A $10$ million litres per day $\text{(MLD)}$ sewage treatment plant $\text{(STP)}$ is based on the Activated Sludge Process $\text{(ASP)}$. First, the sewage undergoes primary treatment and the resulting treated sewage has $\text{BOD5}$ of $140 \mathrm{mg} / \mathrm{L}$ concentration. This is further passed through a $1500 \mathrm{~m}^{3}$ capacity aeration tank $\text{(in ASP)}$, where the mixed liquor volatile suspended solids $\text{(MLVSS)}$ concentration is maintained at $3000 \mathrm{mg} / \mathrm{L}$. The concentration of $\text{BOD 5}$ of the treated sewage is $5 \mathrm{mg} / \mathrm{L}$.
The Food to Microorganisms ratio $\text{(F/M)}$ of the $\text{ASP}$ is___________ day $^{-1}$ $\text{(rounded off to two decimal places)}$.
