# SSC-JE’21 CE: Daily Practices Quiz. 12-July-2021

Quiz: Civil Engineering
Exam: SSC-JE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. The rocks that possess crystalline and compact grains are called:
(a) Siliceous rocks
(b) Stratified rocks
(c) Unstratified rocks
(d) Foliated rocks

Q2. One sq. foot in sq. meter is:
(a) 0.083 sq m
(b) 0.073 sq m
(c) 0.103 sq m
(d) 0.093 sq m

Q3. Determine the amount of yearly depreciation for the property purchased at a cost of Rs 20,00,000 and having useful life of 20 years by straight line method of depreciation-
(a) Rs 60,000
(b) Rs 90,000
(c) Rs 1,00,000
(d) Rs 95,000

Q4. Atterberg limit tests were carried on a certain soil with the following results:
(i) Liquid Limit =40%
(ii) Plastic Limit =25%
(iii) Shrinkage Limit=10%
The value of plasticity index is:
(a) 30%
(b) 15%
(c) 25%
(d) 40%

Q5. What is the mechanical widening required for a pavement of width 7m on a horizontal curve of radius 490m, if the longest wheel base of vehicle expected on the road is 7.0m?
(a) 0.07 m
(b) 0.05 m
(c) 0.49m
(d) 0.1m

Q6. Which of the following represents the value of hourly variation factor?
(a) 1.2
(b) 1.5
(c) 1.7
(d) 2.5

Solutions

S1. Ans.(c)
Sol. Unstratified rocks → These rocks possess crystalline and compact grains. They do not have plane of stratification. Igneous rocks are unstratified rock.
Ex. Granite, marble, trap, Basalt etc.

S2. Ans.(d)
Sol. 1sq. yard = 9 sq. feet
1sq. meter = 1.196 sq. yards
1sq. feet = 0.0929 sq. meter
1 nautical mile = 1853m = 1.853 km.
1 meter = 1.094 yard

S3. Ans.(b)
Sol. Given, Total cost (C) = 20,00,000 Rs.
Design life (n) = 20 years
Scrap value (S) = 10% of total cost = 0.1 × 20,00,000 = 200,000 Rs.
Annual depreciation (D) by straight line method is given by-
D=(C-S)/n
=(20,00,000-200,000)/20
D=90,000 Per year

S4. Ans.(b)
Sol.
w_L=40%
w_P=25%
w_S=10%
Plasticity index (I_P )=w_L-w_p
= 40 – 25
= 15%

S5. Ans.(d)
Sol. Given, length of wheel base (l) = 7m
No. of lanes (n) = 2
mechanical widening W_me = (nl^2)/2R
= (2×(7)^2)/(2×490) = 0.1

S6. Ans.(b)
Sol.
1. Maximum Hourly demand = 1.5 × Average hourly demand
2. Maximum daily demand = 1.8 × Average daily demand

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