Question 1 |

The solution of the equation x\frac{\mathrm{d} y}{\mathrm{d} x}+y=0 passing through the point (1,1) is

x | |

x^{2} | |

x^{-1} | |

x^{-2} |

Question 1 Explanation:

\begin{aligned}x\frac{dy}{dx}+y&=0 \\ x\frac{dy}{dx}&=-y \\ \frac{dy}{y}&=-\frac{dx}{x} \\ \int \frac{1}{y}dy&=\int \frac{-1}{x}dx \\ \ln y&=-\ln x+c\\ \text{when } y&=1, x=1\\ c&=0\\ \Rightarrow \;\; y&=\frac{1}{x}=x^{-1} \end{aligned}

Question 2 |

The graph of a function f(x) is shown in the figure.

For f(x) to be a valid probability density function, the value of h is

For f(x) to be a valid probability density function, the value of h is

0.33 | |

0.66 | |

1 | |

3 |

Question 2 Explanation:

\begin{aligned} \int_{0}^{3}f\left ( x \right )dx &=1 \\ \int_{0}^{1}f\left ( x \right )dx+\int_{1}^{2}f\left ( x \right )dx+\int_{2}^{3}f\left ( x \right )dx &=1 \\ \frac{h}{2}+\frac{2h}{2+\frac{3h}{2}} &=1 \\ 6h &=2 \\ \Rightarrow\; \; h &=\frac{1}{3} \end{aligned}

Question 3 |

A probability distribution with right skew is shown in the figure.

The correct statement for the probability distribution is

The correct statement for the probability distribution is

Mean is equal to mode | |

Mean is greater than median but less than mode | |

Mean is greater than median and mode | |

Mode is greater than median |

Question 3 Explanation:

t_{L}\lt t_{mean}=Curve is skew to right.

Mode \lt mean

i.e., Mean \gt median and mode

Mean is greater than the mode and the median.This is common for a distribution that is skewed to the right [i.e., bunched up toward the left and a 'tail' stretching toward the right].

Question 4 |

All the members of the planar truss (see figure), have the same properties in terms of area of cross-section (A) and modulus of elasticity (E).

For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is:

For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is:

There are 3 members in tension, and 2 members in compression | |

There are 2 members in tension, 2 members in compression, and 1 zero-force member | |

There are 2 members in tension, 1 member in compression, and 2 zero-force members | |

There are 2 members in tension, and 3 zero-force members |

Question 4 Explanation:

Since member BDneither elongate nor contract.

Hence, \quad F_{B D}=0 . So, there are 2 tension members (AB and DC) and 3 zero force members (AD, BD, BC).

Question 5 |

The setting time of cement is determined using

Le Chatelier apparatus | |

Briquette testing apparatus | |

Vicat apparatus | |

Casagrande's apparatus |

Question 5 Explanation:

Vicat apparatus is used to determine the normal consistancy, IST, FST of cement.

Question 6 |

A structural member subjected to compression, has both translation and rotation restrained at one end, while only translation is restrained at the other end. As per IS 456 : 2000, the effective length factor recommended for design is

0.5 | |

0.65 | |

0.7 | |

0.8 |

Question 6 Explanation:

One end is fixed

Other end is pin jointed

Effective length of column (as per IS:456-2000)=0.08L

Other end is pin jointed

Effective length of column (as per IS:456-2000)=0.08L

Question 7 |

A vertical load of 10 kN acts on a hinge located at a distance of L/4 from the roller support Q of a beam of length L (see figure).

The vertical reaction at support Q is

The vertical reaction at support Q is

0.0 kN | |

2.5 kN | |

7.5 kN | |

10.0 kN |

Question 7 Explanation:

Bending moment about hinge point A=0

(consider the right hand side of A )

\begin{aligned} R_{0} \times \frac{L}{4} &=0 \\ R_{0} &=0 \mathrm{kN} \end{aligned}

Question 8 |

A flownet below a dam consists of 24 equipotential drops and 7 flow channels. The difference between the upstream and downstream water levels is 6 m. The length of the flow line adjacent to the toe of the dam at exit is 1 m. The specific gravity and void ratio of the soil below the dam are 2.70 and 0.70, respectively. The factor of safety against piping is

1.67 | |

2.5 | |

3.4 | |

4 |

Question 8 Explanation:

N_{f}=7 \quad N_{d}=24 \quad H=6 \mathrm{m}

Critical Hydraulic Gradient,

i_{c}=\frac{G-1}{1+e}=\frac{2.7-1}{1+0.7}=1

Exit Gradient (i_{\text{exit}})

\begin{aligned} &=\frac{\Delta h}{l}=\frac{\left(\frac{H}{N_{d}}\right)}{l}=\frac{\left(\frac{6}{24}\right)}{1 m}=\frac{1}{4} \\ \text { F.O.S. } &=\frac{i_{c}}{i_{\text {exit }}}=\frac{1}{\left(\frac{1}{4}\right)}=4 \end{aligned}

Critical Hydraulic Gradient,

i_{c}=\frac{G-1}{1+e}=\frac{2.7-1}{1+0.7}=1

Exit Gradient (i_{\text{exit}})

\begin{aligned} &=\frac{\Delta h}{l}=\frac{\left(\frac{H}{N_{d}}\right)}{l}=\frac{\left(\frac{6}{24}\right)}{1 m}=\frac{1}{4} \\ \text { F.O.S. } &=\frac{i_{c}}{i_{\text {exit }}}=\frac{1}{\left(\frac{1}{4}\right)}=4 \end{aligned}

Question 9 |

The contact pressure and settlement distribution for a footing are shown in the figure.

The figure corresponds to a

The figure corresponds to a

rigid footing on granular soil | |

flexible footing on granular soil | |

flexible footing on saturated clay | |

rigid footing on cohesive soil |

Question 9 Explanation:

Rigid footing on granlar soil.

Question 10 |

Which one of the following statements is NOT correct?

When the water content of soil lies between its liquid limit and plastic limit, the soil is said to be in plastic state. | |

Boussinesq's theory is used for the analysis of stratified soil. | |

The inclination of stable slope in cohesive soil can be greater than its angle of internal friction. | |

For saturated dense fine sand, after applying overburden correction, if the Standard Penetration Test value exceeds 15, dilatancy correction is to be applied. |

Question 10 Explanation:

Boussinesq's assumed soil as isotropic hence not applicable for stratified soil.

There are 10 questions to complete.