**Quiz: Civil Engineering **

**Exam: GATE**

**Topic: Miscellaneous**

**Each question carries 2 marks**

**Negative marking: 1/3 mark**

**Time: 12 Minutes**

Q1. The back sight at a station A is 0.865 m and the height of instrument is 561.365 m. the reduced level at A will be (in unit m)

(a) 558.705

(b) 559.260

(c) 560.500

(d) 560.550

Q2. A trapezoidal channel with base of 6 m and side slope of two horizontal to one vertical conveys water at 17 m³/sec with a depth of 1.5 m. The flow situation in the channel is:

(a) Critical

(b) Supercritical

(c) Subcritical

(d) None of the above

Q3. The area between the two isohyets 45 cm and 55 cm is 100 km², and that between 55 cm and 65 cm is 150 km². what is the average depth of annual precipitation over the basin of 250 km²?

(a) 50 cm

(b) 52 cm

(c) 56 cm

(d) 60 cm

Q4. The following data were obtained when a sample of medium sand was tested in a constant head permeameter:

Cross-section area of sample : 100 cm²

Hydraulic gradient : 10

Discharge collected : 10 cc/s

The coefficient of permeability of the sand is

(a) 0.1 m/s

(b) 0.01 m/s

(c) 1 × 10^(-4) m/s

(d) 1 × 10^(-8) m/s

Q5. A crest vertical curve joins two gradients of +3% and -2% for a design speed of 80 km/h and the corresponding stopping sight distance of 120 m. the height of driver’s eye and the object above the road surface are 1.20 m and 0.15 m respectively. The curve length (which is less than stopping sight distance) to be provided is

(a) 120 m

(b) 152 m

(c) 163 m

(d) 240 m

Solutions

S1. Ans.(c)

Sol.

HI=RL_A+BS

RL_A=HI-BS

RL_A=561.365-0.865

▭(RL_A=560.500 m.)

S2. Ans.(c)

Sol. Given,

B = 6m.

Q = 17 m³/s

y = 1.5 m.

2H:1V = mH:1V ⇒ m = 2

T = B + 2my

T = 6 + (2 × 2 × 1.5)

▭(T=12m)

A = (B + my) y

= [6 + (2 × 1.5)] × 1.5

= 13.5 m²

We know,

Froude no.(F)=(Q^2 T)/(gA^3 )

=((17)^2×12)/(9.81×(13.5)^3 )

▭(F=0.14368<1) Hence,flow is Subcritical

S3. Ans.(c)

Sol.

P_1=45 cm.

P_2=55 cm

P_3=65 cm

A_1=100 km^2

A_2=150 km^2

P_avg= ?

Average depth of annual precipitation by Iso-hyetal method is given by →

P_avg=(((P_1+P_2)/2).A_1+((P_2+P_3)/2) A_2)/(A_1+A_2 )

=(((45+55)/2)×100+((55+65)/2)×150)/(100+150)

=(5000+9000)/250

▭(P_avg=56 cm)

S4. Ans.(c)

Sol. Given,

Q = 10 cc/sec.

i = 10

A = 100 cm²

K = ?

We know,

Q = KiA

10 = K × 10 × 100

K=1/100 cm\/sec

▭(K=1×10^(-4) m\/sec.)

S5. Ans.(b)

Sol. Given,

N_1= +3%=0.03

N_2= -2%= -0.02

N=|N_1-N_2 |=|0.03-(-0.02)|=0.05

S=120m

Note→ When, ▭(SSD<Curve length), then curve length is given by –

▭(L=2S-4.4/N)

=(2×120)-4.4/0.05

=240-88

▭(L=152m)