**Quiz: Civil EngineeringExam: GATETopic: Miscellaneous**

**Each question carries 1 markNegative marking: 1/3 markTime: 20 Minutes**

Q1. A beam section is 20 cm wide and 60 cm deep. What is the section modulus of the section?

(a) 12000 cm³

(b) 6000 cm³

(c) 24000 cm³

(d) 6500 cm³

Q2. In a closed loop traverse of 1 km, total length, the closing errors in departure and latitude are 0.3 m and 0.4 m, respectively. The relative precision of this traverse will be

(a) 1: 5000

(b) 1: 4000

(c) 1: 2000

(d) 1: 3000

Q3. If the volume of voids is equal to the volume of solids in a soil mass, then the values of porosity and voids ratio respectively are:

(a) 1.0 and 0.0

(b) 0.0 and 1.0

(c) 0.5 and 1.0

(d) 1.0 and 0.5

Q4. During the flow over length of 30 m, the head loss of 5 m is occurred during the seepage what is the value of hydraulic gradient?

(a) 0.005

(b) 0.045

(c) 0.167

(d) 0.192

Q5. The radius of relative stiffness for a 20 cm. thick slab with E = 3 ×10^5 kg/cm² and poison’s ratio = 0.15, resting on a subgrade having modulus of 5 kg/cm³ is

(a) 320 cm

(b) 120 cm

(c) 80 cm

(d) 10 cm

Q6. A simply supported isotropically reinforced square slab of side 4m. is subjected to a service load of 7 kpa. The thickness of the slab is 120mm. The moment of resistance required as per yield line theory is

(a) 9 kN-m

(b) 9 kN-m/m

(c) 10 kN-m

(d) 10 kN-m/m

Q7. The back sight reading taken from a level at a bench mark is 1.56 m and a fore sight at a point A is taken on an inverted staff is 1.65 m. calculate the reduce level of the point A, If the reduce level of the bench mark is 150 m.

(a) 146.79

(b) 149.91

(c) 152.8

(d) 153.21

Q8. A range of Froude number of a steady jump is

(a) 1.0 to 1.7

(b) 2.5 to 4.5

(c) 1.7 to 2.5

(d) 4.5 to 9.0

Q9. The development length in compression for a 20 mm diameter deformed bar of grade Fe 415 embedded in concrete of grade M 25, whose design bond stress is 1.40 N/mm², is

(a) 1489 mm

(b) 1289 mm

(c) 806 mm

(d) 645 mm

Q10. A rice crop is to be irrigated in a field covering an area of 2400 ha, the duty and base period of rice are given as 860 ha/cumec and 120 days respectively. The volume of water required in the field is nearly:

(a) 500 ha-m

(b) 1400 ha – m

(c) 2000 ha-m

(d) 2880 ha-m

Solutions

S1. Ans.(a)

Sol. Given, B = 20 cm.

D = 60 cm.

Z = ?

We know

Z=I/y=((BD^3)/12)/(D/2)=(BD^2)/6

Z=(20×(60)^2)/6

▭(Z=12,000 cm^3 )

S2. Ans.(c)

Sol.

Given,

ΣD=0.3 , P=1 km.(Perimeter)

ΣL=0.4

Closing error (e)=√((ΣD)^2+(ΣL)^2 )

=√((0.3)^3+(0.4)^2 )

=0.5 m

Relative Precision (Re)=e/p

=0.5m/1km

=1/2000

S3. Ans.(c)

Sol. Let volume of solids = V_s

Volume of voids = V_V

Volume of soil = V

Given,

V_V=V_S

V=V_V+V_S

V=V_S+V_S=2V_S

e=V_V/V_S =V_S/V_S =1 (∵V_V=V_S )

n=V_V/V=V_S/(2V_S )=1/2=0.5

Hence,

▭(n=0.5 & e=1 )

S4. Ans.(c)

Sol. flow length (L) = 30m.

Head loss (h_L )=5m.

Hydraulic gradient (i) = ?

i=h_L/L

=5/30

▭(i =0.167)

S5. Ans.(c)

Sol. Given,

h=20 cm.

E=3×10^5 kg\/cm^2

k=5 kg\/cm^3

μ=0.15

Now, radius of relative stiffness (l)

l= [(E h^3)/(12 k (1-μ)^2 )]^(1\/4)

=[(3×10^5×(20)^3)/(12×5(1-{0.15}^2 ) )]^(1\/4)

=79.98 cm≈80 cm.

S6. Ans. (d)

Sol. Given, service load = 7 kpa = 7 kN/m²

Dead load = 25 × 0.12 = 3 kN/m²

Total factored load on the slab (W_u) = 1.5 × (7+3)

= 15 kN/m²

Now,

Moment of resistance as per yield line theory is given by–

M = (W_u L^2)/24 kN-m/m.

= (15×(4)^2)/24

= (15×16)/24

= 240/24

▭(M=10 kN-m/m.)

S7. Ans.(d)

Sol.

BS=1.56 (at Bench mark)

RL=150 (at Bench mark)

FS=-1.65 (Inverted staff reading )

R.L at point A = ?

H.I.=RL+BS

= 150 + 1.56

= 151.56m

R.L at point A = H.I – F.S

= 151.56 – (-1.65)

= 153.21 m

S8. Ans.(d)

Sol.

Froude number (F) | Types of jump |

1 < F ≤ 1.7 | Undular jump |

1.7 < F ≤ 2.5 | Weak jump |

2.5 < F ≤ 4.5 | Oscillatory jump |

4.5 < F ≤ 9 | Steady jump |

F > 9 | Strong jump |

S9. Ans.(d)

Sol. Given,

ϕ=20 m.

f_ck=25 N\/mm^2

τ_bd= 1.40 N\/mm^2

f_y=415 N\/mm^2

→ for HYSD bars, the bond strength is increased by 60%.

→ for bar in compression, the bond strength is increased by 25%.

Now,

τ_bd=1.4×1.6×1.25

=28 N\/mm^2

Development length (L_d )=(0.87fy.ϕ)/(4τ_bd )

=(0.87×415×20)/(4×2.8)

▭(L_d=644.7 mm≈645 mm)

S10. Ans.(d)

Sol. Given,

B = 120 days

D = 860 ha/cumec.

A = 2400 ha.

∆ =8.64B/D

=(8.64×120)/860

=1.206 m

Water required for 2400 ha. Area =2400×1.206

=2894.4 ha-m