular weir is 28 cm. The length of the crest of the weir end contraction suppressed is 1.27 m. Find the discharge using the Francis formula.
a) 0.346 m3/s
b) 0.556 m3/s
c) 0.788 m3/s
d) 0.225 m3/s

Explanation: Q = m * L * (2*g)0.5 * H1.5
m = ⅔ * Cd
= 0.405 + 0.003/H
= 0.405 + 0.003/0.28
Q = 0.346 m3/s.

7. The head of water over a rectangular weir is 26 cm. The length of the crest of the weir end contraction suppressed is 1.25 m. Find the discharge using the Francis formula.
a) 0.304 m3/s
b) 0.502 m3/s
c) 0.350 m3/s
d) 0.625 m3/s

Explanation: Q = 1.84*L*H1.5
= 0.304 m3/s.

8. The head of water over a rectangular weir is 28 cm. The length of the crest of the weir end contraction suppressed is 1.27 m. Find the discharge using the Francis formula.
a) 0.346 m3/s
b) 0.556 m3/s
c) 0.788 m3/s
d) 0.225 m3/s

Explanation: Q = m * L * (2*g)0.5 * H1.5
m = ⅔ * Cd
= 0.405 + 0.003/H
= 0.405 + 0.003/0.28
Q = 0.346 m3/s.

9. Find the discharge over a cipolletti weir of length 1.5 m when the head over the weir is 0.85 m. Take Cd = 0.61.
a) 2.12 m3/s
b) 1.25 m3/s
c) 2.5 m3/s
d) 1.5 m3/s

Explanation: Q = ⅔ * Cd * L * (2g)0.5 * H1.5
= 2.12 m3/s.

10. Find the discharge over a cipolletti weir of length 1.3 m when the head over the weir is 0.65 m. Take Cd = 0.60.
a) 2.12 m3/s
b) 1.21 m3/s
c) 2.5 m3/s
d) 1.5 m3/s

Explanation: Q = ⅔ * Cd * L * (2g)0.5 * H1.5
= 1.21 m3/s.

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