Question 1 |

The partial differential equation \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\frac{\partial^2 u}{\partial x^2}
is a

linear equation of order 2 | |

non-linear equation of order 1 | |

linear equation of order 1 | |

non-linear equation of order 2 |

Question 1 Explanation:

A differential equation in the form \frac{d y}{d x}+P y=Q

where, P and Q are functions of x i.e., f(x) is said

to be linear equation.

\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}

The given equation is not complying with the definition of linear equation, therefore it is a nonlinear equation of order 2.

where, P and Q are functions of x i.e., f(x) is said

to be linear equation.

\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}

The given equation is not complying with the definition of linear equation, therefore it is a nonlinear equation of order 2.

Question 2 |

The eigenvalues of a symmetric matrix are all

complex with non-zero positive imaginary part. | |

complex with non-zero negative imaginary part. | |

real. | |

pure imaginary. |

Question 2 Explanation:

(i) The Eigen values of symmetric matrix

\left[A^{T}=A\right] are purely real

(ii) The Eigen value of skew-symmetric matrix

\left[A^{T}=-A\right] are either purely imaginary or zeros

\left[A^{T}=A\right] are purely real

(ii) The Eigen value of skew-symmetric matrix

\left[A^{T}=-A\right] are either purely imaginary or zeros

Question 3 |

Match the CORRECT pairs.

P-2, Q-1, R-3 | |

P-3, Q-2, R-1 | |

P-1, Q-2, R-3 | |

P-3, Q-1, R-2 |

Question 4 |

A rod of length L having uniform cross-sectional area A is subjected to a tensile force P as shown in the figure below. If the Young's modulus of the material varies linearly from E_{1}
to E_{2}
along the length of the rod, the normal stress developed at the section-SS is

\frac{P}{A} | |

\frac{P(E_{1}-E_{2})}{A(E_{1}+E_{2})} | |

\frac{PE_{2}}{AE_{1}} | |

\frac{PE_{1}}{AE_{2}} |

Question 4 Explanation:

Normal stress at any section is independent of modulus of elasticity.

Question 5 |

Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 4 times that of bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is

16 | |

24 | |

36 | |

48 |

Question 5 Explanation:

Given:

\begin{aligned} P_{1}&=P_{2}=P\\ &\text{(identical tensile load on bolt A \& B )} \\ &\qquad \text{(same length)}\\ L_{1} &=L_{2}=L \\ d_{A} &=12 \mathrm{mm} \\ U_{A} &=\text { strain energy in bolt } A \\ U_{B} &=\text { Strain energy in bolt } B \\ U_{A} &=4 U_{B}(\text { Given }) \qquad \cdots(i)\\ \therefore \quad d_{B} &=? \end{aligned}

Strain energy is given by

U_{A}=\frac{1}{2} P \times \delta=\frac{1}{2} \frac{P^{2} L}{A E}

\therefore Eq. (i)

\begin{aligned} \frac{1 P_{1}^{2} L_{1}}{2} &=4 \times \frac{1}{2} \frac{P_{2}^{2} L_{2}}{A_{2} E} \\ \frac{1}{A_{1}} &=4 \times \frac{1}{A_{2}} \\ \frac{4}{\pi(12)^{2}} &=4 \times \frac{4}{\pi\left(d_{B}\right)^{2}} \\ d_{B}^{2} &=576 \\ d_{B} &=24 \mathrm{mm} \end{aligned}

\begin{aligned} P_{1}&=P_{2}=P\\ &\text{(identical tensile load on bolt A \& B )} \\ &\qquad \text{(same length)}\\ L_{1} &=L_{2}=L \\ d_{A} &=12 \mathrm{mm} \\ U_{A} &=\text { strain energy in bolt } A \\ U_{B} &=\text { Strain energy in bolt } B \\ U_{A} &=4 U_{B}(\text { Given }) \qquad \cdots(i)\\ \therefore \quad d_{B} &=? \end{aligned}

Strain energy is given by

U_{A}=\frac{1}{2} P \times \delta=\frac{1}{2} \frac{P^{2} L}{A E}

\therefore Eq. (i)

\begin{aligned} \frac{1 P_{1}^{2} L_{1}}{2} &=4 \times \frac{1}{2} \frac{P_{2}^{2} L_{2}}{A_{2} E} \\ \frac{1}{A_{1}} &=4 \times \frac{1}{A_{2}} \\ \frac{4}{\pi(12)^{2}} &=4 \times \frac{4}{\pi\left(d_{B}\right)^{2}} \\ d_{B}^{2} &=576 \\ d_{B} &=24 \mathrm{mm} \end{aligned}

Question 6 |

A link OB is rotating with a constant angular velocity of 2 rad/s in counter clockwise direction and a block is sliding radially outward on it with an uniform velocity of 0.75 m/s with respect to the rod, as shown in the figure below. If OA = 1 m, the magnitude of the absolute acceleration of the block at location A in m/s^{2}
is

3 | |

4 | |

5 | |

6 |

Question 6 Explanation:

Acceleration of the block at A=\vec{a}_{c r}+\vec{a}_{r}

\begin{aligned} &=\overrightarrow{2 \omega V}+\overrightarrow{\omega^{2} r} \\ \Rightarrow \quad r&=O A\\ \Rightarrow \quad 2 \omega V&=2 \times 2 \times 0.75=3 \mathrm{m} / \mathrm{s}^{2}\\ \Rightarrow \quad \omega^{2} r&=2^{2} \times 1=4 \mathrm{m} / \mathrm{s}^{2}\\ \end{aligned}

As coriolis component and radial component are perpendicular to each other.

So, \left|\vec{a}_{a b s}\right|=\sqrt{3^{2}+4^{2}}=5 \mathrm{m} / \mathrm{s}

\begin{aligned} &=\overrightarrow{2 \omega V}+\overrightarrow{\omega^{2} r} \\ \Rightarrow \quad r&=O A\\ \Rightarrow \quad 2 \omega V&=2 \times 2 \times 0.75=3 \mathrm{m} / \mathrm{s}^{2}\\ \Rightarrow \quad \omega^{2} r&=2^{2} \times 1=4 \mathrm{m} / \mathrm{s}^{2}\\ \end{aligned}

As coriolis component and radial component are perpendicular to each other.

So, \left|\vec{a}_{a b s}\right|=\sqrt{3^{2}+4^{2}}=5 \mathrm{m} / \mathrm{s}

Question 7 |

For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, the pressure drop \Delta p
over a length L and the wall shear stress \tau _{w}
are reated by

\tau _{w}=\frac{\Delta p D}{4 L} | |

\tau _{w}=\frac{\Delta p D^{2}}{4 L^{2}} | |

\tau _{w}=\frac{\Delta p D}{2 L} | |

\tau _{w}=\frac{4\Delta p L}{ D} |

Question 7 Explanation:

Shear stress on wall,

\begin{aligned} \tau_{w} &=-\frac{\partial p}{\partial x} \frac{R}{2} \\ \text { where } \quad-\frac{\partial p}{\partial x} &=\frac{\Delta p}{L} \\ \Delta p &=\text { Pressure drop }\\ \text{and }\quad R&=\frac{D}{2}\\ \therefore \tau_{w}&=\frac{\Delta p}{L} \times \frac{D}{2 \times 2}=\frac{\Delta p D}{4 L} \end{aligned}

\begin{aligned} \tau_{w} &=-\frac{\partial p}{\partial x} \frac{R}{2} \\ \text { where } \quad-\frac{\partial p}{\partial x} &=\frac{\Delta p}{L} \\ \Delta p &=\text { Pressure drop }\\ \text{and }\quad R&=\frac{D}{2}\\ \therefore \tau_{w}&=\frac{\Delta p}{L} \times \frac{D}{2 \times 2}=\frac{\Delta p D}{4 L} \end{aligned}

Question 8 |

The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30^{\circ}C and 70%, respectively. If the saturated steam pressure at 30^{\circ}C is 4.25 kPa, the specific humidity of the room air in kg water vapour/kg dry air is

0.0083 | |

0.0101 | |

0.0191 | |

0.0232 |

Question 8 Explanation:

\begin{aligned} p_{\mathrm{atn}} &=1 \mathrm{bar}=100 \mathrm{kPa} \\ \mathrm{DBT} &=30^{\circ} \mathrm{C} \\ \phi &=70 \%=0.7 \\ p_{\mathrm{vs}} &=4.25 \mathrm{kPa} \end{aligned}

Specific humidity: \omega=?

\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.7 &=\frac{p_{v}}{4.25} \\ p_{v} &=2.975 \mathrm{kPa} \end{aligned}

Specific humidity,

\begin{aligned} \omega&=0.622 \times \frac{p_{v}}{p-p_{v}}\\ &=0.622 \times \frac{2.975}{100-2.975} \\ &=00.0191 \end{aligned}

Specific humidity: \omega=?

\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.7 &=\frac{p_{v}}{4.25} \\ p_{v} &=2.975 \mathrm{kPa} \end{aligned}

Specific humidity,

\begin{aligned} \omega&=0.622 \times \frac{p_{v}}{p-p_{v}}\\ &=0.622 \times \frac{2.975}{100-2.975} \\ &=00.0191 \end{aligned}

Question 9 |

Consider one-dimensional steady state heat conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The conductivity of the wall is given by k=k_{0}+bT; where k_{0}
and b are positive constants, and is T temperature.

As x increases, the temperature gradient ( dT/dx ) will

As x increases, the temperature gradient ( dT/dx ) will

remain constant | |

be zero | |

increase | |

decrease |

Question 9 Explanation:

Q=-k A \frac{d T}{d x} \ldots(i)

where \quad k=k_{0}+b T

As x increases, T increases, then k also increases.

i.e., k is +ve

Equation (i), indicates \frac{d T}{d x} decreases.

Question 10 |

In a rolling process, the state of stress of the material undergoing deformation is

pure compression | |

pure shear | |

compression and shear | |

tension and shear |

Question 10 Explanation:

Most metal rolling operations are similar in that the work material is plastically deformed by compressive forces between two constantly spinning rolls. Thus in a Rolling process, the material undergoing deformation is in the state of pure biaxial compression

There are 10 questions to complete.