21. To determine the BOD5 of a waste water sample, 5, 10 and 50 ml aliquots of the waste water were diluted to 300 ml and incubated at 20°C in BOD bottles for 5 days.
    
    Based on the data, the average BOD₅ of the wastewater is equal to

1. 1. 139.5 mg/l

   
   
   

   2. 126.5 mg/l

   
   
   

   3. 109.8 mg/l

   
   
   

   4. 72.2 mg/l

   
   
   

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1. View Hint View Answer Discuss in Forum

   BOD₅ = Oxygen consumed × Dilution factor
   Sample 1: BOD₅ = (9.2 – 6.9) × 300/5 = 138 mg/l
   Sample 2: BOD₅ = (9.1 – 4.4) × 300/10 = 141 mg/l
   Sample 3: BOD₅ = (8.4 – 0) × 300/50 = 50.4 mg/l
   

   Average BOD =	138 + 141 + 50.4	= 109.8 mg/l
   	3

   Correct Option: C

   BOD₅ = Oxygen consumed × Dilution factor
   Sample 1: BOD₅ = (9.2 – 6.9) × 300/5 = 138 mg/l
   Sample 2: BOD₅ = (9.1 – 4.4) × 300/10 = 141 mg/l
   Sample 3: BOD₅ = (8.4 – 0) × 300/50 = 50.4 mg/l
   

   Average BOD =	138 + 141 + 50.4	= 109.8 mg/l
   	3

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22. 50 g of CO₂ and 25 g of CH₄ are produced from the decomposition of municipal solid waste (MSW) with a formula weight of 120 g. What is the average per capita green house gas production in a city of 1 million people with a MSW production rate of 500 ton/day?

1. 1. 104 g/day

   
   
   

   2. 120 g/day

   
   
   

   3. 208 g/day

   
   
   

   4. 313 g/day

   
   
   

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1. View Hint View Answer Discuss in Forum

   MSW percapita =	500 × 106	= 500 g/capita.
   	1 × 106

   
   120 g MSW produces 75 g green house gases (50g CO₂, 25g CH₄)
   Therefore, 500 g MSW will produce 75/120 × 500 = 313 g green house gases.

   Correct Option: D

   MSW percapita =	500 × 106	= 500 g/capita.
   	1 × 106

   
   120 g MSW produces 75 g green house gases (50g CO₂, 25g CH₄)
   Therefore, 500 g MSW will produce 75/120 × 500 = 313 g green house gases.

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23. A wastewater sample contains 10^–5.6 m mol/l of OH– ions at 25°C. The pH of this sample is

1. 1. 8.6

   
   
   

   2. 8.4

   
   
   

   3. 5.6

   
   
   

   4. 5.4

   
   
   

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1. View Hint View Answer Discuss in Forum

   [H^–1] [OH^–] = 10^–14
   [OH^–] = 10^–5.6 m mol/l
   = 10^–5.6 × 10^–3 mol/l
   = 10^–8.6 mol/l
   

   ∴ [H+] =	10–14	= 10–5.4mol/l
   	10–8.6

   
   pH = –log[H⁺] = – log[10^–5.4] = 5.4.

   Correct Option: D

   [H^–1] [OH^–] = 10^–14
   [OH^–] = 10^–5.6 m mol/l
   = 10^–5.6 × 10^–3 mol/l
   = 10^–8.6 mol/l
   

   ∴ [H+] =	10–14	= 10–5.4mol/l
   	10–8.6

   
   pH = –log[H⁺] = – log[10^–5.4] = 5.4.

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24. The 5-day BOD of a waste water sample is obtained as 190 mg/l (with k = 0.01h^–1). The ultimate oxygen demand (mg/l) of the sample will be

1. 1. 3800

   
   
   

   2. 475

   
   
   

   3. 271

   
   
   

   4. 190

   
   
   

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1. View Hint View Answer Discuss in Forum

   BOD₅ = BOD₄[1 – 10^–kD.t]
   

   BOD4 =	BOD5
   	[1 - 10–kD.t]

   
   k_D = 0.434 k = .434 × 0.01 h^–1
   k_Dt = .434 × .01 × 5 × 24 = 0.5208
   

   BOD4 =	190
   	[1 - 10-0.5208]

   
   = 271 mg/l.

   Correct Option: C

   BOD₅ = BOD₄[1 – 10^–kD.t]
   

   BOD4 =	BOD5
   	[1 - 10–kD.t]

   
   k_D = 0.434 k = .434 × 0.01 h^–1
   k_Dt = .434 × .01 × 5 × 24 = 0.5208
   

   BOD4 =	190
   	[1 - 10-0.5208]

   
   = 271 mg/l.

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25. A horizontal flow primary clarifier treats waste water in which 10%, 60% and 30% of particles have settling velocities of 0.1 mm/s, 0.2 mm/s and 1.0 mm/s respectively. What would be the total percentage of particles removed if clarifier operates at a surface overflow rate (SOR) of 43.2 m³/m². d?

1. 1. 43%

   
   
   

   2. 56%

   
   
   

   3. 86%

   
   
   

   4. 100%

   
   
   

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1. View Hint View Answer Discuss in Forum

   SOR = 43.2 m³/m²d
   

   =	43.2 × 10³	mm/s
   	24 × 60 × 60

   
   = 0.5 mm/s
   30% particles has settling velocity more than SOR. So, it will settle particles with velocity lower than SOR, that will settle equals to its ratio with SOR.
   

   ∴ % particles removed = 30% + 60% ×	0.2	+ 10 % ×	0.1	= 56 %
   	0.5		0.5

   Correct Option: B

   SOR = 43.2 m³/m²d
   

   =	43.2 × 10³	mm/s
   	24 × 60 × 60

   
   = 0.5 mm/s
   30% particles has settling velocity more than SOR. So, it will settle particles with velocity lower than SOR, that will settle equals to its ratio with SOR.
   

   ∴ % particles removed = 30% + 60% ×	0.2	+ 10 % ×	0.1	= 56 %
   	0.5		0.5

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