6. The potable water is prepared from turbid surface water by adopting the following treatment sequence.

1. 1. Turbid surface water → Coagulation → Flocculation → Sedimentation → Filtration → Disinfection → Storage & Supply

   
   
   

   2. Turbid surface water → Disinfection → Flocculation → Sedimentation → Filtration → Coagulation → Storage & Supply

   
   
   

   3. Turbid surface water → Filtration → Sedimentation → Disinfection → Flocculation → Coagulation → Storage & Supply

   
   
   

   4. Turbid surface water → Sedimentation → Flocculation → Coagulation → Disinfection → Filtration → Storage & Supply

   
   
   

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1. View Hint View Answer Discuss in Forum

   Coagulation – flocculation process is always followed by Sedimentation and then filtration. Disinfection is the last process.

   Correct Option: D

   Coagulation – flocculation process is always followed by Sedimentation and then filtration. Disinfection is the last process.

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7. The dominating microorganisms in an activated sludge process reactor are

1. 1. aerobic heterotrophs

   
   
   

   2. anaerobic heterotrophs

   
   
   

   3. autotrophs

   
   
   

   4. phototrophs

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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8. An ideal horizontal flow setting basin in 3m deep having surface area 900 m². Water flows at the rate of 8000 m³ /d, at water temperature 20°C (m = 10³ kg/m.s and p = 1000 kg/m³). Assuming Stokes law to be valid, the proportion (percentage) of spherical sand particles (0.01 mm in diameter with specific gravity 2.65), that will be removed is

1. 1. 32.5

   
   
   

   2. 67

   
   
   

   3. 87.5

   
   
   

   4. 95.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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9. Results of a water sample analysis are as follows:
   
   (milliequivalent weight of CaCO₃ = 50 mg/meq).
   Harness of the water sample in mg/L as CaCO₃ is

1. 1. 44.8

   
   
   

   2. 89.5

   
   
   

   3. 179

   
   
   

   4. 358

   
   
   

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1. View Hint View Answer Discuss in Forum

   Hardness = Mg2+ ×	equivalent wt of CaCO3	+ Ca2+ ×	equivalent wt of CaCO3
   	equivalent wt of Mg2+		equivalent wt of Ca2+

   
   

   = 20 ×	50	+ 55 ×	50
   	12.2		20

   
   = 179 mg/l of CaCO₃.

   Correct Option: C

   Hardness = Mg2+ ×	equivalent wt of CaCO3	+ Ca2+ ×	equivalent wt of CaCO3
   	equivalent wt of Mg2+		equivalent wt of Ca2+

   
   

   = 20 ×	50	+ 55 ×	50
   	12.2		20

   
   = 179 mg/l of CaCO₃.

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10. The theoretical oxygen dmenad of a 0.001 mol/L glucose solution is

1. 1. 180 mg/L

   
   
   

   2. 192 mg/L

   
   
   

   3. 90 mg/L

   
   
   

   4. 96 mg/L

   
   
   

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1. View Hint View Answer Discuss in Forum

   Glucose: C₆H₁₂O₆
   
   0.001 mol/L of glucose = 180 × 0.001 g/litre
   = 180 mg/L glucose
   From the equation 180 parts glucose need 192 parts oxygen.
   ∴ 180 mg/L glucose need 192 mg/L oxygen

   Correct Option: B

   Glucose: C₆H₁₂O₆
   
   0.001 mol/L of glucose = 180 × 0.001 g/litre
   = 180 mg/L glucose
   From the equation 180 parts glucose need 192 parts oxygen.
   ∴ 180 mg/L glucose need 192 mg/L oxygen

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