Engineering Jobs   »   Civil Engineering quizs   »   NHPC-JE

# NHPC-JE’21 CE: Daily Practices Quiz. 01-Nov-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. What should be the ratio of jet speed to blade speed for maximum efficiency of a Pelton wheel?
(a) 1/6
(b) 3/4
(c) 1/4
(d) 1/2

Q2. A circular shaft can transmit a torque of 5 kNm. If the torque is reduced to 4 kNm, then the maximum value of bending moment that can be applied to the shaft is
(a) 1 kNm
(b) 2 kNm
(c) 3 kNm
(d) 4 kNm

Q3. California bearing ratio (CBR)
(a) is a measure of soil strength
(b) is a procedure for designing flexible pavements
(c) is a method of soil identification
(d) is a measure to indicate the relative strength of paving materials

Q4. The minimum size of one side or diameter of column in reinforced cement concrete structure to make it earthquake resistant should not be less than:
(a) 200 mm
(b) 250 mm
(c) 300 mm
(d) 350 mm

Q5. Efficiency of a riveted joint, having the minimum pitch as per IS:800
(a) 40%
(b) 50%
(c) 60%
(d) 70%

Q6. For 100 sq. m cement concrete (1: 2: 4) 4 cm thick floor, the quantity of cement required is…………..
(a) 0.90 m³
(b) 0.94 m³
(c) 0.98 m³
(d) 1.00 m³

Solutions

S1. Ans.(d)
Sol. For maximum efficiency
u=V/2
u – Rotational speed
V – Jet velocity

S2. Ans.(c)
Sol. T_e=5 KN.m.
T= 4 KN.m
M= ?
Equivalent Torque (Te) = √(T^2+M²)
5=√(4^2+M²)
M= 3 KN.m

S3. Ans.(a)
Sol. California Bearing ratio (CBR) is a measure of arbitrary soil strength. IRC consider CBR method for designing of flexible pavement.

S4. Ans.(a)
Sol. Minimum size of column for reinforced cement concrete is 200 mm. if size of column is less than 200 mm then plain cement concrete used.

S5. Ans.(c)
Sol. Efficiency of a rivet joint per pitch length =
(P-d)/P×100
Minimum pitch (P) = 2.5d
Efficiency (η) = (2.5d-d)/2.5d
=1.5/2.5×100
=60%

S6. Ans.(a)
Sol. Volume of concrete = 100×0.04 = 4m^3
Note-Volume of dry concrete is 1.54 times the volume of wet concrete.
Hence,
Quantity of dry concrete required = 1.54 × 4 = 6.16 m³
For 1: 2: 4 concrete, quantity of cement in 6.16 m³ Concrete = 1/((1+2+4) )×6.16
= 0.88 ≃ 0.90 m³.

Sharing is caring!