Waste Water Engineering For the survival of fish in a river stream, the minimum dissolved oxygen is prescribed 4 ppm 3 ppm 10 ppm 5 ppm 4 ppm 3 ppm 10 ppm 5 ppm ANSWER DOWNLOAD EXAMIANS APP
Waste Water Engineering If a 2% solution of sewage sample is incubated for 5 days at 20°C and the dissolved oxygen depletion was found to be 8 mg/l. The BOD of the sewage is 100 mg/l 300 mg/l 400 mg/l 200 mg/l 100 mg/l 300 mg/l 400 mg/l 200 mg/l ANSWER DOWNLOAD EXAMIANS APP
Waste Water Engineering At the junction of sewers Tops of both the sewers are at the same level Top of smaller sewer is kept lower None of these Top of larger sewer is kept lower Tops of both the sewers are at the same level Top of smaller sewer is kept lower None of these Top of larger sewer is kept lower ANSWER DOWNLOAD EXAMIANS APP
Waste Water Engineering To prevent settling down of sewage both at the bottom and on the sides of a large sewer, self-cleaning velocity recommended for Indian conditions, is 0.50 m/sec 0.75 m/sec 0.60 m/sec 0.70 m/sec 0.50 m/sec 0.75 m/sec 0.60 m/sec 0.70 m/sec ANSWER DOWNLOAD EXAMIANS APP
Waste Water Engineering If the length of overland flow from the critical point to the mouth of drain is 13.58 km and difference in level between the critical point and drain mouth is 10 m, the inlet time is 4 hours 2 hours 8 hours 6 hours 4 hours 2 hours 8 hours 6 hours ANSWER DOWNLOAD EXAMIANS APP
Waste Water Engineering The settling velocity of a spherical particle of diameter less than 0.1 mm as per Stock’s law, is Vs = 418 (Gs – Gw)d² [(3T + 70)/100] Vs = 218 (Gs – Gw)d [(3T + 70)/100] Vs = 418 (Gs – Gw) d [(3T + 70)/100] Vs = 218 (Gs – Gw)d² [(3T + 70)/100] Vs = 418 (Gs – Gw)d² [(3T + 70)/100] Vs = 218 (Gs – Gw)d [(3T + 70)/100] Vs = 418 (Gs – Gw) d [(3T + 70)/100] Vs = 218 (Gs – Gw)d² [(3T + 70)/100] ANSWER DOWNLOAD EXAMIANS APP
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