Question 1 |

The minimum value of function y=x^2 in the interval [1,5] is

0 | |

1 | |

25 | |

Undefined |

Question 1 Explanation:

\begin{aligned} &Given \quad &y=x^{2}\\ &At\quad &x=1 ; \quad y=1\\ &at\quad &x=5 ; \quad y=25\\ \end{aligned}

Minimum value of function is 1

Minimum value of function is 1

Question 2 |

If a square matrix A is real and symmetric, then the Eigenvalues

are always real | |

are always real and positive | |

are always real and non-negative | |

occur in complex conjugate pairs |

Question 2 Explanation:

The eigen values of any real and symmetric matrix is always real.

Question 3 |

If \varphi (x,y)
and \Psi (x,y)
are functions with continuous second derivatives, then \varphi (x,y) + i \Psi (x,y)
can be expressed as an analytic function of x + i y \; (i=\sqrt{-1})
, when

\frac{\partial \varphi }{\partial x}= -\frac{\partial \Psi }{\partial x}; \frac{\partial \varphi }{\partial y}= \frac{\partial \Psi }{\partial y} | |

\frac{\partial \varphi }{\partial y}= -\frac{\partial \Psi }{\partial x}; \frac{\partial \varphi }{\partial x}= \frac{\partial \Psi }{\partial y} | |

\frac{\partial^2 \varphi }{\partial x^2}+\frac{\partial^2 \varphi }{\partial y^2}=\frac{\partial^2 \Psi }{\partial x^2}+\frac{\partial^2 \Psi }{\partial y^2}=1 | |

\frac{\partial \varphi }{\partial x}+\frac{\partial \varphi }{\partial y}=\frac{\partial \Psi }{\partial x}=\frac{\partial \Psi }{\partial y}= 0 |

Question 3 Explanation:

The necessary condition for a function

f(z)=\varphi(x, y)+i \psi(x, y) to be analytic

\left.\begin{array}{l}\text { (i) } \frac{\partial \varphi}{\partial x}=\frac{\partial \psi}{\partial y} \\ \text { (ii) } \frac{\partial \varphi}{\partial y}=-\frac{\partial \psi}{\partial x}\end{array}\right\} are known as Cauchy Reiman equations

Provided \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \psi}{\partial x}, \frac{\partial \psi}{\partial y} exist.

f(z)=\varphi(x, y)+i \psi(x, y) to be analytic

\left.\begin{array}{l}\text { (i) } \frac{\partial \varphi}{\partial x}=\frac{\partial \psi}{\partial y} \\ \text { (ii) } \frac{\partial \varphi}{\partial y}=-\frac{\partial \psi}{\partial x}\end{array}\right\} are known as Cauchy Reiman equations

Provided \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \psi}{\partial x}, \frac{\partial \psi}{\partial y} exist.

Question 4 |

The partial differential equation \frac{\partial^2\varphi }{\partial x^2}+\frac{\partial^2\varphi }{\partial y^2}+\left ( \frac{\partial \varphi }{\partial x} \right )+\left ( \frac{\partial \varphi }{\partial y} \right )= 0
has

degree 1 order 2 | |

degree 1 order 1 | |

degree 2 order 1 | |

degree 2 order 2 |

Question 4 Explanation:

**Order**: The order of a differential equation is the order of the highest derivative appears in the equation

**Degree**:The degree of a differential equation is the degree of the highest order differential coefficient or derivative, when the differential coefficient are free from radicals and fraction. The general solution of a differential equation of order 'n' must involve 'n' arbitrary constant.

Question 5 |

Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)?

\delta Q = dU + \delta W | |

T dS = dU + pdV | |

T dS = dU + \delta W | |

\delta Q = dU + p dV |

Question 5 Explanation:

\delta Q=d U+p d V

This equation holds good for a closed system when only pdV work is present. This is true only for a reversible (quasistatic) process.

This equation holds good for a closed system when only pdV work is present. This is true only for a reversible (quasistatic) process.

Question 6 |

Water has a critical specific volume of 0.003155 m^3/kg. A closed and rigid steel tank of volume 0.025 m^3 contains a mixture of water and steam at 0.1 MPa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level inside the tank

will rise | |

will fall | |

will remain constant | |

may rise or fall depending on the amount of heat transferred |

Question 6 Explanation:

Given data:

Critical specific volume of water

v_{c}=0.003155 \mathrm{m}^{3} / \mathrm{kg}

Volume of steel tank,

V=0.025 \mathrm{m}^{3}

Pressure of mixture of water and steam,

p=0.1 \mathrm{MPa}=1 \mathrm{bar}

Mass of mixture of water and steam,

m=10 \mathrm{kg}

Specific volume of mixture,

\begin{aligned} v &=\frac{V}{m}=\frac{0.025}{10} \\ &=0.0025 \mathrm{m}^{3} / \mathrm{kg} \end{aligned}

As v \lt v_{c^{\prime}} the condition of steam lies near to saturated liquid line and the liquid level inside the tank will rise with heating.

Critical specific volume of water

v_{c}=0.003155 \mathrm{m}^{3} / \mathrm{kg}

Volume of steel tank,

V=0.025 \mathrm{m}^{3}

Pressure of mixture of water and steam,

p=0.1 \mathrm{MPa}=1 \mathrm{bar}

Mass of mixture of water and steam,

m=10 \mathrm{kg}

Specific volume of mixture,

\begin{aligned} v &=\frac{V}{m}=\frac{0.025}{10} \\ &=0.0025 \mathrm{m}^{3} / \mathrm{kg} \end{aligned}

As v \lt v_{c^{\prime}} the condition of steam lies near to saturated liquid line and the liquid level inside the tank will rise with heating.

Question 7 |

Consider an incompressible laminar boundary layer flow over a flat plate of length L, aligned with the direction of an oncoming uniform free stream. If F is the ratio of the drag force on the front half of the plate to the drag force on the rear half, then

F \lt 1/2 | |

F = 1/2 | |

F = 1 | |

F \gt 1 |

Question 7 Explanation:

Drag force,

\begin{aligned} F_{D} &=C_{f} \times \frac{1}{2} \rho A V^{2}=\frac{1.328}{\sqrt{\text{Re}_{L}}} \times \frac{1}{2} \rho A V^{2} \\ &=\frac{1.328}{\sqrt{\frac{\rho L V}{\mu}}} \times \frac{1}{2} \rho \times A \times L \times V^{2}\\ \text{i.e. }F_{D} &\propto \sqrt{L} \\ \end{aligned}

Now Drag force on front half,

\begin{aligned} \quad F_{D R} &\propto \sqrt{\frac{L}{2}} \\ \therefore F_{D / 2}&=\frac{F_{D}}{\sqrt{2}} \end{aligned}

Drag force on rear half,

\begin{aligned} F_{D / 2}^{\prime}&=F_{D}-F_{D / 2}=\left(1-\frac{1}{\sqrt{2}}\right) F_{D}\\ \text{Now} F&=\frac{F_{D / 2}}{F_{D / 2}^{\prime}}=\frac{\frac{F_{D}}{\sqrt{2}}}{\left(1-\frac{1}{\sqrt{2}}\right) F_{D}}=\frac{1}{\sqrt{2}-1}>1\\ \therefore \quad F&>1 \end{aligned}

\begin{aligned} F_{D} &=C_{f} \times \frac{1}{2} \rho A V^{2}=\frac{1.328}{\sqrt{\text{Re}_{L}}} \times \frac{1}{2} \rho A V^{2} \\ &=\frac{1.328}{\sqrt{\frac{\rho L V}{\mu}}} \times \frac{1}{2} \rho \times A \times L \times V^{2}\\ \text{i.e. }F_{D} &\propto \sqrt{L} \\ \end{aligned}

Now Drag force on front half,

\begin{aligned} \quad F_{D R} &\propto \sqrt{\frac{L}{2}} \\ \therefore F_{D / 2}&=\frac{F_{D}}{\sqrt{2}} \end{aligned}

Drag force on rear half,

\begin{aligned} F_{D / 2}^{\prime}&=F_{D}-F_{D / 2}=\left(1-\frac{1}{\sqrt{2}}\right) F_{D}\\ \text{Now} F&=\frac{F_{D / 2}}{F_{D / 2}^{\prime}}=\frac{\frac{F_{D}}{\sqrt{2}}}{\left(1-\frac{1}{\sqrt{2}}\right) F_{D}}=\frac{1}{\sqrt{2}-1}>1\\ \therefore \quad F&>1 \end{aligned}

Question 8 |

In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by v = u_{0}\left ( 1+\frac{3x}{L} \right )i, where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit lane of the nozzle is

\frac{L}{u_{0}} | |

\frac{L}{3u_{0}}ln4 | |

\frac{L}{4u_{0}} | |

\frac{L}{2.5u_{0}} |

Question 8 Explanation:

v=u_{o}\left(1+\frac{3 x}{L}\right)

We know that

Velocity: \quad v=\frac{d x}{d t}

\therefore \quad \frac{d x}{d t}=u_{0}\left(1+\frac{3 x}{L}\right)

\text{or }\quad u_{0}. d t=\frac{d x}{\left(1+\frac{3 x}{L}\right)}

Integrating both sides, we get

\begin{aligned} u_{0} \int_{0}^{t} d t &=\int_{0}^{L} \frac{d x}{\left(1+\frac{3 x}{L}\right)} \\ u_{0} t &=\frac{L}{3}\left[\ln \left(1+\frac{3 x}{L}\right)\right]_{0}^{L}=\frac{L}{3} \ln 4 \\ \therefore \quad t &=\frac{L}{3 u_{0}} \ln 4 \end{aligned}

Question 9 |

Consider steady laminar incompressible axi-symmetric fully developed viscous flow through a straight circular pipe of constant cross - sectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is

5 | |

1/5 | |

0 | |

\infty |

Question 9 Explanation:

Reynolds number, \mathrm{Re}=\frac{\text { Inertia force }}{\text { viscous force }}=5

Question 10 |

In a simply - supported beam loaded as shown below, the maximum bending moment in Nm is

25 | |

30 | |

35 | |

60 |

Question 10 Explanation:

Resultant beam diagram

R_{A}+R_{B}=100 \quad \cdots(1)

Taking moment about A

-100 \times 0.5-10+R_{B} \times 1=0

\therefore R_{B}=60 N

\text{and }\quad R_{A}=40 N

Bending moment diagram

Maximum bending movement

=40\times0.5+1.=30\text{Nm}

R_{A}+R_{B}=100 \quad \cdots(1)

Taking moment about A

-100 \times 0.5-10+R_{B} \times 1=0

\therefore R_{B}=60 N

\text{and }\quad R_{A}=40 N

Bending moment diagram

Maximum bending movement

=40\times0.5+1.=30\text{Nm}

There are 10 questions to complete.