Question 1 |

In the following partial differential equation, \theta
is a function of t and z, and D and K are
functions of \theta

D(\theta )\frac{\partial^2 \theta }{\partial z^2}+\frac{\partial K(\theta )}{\partial z}-\frac{\partial \theta }{\partial t}=0

The above equation is

D(\theta )\frac{\partial^2 \theta }{\partial z^2}+\frac{\partial K(\theta )}{\partial z}-\frac{\partial \theta }{\partial t}=0

The above equation is

a second order linear equation | |

a second degree linear equation | |

a second order non-linear equation | |

a second degree non-linear equation |

Question 1 Explanation:

\because \;\;1^{st}
term of given D. Equation contains product of dependent variable with it's derivative, so it is non-linear and also we have 2nd order derivative so it's order is two

i.e., 2nd order non linear equation.

i.e., 2nd order non linear equation.

Question 2 |

The value of \lim_{x \to \infty }\frac{x^2-5x+4}{4x^2+2x}

0 | |

\frac{1}{4} | |

\frac{1}{2} | |

1 |

Question 2 Explanation:

It is in \left (\frac{\infty }{\infty } \right )
from so by L-Hospital Rule

\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}

\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}

Question 3 |

The true value of ln(2) is 0.69. If the value of ln(2) is obtained by linear interpolation
between ln(1) and ln(6), the percentage of absolute error (round off to the nearest integer),
is

35 | |

48 | |

69 | |

84 |

Question 3 Explanation:

True value of \ln 2=0.69=T

\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}

Divided differentiation

\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}

\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}

Divided differentiation

\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}

Question 4 |

The area of an ellipse represented by an equation \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is

\frac{\pi a b}{4} | |

\frac{\pi a b}{2} | |

\pi a b | |

\frac{4 \pi a b}{3} |

Question 4 Explanation:

\begin{aligned} \text{Area} &=\int \int (1)dydx \\ &=\int_{x=-a}^{a}\int_{y=-\frac{b}{a}}^{+\frac{b}{a}}(1)dydx \\ &=4\int_{x=0}^{a} \int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}}(1)dydx\\ &= 4 \int_{x=0}^{a}\int_{y=0}^{\frac{b}{a}\sqrt{a^2-x^2}} dx\\ &= \pi ab \end{aligned}

Question 5 |

Consider the planar truss shown in the figure (not drawn to the scale)

Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is

Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is

6 | |

7 | |

8 | |

9 |

Question 5 Explanation:

As \Delta _{AB}=0, hence F _{AB}=0

Total number of zero force member = 8

Question 6 |

A reinforcing steel bar, partially embedded in concrete, is subjected to a tensile force P. The figure that appropriately represents the distribution of the magnitude of bond stress
(represented as hatched region), along the embedded length of the bar, is

A | |

B | |

C | |

D |

Question 7 |

In a two-dimensional stress analysis, the state of stress at a point P is

[\sigma ]=\begin{bmatrix} \sigma _{xx} &\tau _{xy} \\ \tau _{xy}& \sigma _{yy} \end{bmatrix}

The necessary and sufficient condition for existence of the state of pure shear at the point P, is

[\sigma ]=\begin{bmatrix} \sigma _{xx} &\tau _{xy} \\ \tau _{xy}& \sigma _{yy} \end{bmatrix}

The necessary and sufficient condition for existence of the state of pure shear at the point P, is

\sigma _{xx}\sigma _{yy} -\tau^2 _{xy}=0 | |

\tau _{xy}=0 | |

\sigma _{xx}+\sigma _{yy}=0 | |

(\sigma _{xx}-\sigma _{yy})^2 +4\tau^2 _{xy}=0 |

Question 7 Explanation:

In pure shear condition \sigma _x=0, \sigma _y=0, \tau _{xy}=\tau

For this condition \sigma _{xx}+ \sigma _{yy}=0 is true.

Question 8 |

During the process of hydration of cement, due to increase in Dicalcium Silicate (C_2S)
content in cement clinker, the heat of hydration

increases | |

decreases | |

initially decreases and then increases | |

does not change |

Question 8 Explanation:

Due to increase in C_2S heat of hydration decreases.

Question 9 |

The Los Angeles test for stone aggregates is used to examine

abrasion resistance | |

crushing strength | |

soundness | |

specific gravity |

Question 9 Explanation:

Los Angles abrasion test is carried out to
examine the hardness i.e., abrasion resistance
property of aggregate

Question 10 |

Which one of the following statements is NOT correct?

A clay deposit with a liquidity index greater than unity is in a state of plastic consistency. | |

The cohesion of normally consolidated clay is zero when tri-axial test is conducted
under consolidated undrained condition. | |

The ultimate bearing capacity of a strip foundation supported on the surface of sandy
soil increase in direct proportion to the width of footing. | |

In case of a point load, Boussinesq's equation predicts higher value of vertical stress
at a point directly beneath the load as compared to Westergaard's equation. |

Question 10 Explanation:

A clay deposit with liquidty index greater then 1, will be in liquid stage of consistency.

\because \;\;I_L=\frac{w_n-w_p}{w_l-w_p} \gt 1

\therefore \;\; w_n \gt w_L

\because \;\;I_L=\frac{w_n-w_p}{w_l-w_p} \gt 1

\therefore \;\; w_n \gt w_L

There are 10 questions to complete.