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

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

Q1. If principal stresses in a two-dimensional case are -20 MPa and 30 MPa respectively, then maximum shear stress at the point is

(a) 10 MPa

(b) 15 MPa

(c) 25 MPa

(d) 30 MPa

Q2. The four cross – section shown below are required to be ordered in the increasing order of their respective shape factor

Which of the following order is correct?

(a) iv, ii, iii, i

(b) i, ii, iii, iv

(c) ii, iv, iii, i

(d) iii, I, iv, ii

Q3. The sequent depth ratio of a hydraulic jump in a rectangular horizontal channel is 10.32. the Froude number at the beginning of the jump is

(a) 7.64

(b) 5.64

(c) 13.61

(d) 8.05

Q4. If tomato juice is having a pH of 4.2, the hydrogen ion concentration will be

(a) 9.94 × 10^(-5) mol/l

(b) 8.94 × 10^(-5) mol/l

(c) 10.31 × 10^(-5) mol/l

(d) 6.31 × 10^(-5) mol/l

Q5. A 1:50 scale model of a spillway is to be tested in the laboratory. The discharge in the prototype is 1010 m³/sec. The discharge to be maintained in the model test is

(a) 5.7 m³/sec

(b) 0.057 m³/sec

(c) 0.57 m³/sec

(d) 57 m³/sec

Q6. The following data was obtained from a liquid test conducted on a soil sample

Number of blows | 17 | 22 | 25 | 28 | 34 |

Water content (%) | 53.8 | 53.1 | 51.9 | 50.6 | 50.5 |

The liquid limit of the soil is

(a) 53.1%

(b) 52.8%

(c) 51.9%

(d) 50.6%

Q7. A metal bar of length 150 mm is inserted between two rigid supports and its temperature is increased by 10°C. if the coefficient of thermal expansion is 12 × 10^(-6)/°C and young’s modulus is 2 × 10^5 MPa, the stress in the bar is

(a) 240 MPa

(b) 2.4 MPa

(c) Zero

(d) 24 MPa

Q8. A levelling is carried out to establish the reduces levels (RL) of point R with respect to bench mark (BM) at P. the staff reading taken are given below:

Staff Station | BS | IS | FS | RL |

P | 1.656 m | 100.000m | ||

Q | -0.951m. | -1.510m | ||

R | 0.751m |

If RL of P is +100.000m, then RL (in m) of R is

(a) 103.464

(b) 101.564

(c) 100.464

(d) 101.464

Q9. A fixed beam is loaded as in figure. The fixed end moment at support B.

(a) (WL^2)/20

(b) (WL^2)/30

(c) (WL^2)/8

(d) (WL^2)/10

Q10. 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

Solutions

S1. Ans.(c)

Sol. given σ_1=30 MPa ,σ_2= -20MPa

Maximum shear stress (τ) = (σ_1-σ_2)/2

=(30—(-20))/2

= 25 MPa

S2. Ans.(a)

Sol.

S.N. | Shape | Shape factor |

1 | I-Section | 1.12-1.15 |

2. | Rectangle | 1.5 |

3. | Circle | 1.7 |

4. | Diamond | 2.0 |

S3. Ans.(a)

Sol. Sequent depth ratio

y_2/y_1 =1/2 [-1+√(1+8 f_(r_1)^2 )]

Given, y_2/y_1 =10.32,〖 f〗*(r_1 )= ? ⇒10.32=1/2 [-1+√(1+8 f*(r_1)^2 )]

⇒20.64= -1+ √(1+8 f_(r_1)^2 )

⇒(21.64)^2=(1+8 f_(r_1)^2 )

By solving the above equation, we get

▭(〖f_r〗_1=7.64)

S4. Ans.(d)

Sol. ▭(pH= -log_10 [H^+ ] )

Given, pH = 4.2

4.2= -log_10 [H^+ ]

[H^+ ]=10^(-4.2)

▭([H^+ ]=6.31×10^(-5) mol\/l)

S5. Ans.(b)

Sol. Froude model law will be applicable in this case.

Q_m/Q_p =(L_m/L_p )^(5\/2)

Given,

(L_m/L_P )=(1/50)

Q_P=1010 m^3 \/sec.

Now,

Q_m/1010=(1/50)^(5\/2)

Q_M=1010×(1/50)^2.5

▭(Q_m=0.057 m^3 \/sec)

S6. Ans.(c)

Sol. the liquid limit is the water content corresponding to 25 number of blows

W_L=51.9%

S7. Ans.(d)

Sol. Given, Length (L) = 150 mm.

Coefficient of thermal expansion (α) = 12×10^(-6)/°C

Young’s modulus of elasticity (E) = 2×10^5 MPa

Change in temp. (∆T)=10°C

Thermal stress (σ_th )= ?

Now,

Thermal stress(σ_th ) = E α ∆T

=2×10^5×12×10^(-6)×10

=24 MPa

S8. Ans.(d)

Sol

Staff station | BS | IS | FS | HI | RL |

P | 1.656 | – | – | 101.656 | 100.000 |

Q | -0.951 | – | -1.51 | 102.215 | 103.166 |

R | – | – | 0.751 | – | 101.464 |

Hence,RL of R=101.464

S9. Ans. (b)

Sol. fixed end moment for beam loaded with uniformly varying load.

S10. 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.)