The average sewage from a city is 90 million litres per day and the average 5 day Biochemical Oxygen Demand (BOD₅) is 300 mg/l. Average standard BOD₅ of the domestic sewage is 0.08 kg/day/person. The population equivalent of the city is

The correct option is 337500.

To find the population equivalent of the city, we first need to determine the total daily Biochemical Oxygen Demand (BOD₅) load produced by the city's sewage.

The average daily sewage flow ($Q$) is given as 90 million litres per day (MLD):
$Q=90\times {10}^6\text{litres/day}$

The average 5-day BOD (BOD₅) concentration of the sewage is:
${\text{BOD}}_5=300\text{mg/l}$

The total daily BOD₅ load generated by the city can be calculated by multiplying the sewage flow rate by the BOD₅ concentration:
$\text{Total BOD load}=Q\times {\text{BOD}}_5$
Since $1\text{mg/l}={10}^{-6}\text{kg/l}$, we can write:
$\text{Total BOD load}=90\times {10}^6\text{l/day}\times 300\times {10}^{-6}\text{kg/l}$
$\text{Total BOD load}=90\times 300\text{kg/day}=27000\text{kg/day}$

The average standard domestic BOD₅ contribution per person per day is:
$\text{Standard BOD load per person}=0.08\text{kg/day/person}$

The population equivalent is defined as the ratio of the total BOD₅ load to the standard BOD₅ load contributed by a single person:
$\text{Population Equivalent}=\frac{\text{Total BOD load}}{\text{Standard BOD load per person}}$
$\text{Population Equivalent}=\frac{27000}{0.08}$
$\text{Population Equivalent}=337500$

Therefore, the population equivalent of the city is 337500.