RSMSSB-JE'21 CE: Daily Practices Quiz. 06-Aug-2021 |_00.1
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RSMSSB-JE’21 CE: Daily Practices Quiz. 06-Aug-2021

Quiz: Civil Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 8 Minutes

Q1. Find value of the building based on following data:
Type of building: single storied
Built up area available = 150 m².
Area of court yard = 25 m².
The plinth area rate = Rs. 5000/m².
(a) 625,000
(b) 525,000
(c) 425,000
(d) 325,000

Q2. The rocks that possess crystalline and compact grains are called:
(a) Siliceous rocks
(b) Stratified rocks
(c) Unstratified rocks
(d) Foliated rocks

Q3. What should be the height (m) of a lighthouse, so that it can be visible from a distance of 3 km?
(a) 0.101
(b) 0.605
(c) 0.673
(d) 0.707

Q4. The factor of safety for steel as compared to concrete is ……………
(a) Higher
(b) Same
(c) Lower
(d) None of these

Q5. A vehicle has a wheel base of 5.5 m. What is the off-tracking while negotiating a curved path with a mean radius 30 m?
(a) 0.6
(b) 0.7
(c) 0.4
(d) 0.5

Q6. The relationship between air content of soil (a_c) and its degree of saturation (S_r) is expressed as
(a) a_c=1+S_r
(b) a_c=S_r-1
(c) a_c=1-S_r
(d) None of these

Solutions

S1. Ans.(a)
Sol. Plinth area rate = 5000 Rs/m²
Courtyard area = 25m²
Built up area = 150m²
Plinth area = Built up area – courtyard area
= 150 – 25
= 125m²
Total Cost = 125×5000
= 625,000 Rs.

S2. 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.

S3.Ans (b)
Sol. distance (d) = 3 km
” Height of lighthouse “(h)=0.0673d^2
=0.0673×3^2
=0.6057m

S4. Ans.(c)
Sol. factor of safety of steel ⇒1.15
Factor of safety of concrete ⇒ 1.5
→ Due to variability in strength of concrete due to poor quality control, poor work man ship etc, higher factor of safety is given for concrete in comparison to steel.

S5. Ans.(d)
Sol. Given, length of wheel base (l) = 5.5 m
Radius (R) = 30 m.
Off tracking = ?
Now,
Off Tracting =l^2/2R
=(5.5)^2/(2×30)
=0.50

S6. Ans.(c)
Sol. ▭(S+a_c=1)
S → degree of saturation
a_c → air content

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