Quiz: Civil Engineering

Exam: DFCCIL-JE

Topic: MISCELLANEOUS

Each question carries 1 mark

Negative marking: 1/4 mark

Time: 8 Minutes

Q1. Coefficient of discharge of an ogee spillway

(a) Remains constant

(b) Depends on depth of approach and upstream slope

(c) Depends on downstream apron interference and downstream submergence

(d) Both (2) and (3)

Q2. The water stored in the reservoir below the minimum pool level is called

(a) Useful storage

(b) Dead storage

(c) Valley storage

(d) Surcharge storage

Q3. As per Lacey’s regime theory, the flow velocity is proportional to

(a) (Qf²)^(1/6)

(b) Q/f²

(c) (Q/f²)^(1/4)

(d) (Qf)^(1/3)

Q4. By doubling the diameter of a well in an unconfined aquifer the discharge according to Dupit’s formula will be increasing by

(a) 28%

(b) 42%

(c) 66%

(d) 87%

Q5. The length of a base line is measured on ground at an elevation of 300 metres above means sea level is 2250 metres. The required correction to reduces to sea level length (given Radius of earth = 6370 km) will be

(a) 106 mm

(b) 206 mm

(c) 306 mm

(d) 212 mm

Q6. Every 20 m chain should be accurate to within

(a) ± 8 mm

(b) ± 2 mm

(c) ± 5mm

(d) ± 20 mm

SOLUTION

S1. Ans.(d)

Sol. Coefficient of discharge of an ogee spillway is depends upon depth of approach and upstream slope as well as on downstream apron interference and downstream submergence.

S2. Ans.(b)

Sol. The water stored in the reservoir below the minimum pool level is called dead storage. While Useful storage is the volume of water stored in a reservoir between the minimum pool and normal pool level.

S3. Ans.(a)

Sol. As per Lacey’s regime theory the flow velocity (V_f ) is given by –

▭(V_f=((Qf^2)/140)^(1\/6) )

▭(V_f α (Qf^2 )^(1\/6) )

Where Q = discharge

f = Lacey’s silt factor

S4. Ans.(c)

Sol. According to Dupit’s, the discharge in an unconfined aquifer is given by-

▭(Q_1=πK(H_1^2-H_1^2 )/(2.303 log (R/r) ))

Now, D=2D

▭(R^’=2D=4R)

Q_2= πK(H_1^2-H_1^2 )/(2.303 log (4R/r) )

▭(Q_2=1.66 Q_1 )

Percentage increase in discharge =(Q_2-Q_1)/Q_1 ×100

=(1.66Q_1-Q_1)/Q_1 ×100

=66%

S5. Ans.(a)

Sol. Given, Length of base line (L) = 300.

Length above mean sea level (h) =- 2250 m.

Radius of earth (R) = 6370 km.

Now,

Mean sea level correction = Lh/R

=(300×2250)/(6370×1000)

=0.1059 m≃106 mm

S6. Ans.(c)

Sol.

Chain Tolerance

10 m. chain————— ± 3

20 m. chain —————± 5

30 m. chain —————± 8