Question 1 
In the context of water and wastewater treatments, the correct statements are:
particulate matter may shield microorganisms during disinfection  
ammonia decreases chlorine demand  
phosphorous stimulates algal and aquatic growth  
calcium and magnesium increase hardness and total dissolved solids 
Question 1 Explanation:
Total dissolved solids comprise of inorganic salts, principally calcium, magnesium, potassium, sodium bicarbonates, chlorides, sulfates and some small amount of organic matter that are dissolved in water.
Hardness is due to presence of multivalent cations. In most cases, it is due to the presence of soluble bicarbonates, chlorides and sulfates of calcium and magnesium.
Too much phosphorous can cause increased growth of algae and large aquatic plants. It can result in decreased levels of dissolved oxygen and the process is called 'Eutrophication'.
Ammonia creates a large chlorine demand, as the ammonia is converted to nitrogen gas and the chlorine becomes chloride during disinfection of water by chlorination.
Presence of particulate matter decreases the effectiveness of UV disinfection or chlorine disinfection by shielding the targeted microorganisms.
Hardness is due to presence of multivalent cations. In most cases, it is due to the presence of soluble bicarbonates, chlorides and sulfates of calcium and magnesium.
Too much phosphorous can cause increased growth of algae and large aquatic plants. It can result in decreased levels of dissolved oxygen and the process is called 'Eutrophication'.
Ammonia creates a large chlorine demand, as the ammonia is converted to nitrogen gas and the chlorine becomes chloride during disinfection of water by chlorination.
Presence of particulate matter decreases the effectiveness of UV disinfection or chlorine disinfection by shielding the targeted microorganisms.
Question 2 
The wastewater inflow to an activated sludge plant is 0.5 \mathrm{~m}^{3} / \mathrm{s}, and the plant is to be operated with a food to microorganism ratio of 0.2 \mathrm{mg} / \mathrm{mg}\mathrm{d}. The concentration of influent biodegradable organic matter of the wastewater to the plant (after primary settling) is 150 \mathrm{mg} / \mathrm{L}, and the mixed liquor volatile suspended solids concentration to be maintained in the plant is 2000 \mathrm{mg} / \mathrm{L}. Assuming that complete removal of biodegradable organic matter in the
tank, the volume of aeration tank (in \mathrm{~m}^{3}
, in integer)
required for the plant is _____.
16200  
14752  
23546  
54231 
Question 2 Explanation:
Given,
\begin{aligned} Q_1&=6.5 m^3/s \\ &=0.5 \times 24 \times 60 \times 60 \\ &=43200 m^3/d\\ \frac{F}{M} & =0.2 (mg/mgd)\\ \mathrm{S}_0 & =150 \mathrm{mg} / \mathrm{l} \\ \mathrm{X} & =2000 \mathrm{mg} / \mathrm{l} \\ \because \quad \frac{F}{M} & =\frac{\mathrm{Q}_0 \mathrm{~S}_0}{\mathrm{VX}} \\ 0.2 & =\frac{43200 \times 150}{\mathrm{~V} \times 2000} \\ V & =16200 \mathrm{~m}^3\end{aligned}
\begin{aligned} Q_1&=6.5 m^3/s \\ &=0.5 \times 24 \times 60 \times 60 \\ &=43200 m^3/d\\ \frac{F}{M} & =0.2 (mg/mgd)\\ \mathrm{S}_0 & =150 \mathrm{mg} / \mathrm{l} \\ \mathrm{X} & =2000 \mathrm{mg} / \mathrm{l} \\ \because \quad \frac{F}{M} & =\frac{\mathrm{Q}_0 \mathrm{~S}_0}{\mathrm{VX}} \\ 0.2 & =\frac{43200 \times 150}{\mathrm{~V} \times 2000} \\ V & =16200 \mathrm{~m}^3\end{aligned}
Question 3 
Identify the waterborne diseases caused by viral
pathogens:
Acute anterior poliomyelitis  
Cholera  
Infectious hepatitis  
Typhoid fever

Question 3 Explanation:
Water Borne Disease caused by
(i) Bacterial  (a) Typhoid (b) Cholera
(ii) Virus  (a) Jaundice (Hapatitis virus) (b) Poliomyetitis
(iii) Protozoa  (a) Amoebic dysentry
(i) Bacterial  (a) Typhoid (b) Cholera
(ii) Virus  (a) Jaundice (Hapatitis virus) (b) Poliomyetitis
(iii) Protozoa  (a) Amoebic dysentry
Question 4 
A 2% sewage sample (in distilled water) was incubated for 3 days at 27^{\circ}C
temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was
recorded. The biochemical oxygen demand (BOD) rate constant at 27^{\circ}C was
found to be 0.23 \; day^{1} (at base e).
The ultimate BOD (in mg/L) of the sewage will be_____________. (round off to the nearest integer)
The ultimate BOD (in mg/L) of the sewage will be_____________. (round off to the nearest integer)
825  
1254  
1003  
1502 
Question 4 Explanation:
Depletion of dissolved oxygen after 3 days
incubation = 10 mg/L
\text{Dilution factor }=\frac{\text{Vol. of dilutedsample}}{\text{Vol.of undiluted sewage sample}}=\frac{100}{2}50
BOD_3= \text{DO consumed} \times \text{dilution factor}=10 \times 50=500 mg/L
\begin{aligned} BOD_3&= BOD_u(1e^{kt}) \\ 500 &=BOD_u(1e^{0.23 \times 3}) \\ BOD_u &= 1003.162 mg/L \end{aligned}
\text{Dilution factor }=\frac{\text{Vol. of dilutedsample}}{\text{Vol.of undiluted sewage sample}}=\frac{100}{2}50
BOD_3= \text{DO consumed} \times \text{dilution factor}=10 \times 50=500 mg/L
\begin{aligned} BOD_3&= BOD_u(1e^{kt}) \\ 500 &=BOD_u(1e^{0.23 \times 3}) \\ BOD_u &= 1003.162 mg/L \end{aligned}
Question 5 
A grit chamber of rectangular crosssection is to be designed to remove particles with diameter of 0.25 mm and specific gravity of 2.70. The terminal settling velocity of the particles is estimated as 2.5 cm/s. The chamber is having a width of 0.50 m and has to carry a peak wastewater flow of 9720 \mathrm{~m}^{3} / \mathrm{d} giving the depth of flow as 0.75 m. If a flowthrough velocity of 0.3 m/s has to be maintained using a proportional weir at the outlet end of the chamber, the minimum length of the chamber (in m,in integer) to remove 0.25 mm particles completely is _________
5  
7  
9  
10 
Question 5 Explanation:
Minimum length of chamber
\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}
\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}
There are 5 questions to complete.