Quiz: Civil Engineering
Exam: UKSSSC-JE
Topic: Environmental Engineering
Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes
Q1. If the pH of the water is more than 8.5, then which of the following chloramines will be
(a) Only monochloramine
(b) Only dichloramine
(c) Only trichloramine
(d) Both monochloramine and dichloramine
L1Difficulty 2
QTags Environmental Engineering
QCereator Anubhav Gupta
Q2. Which of the following is the process of removal of permanent hardness of water?
(a) Boiling of water
(b) Zeolite process
(c) Filtration process
(d) Lime treatment
L1Difficulty 2
QTags Environmental Engineering
QCereator Anubhav Gupta
Q3. The most common cause of acidity in water is
(a) Nitrogen
(b) Sodium hydroxide
(c) Carbon dioxide
(d) Calcium
L1Difficulty 3
QTags Environmental Engineering
QCereator Anubhav Gupta
Q4. The acceptable limit of total hardness (CaCO₃) as per the Indian standard is:
(a) 100
(b) 250
(c) 150
(d) 200
L1Difficulty 2
QTags Environmental Engineering
QCereator Anubhav Gupta
Q5. The maximum permissible limit of Chlorides (mg/l) in potable water, in absence of an alternate source is:
(a) 1000
(b) 750
(c) 250
(d) 500
L1Difficulty 2
QTags Environmental Engineering
QCereator Anubhav Gupta
Q6. Which of the following is an objective of aeration of water?
(a) To increase the carbon dioxide content of water
(b) To preserve hydrogen sulphide
(c) To remove tastes and odours from the gases produced due to organic decomposition
(d) to decrease the dissolved oxygen content of the water
L1Difficulty 3
QTags Environmental Engineering
QCereator Anubhav Gupta
Q7. The alum added as coagulant in water treatment functions better when the raw water is:
(a) Acidic with high turbidity.
(b) Alkaline with high turbidity.
(c) Neutral with low turbidity.
(d) Acidic with low turbidity.
L1Difficulty 3
QTags Environmental Engineering
QCereator Anubhav Gupta
Q8. A water treatment plant treats 6000 m³ of water per day. If it consumes 20 kg chlorine per day, then the chlorine dosage would be
(a) 3.00 mg/l
(b) 3.75 mg/l
(c) 4.25 mg/l
(d) 3.33 mg/l
L1Difficulty 4
QTags Environmental Engineering
QCereator Anubhav Gupta
Q9. Mostly used coagulant in treatment of water is:
(a) Chlorine
(b) Alum
(c) Lime
(d) Bleaching powder
L1Difficulty 2
QTags Environmental Engineering
QCereator Anubhav Gupta
Q10. Calculate detention time of the tank whose breadth is 2 m, length is 4 m and depth is 2.5m. the rate of flow is 4 x 10³ liters per hour
(a) 5 hours
(b) 4 hours
(c) 5.5 hours
(d) 2.5 hours
L1Difficulty 4
QTags Environmental Engineering
QCereator Anubhav Gupta
Solutions
S1. Ans.(a)
Sol.
pH value Chloramine
> 7 Monochloramine
4-7 Dichloramine
1-3 Tri-chloramine
S2. Ans.(b)
Sol. permanent hardness of water is removed by zeolite process. zeolite is a natural or synthetic cation or base exchange hydrated silicate of aluminium & sodium.
S3. Ans.(c)
Sol. the acidity in water due to the presence of carbon di-oxide
S4. Ans.(d)
Sol. the acceptable limit of total hardness as per Indian standard is 200 mg/l and cause of rejection is 600 mg/l.
S5. Ans.(a)
Sol. the permissible limit of chlorides content in drinking water is 250 mg/l. and maximum permissible limit is 1000 mg/l.
S6. Ans.(c)
Sol. in aeration method, the water is brought in intimate contact with air which removes undesirable gases (like CO_2,H_2 S), tastes, odurs, volatile liquid and dissolved mineral (like iron and manganese) from water.
S7. Ans.(b)
Sol. the alum is added when the raw water is alkaline with high turbidity. It reacts with alkaline present in water and leads to the formation of sticky gelatinous precipitate of aluminium hydroxide.
S8. Ans.(d)
Sol. Quantity of water treated per day = 6000 m³
Quantity of chlorine consumed per day = 20 kg.
Chlorine dosage = (Consumed chlorine per day)/(Quantity of water treated per day)
= (20×10^6 mg.)/(6000× 10^3 l.)
= 3.33 mg/l.
S9. Ans.(b)
Sol. Alum is mostly used coagulant in treatment of water. It reacts with alkalinity present in water and leads to the formation of sticky gelatinous precipitate of aluminium hydroxide.
S10. Ans.(a)
Sol. Given,
Length (L) = 4m.
Breadth (B) = 2 m.
Depth (D) = 2.5 m.
Discharge (Q) = 4 × 10³ l/hour
Detention time of tank (t_d) =?
t_d = (Volume of tank (V))/(Discharge (Q) )
= (4×2×2.5 (m^3 ))/(4×10^3×10^(-3) (m^3 \/hour) )
= 5 hours.
