SSC JE Civil Engineering 30-Oct-2020 Shift 1 Paper

Question : 137 of 200

Marks: +1, -0

The following observations were made on 2% dilution of wastewater:
Dissolved oxygen of aerated water used for dilution = 0 mg/l
Dissolved oxygen of diluted sample after 5 days = 0.6 mg/l
Dissolved oxygen of original sample = 2.0 mg/l
The BOD of 5 days of the sample is:

80 mg/l
60 mg/l
100 mg/l
70 mg/l

Validate

Solution:

Concept:
Standard BOD is determined at 5 days incubation at 20° C

Dilution factor

={\text" Volume of dilution sample "}/{\text " Volume of initial water sample "}

={x+y}/{x}

{\BOD}_{{5} @ 20°{\C}}=({\D.\O}_{{i}}-{\D.\O} _{{\f}}) ×{\D.\F}

Where,

{\D.\O }_\i

= Initial Dissolve Oxygen,

{\D.\O} _{{\f}}

= Final Dissolve Oxygen,

\D.\F

= Dilution factor
Calculations:
BOD = (2 - 0.6) ×

100/2

= 70 mg/l

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