36. Carbonate hardness (mg/l as CaCO₃) present in the above water sample is

1. 1. 205

   
   
   

   2. 250

   
   
   

   3. 275

   
   
   

   4. 289

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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37. Total hardness (mg/l as CaCO₃) present in the above water sample is

1. 1. 205

   
   
   

   2. 250

   
   
   

   3. 275

   
   
   

   4. 308

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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38. Alkalinity present in the water in mg/l as CaCO₃ is

1. 1. 250

   
   
   

   2. 1500

   
   
   

   3. 1750

   
   
   

   4. 5000

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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39. Ultimate BOD of a river water sample is 20 mg/L. BOD rate constant (natural log) is 0.15 day^-1. The respective values of BOD (in%) exerted and remaining after 7 days are:

1. 1. 45 and 55

   
   
   

   2. 55 and 45

   
   
   

   3. 65 and 35

   
   
   

   4. 75 and 25

   
   
   

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1. View Hint View Answer Discuss in Forum

   Y_u = 20 mg/L
   Y₇ = Y_u × e^–kt
   = 20 × e^{–0.15 × 7} = 7
   

   % remaining =	7	× 100 = 35%
   	20

   
   % exerted = 100 – 35 = 65%

   Correct Option: C

   Y_u = 20 mg/L
   Y₇ = Y_u × e^–kt
   = 20 × e^{–0.15 × 7} = 7
   

   % remaining =	7	× 100 = 35%
   	20

   
   % exerted = 100 – 35 = 65%

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40. A waste water stream (flow = 2m³/s, ultimate BOD = 90 mg/l) is joining a small river (flow = 12 m³/s, ultimate BOD = 5 mg/ l). Both water streams get mixed up instantaneously. Crosssectional area of the river is 50 m². Assuming the deoxygenation rate constant, k = 0.25.day, the BOD (in mg/l) of the river water, 10 km downstream of the mixing point is

1. 1. 1.68

   
   
   

   2. 12.63

   
   
   

   3. 15.46

   
   
   

   4. 1.37

   
   
   

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1. View Hint View Answer Discuss in Forum

   BOD of mixture, Y =	(2 × 90) + (12 × 5)	= 17.143 g/m³
   	2 × 12

   
   K_D = 0.434 × k¹
   = 0.434 × 25 = 0.1085
   

   Velocity of stream =	14	= 0.28 m/s
   	50

   
   

   Time taken =	10 km	= 9.921 hours
   	0.28 m/s

   
   Y_t = Y_o (1 – 10^–kt)
   = 17.143 (1 – 10^{–0.1085 × 9.921}) = 15.7 mg/l

   Correct Option: C

   BOD of mixture, Y =	(2 × 90) + (12 × 5)	= 17.143 g/m³
   	2 × 12

   
   K_D = 0.434 × k¹
   = 0.434 × 25 = 0.1085
   

   Velocity of stream =	14	= 0.28 m/s
   	50

   
   

   Time taken =	10 km	= 9.921 hours
   	0.28 m/s

   
   Y_t = Y_o (1 – 10^–kt)
   = 17.143 (1 – 10^{–0.1085 × 9.921}) = 15.7 mg/l

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