GATE CE topic:water-and-wastewater-treatment PYQs & Answers — Page 2

Questions 11–20 of 81. Imported answers require independent review unless marked reviewed.

The maximum demand at a water purification plant has been estimated as 12 million litres per day. For the raw supplies, a rectangular sedimentation tank is to be designed with mechanical sludge removal arrangement. Consider depth of the tank as 4 m, detention period as 6 hours, and velocity of flow as 0.003 m/s.

The width (in m) of the detention tank is ________ (rounded off to two decimal places).

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Fecal Coliform (FC) concentration in river water was measured as 10780 cfu/100 ml. The FC concentration after the conventional water treatment, but before chlorination, was measured as 23 cfu/100 ml. The ‘Log Kill’ (inactivation) of FC due to the conventional water treatment is closest to

Your answerChoose one option.

4.00

2.50

2.67

3.00

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The Surface Overflow Rate (SOR) in a rectangular sedimentation tank is 45 m³/m²/d. Minimum diameters of spherical inorganic and organic particles expected to be completely removed in this tank are calculated. Assume that Stoke’s law is applicable. Which of the following options is/are correct:

Specific gravity of inorganic particles = 2.65

Specific gravity of organic particles = 1.20

Acceleration due to gravity (g) = 9.81 m/s²

Kinematic viscosity (ν) = 1×10⁻⁶ m²/s

Your answerSelect all correct options. You can choose more than one.

Minimum diameter of inorganic particles is 24 μm

Minimum diameter of organic particles is 69 μm

Minimum diameter of inorganic particles is 15 μm

Minimum diameter of organic particles is 55 μm

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Aeration is employed as a treatment option for the removal of several pollutants from contaminated water.

Identify the pollutant(s), where aeration is employed as a part of their removal:

Your answerSelect all correct options. You can choose more than one.

Iron

Cadmium

Manganese

Zinc

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Free residual chlorine concentration in water was measured to be 2 mg/l (as Cl

). The pH of water is 8.5. By using the chemical equation given below, the HOCl concentration (in

moles/l) in water is ______________ (round off to one decimal place).

HOCl ⇌ H

+ OCl

, pK = 7.50

Atomic weight: Cl (35.5)

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MgCl₂ and CaSO₄ salts are added to 1 litre of distilled deionized water and mixed until completely dissolved. Total Dissolved Solids (TDS) concentration is 500 mg/l, and Total Hardness (TH) is 400 mg/l (as CaCO₃). The amounts of MgCl₂ and CaSO₄ added are calculated (rounded off to the nearest integer). Which of the following options is/are true:

Atomic weights: Ca(40), Mg(24), S(32), O(16), Cl(35.5), C(12)

Your answerSelect all correct options. You can choose more than one.

Amount of MgCl₂ added is 143 mg

Amount of CaSO₄ added is 357 mg

Amount of MgCl₂ added is 103 mg

Amount of CaSO₄ added is 397 mg

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A facultative pond system is employed for wastewater treatment. Which of the following statements is/are true:

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The dissolved oxygen concentration will be high during daytime compared to night-time

The pH will be high during daytime compared to night-time

The dissolved oxygen concentration will be low during daytime compared to night-time

The pH will be low during daytime compared to night-time

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A compound has a general formula CₐHᵦO꜀Nᵈ and molecular weight 187. A 935 mg/l solution of the compound is prepared in distilled deionized water. The Total Organic Carbon (TOC) is measured as 360 mg/l (as C). The Chemical Oxygen Demand (COD) and the Total Kjeldahl Nitrogen (TKN) are determined as 600 mg/l (as O₂) and 140 mg/l (as N), respectively (as per the chemical equation given below). Which of the following options is/are CORRECT?

CₐHᵦO꜀Nᵈ + ((4a + b − 2c − 3d)/4) O₂ → aCO₂ + ((b − 3d)/2) H₂O + dNH₃

Atomic weight: C(12), H(1), O(16), N(14)

Your answerSelect all correct options. You can choose more than one.

a = 6

b = 7

c = 5

d = 3

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The analyses results of a water sample are given below. The non-carbonate hardness of the water (in mg/L) as CaCO

is ______ (in integer).

= 150 mg/L as CaCO

= 40 mg/L as CaCO

= 10 mg/L as CaCO

= 50 mg/L as CaCO

= 10 mg/L as CaCO

= 120 mg/L as CaCO

HCO

= 30 mg/L as CaCO

= 50 mg/L as CaCO

; Other anions were not analysed.

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Which one of the following products is NOT obtained in anaerobic decomposition of glucose?

Your answerChoose one option.

CO₂

CH₄

H₂S

H₂O

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GATE CE topic:water-and-wastewater-treatment PYQs

Question 65

Question 21

Question 31

Question 32

Question 34

Question 44

Question 45

Question 49

Question 61

Question 26