NHPC-JE'21 CE: Daily Practices Quiz. 02-Dec-2021

Know your strengths and practice your concepts with this quiz on NHPC JE Recruitment 2021. This quiz for NHPC JE Recruitment 2021 is designed specially according to NHPC Syllabus 2021.

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

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

Q1. When used in road work, the coefficient of hardness of a stone should be greater than:
(a) 17
(b) 10
(c) 15
(d) 12

Q2. If the focus length of the object glass is 25cm and the distance from object glass to the trunnion axis is 15 cm, the additive constant is
(a) 0.1
(b) 0.4
(c) 0.6
(d) 1.33

Q3. The size of diameter of large knot is greater than
(a) 40 mm
(b) 50 mm
(c) 45 mm
(d) 30 mm

Q4. The length of a curve connecting two uniform gradients of +0.8% and -0.6%, the rate of change of grade being 0.1% per 30 m, will be
(a) 220 m
(b) 320 m
(c) 420 m
(d) 520 m

Q5. Which one of the following statements is correct? An indeterminate building frame may be converted to a determinate one by assuming
(a) hinges at mid-height of columns
(b) hinges at mid-span of the beams
(c) hinges at both mid-height of columns and midspan of beams
(d) one support as fixed at base and other support on rollers

Q6. If the shrinkage limit of a soil is 21%. What will be the void ratio in dry state? Take G=2.70
(a) 0.79
(b) 0.29
(c) 0.49
(d) 0.57

Solutions

S1. Ans.(a)
Sol. The coefficient of hardness of a stone should be greater than 17.
▭(Coefficent of Hardness=(20- Loss in weight (in gm))/3)

S2. Ans.(b)
Sol. f = 25 cm.
d = 15 cm.
Additive constant =?
In tacheometer. ▭(D=kS+c)
K ⇒ Multiplying constant
▭(K=f/i)
C ⇒ Additive constant
▭(C=f+d)
C = 25+15
▭(C=40 cm)=0.40 m.

S3. Ans.(a)
Sol.

Knot	Diameter of knot
Pin knot	Up to 6.50 mm dia.
Small knot	6.50 mm – 20 mm
Medium knot	20 mm – 40 mm
Large knot	Greater than 40 mm

S4. Ans.(c)
Sol.

Length of curve =(0.8-(-0.6))/0.1 ×30
▭(L=420m.)

S5. Ans.(c)
Sol. An indeterminate building frame is statically indeterminate to third degree hence to make it determinate we assume hinges at both mid – height of column and midspan of beams.

S6. Ans.(d)
sol.
w=21%
G=2.70
Se = wG
1×e=0.21×2.70
▭(e=0.567)