Question 1 |
In matrix equation [A]{X}={R},
[A]=\begin{bmatrix} 4 & 8 & 4\\ 8& 16 & -4\\ 4& -4 & 15 \end{bmatrix}, \{X\}=\begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} and \{R\}=\begin{Bmatrix} 32\\ 16\\ 64 \end{Bmatrix}.
One of the eigenvalues of matrix [A] is
[A]=\begin{bmatrix} 4 & 8 & 4\\ 8& 16 & -4\\ 4& -4 & 15 \end{bmatrix}, \{X\}=\begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} and \{R\}=\begin{Bmatrix} 32\\ 16\\ 64 \end{Bmatrix}.
One of the eigenvalues of matrix [A] is
4 | |
8 | |
15 | |
16 |
Question 1 Explanation:
Given that AX=R
\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=\left[\begin{array}{l} 32 \\ 16 \\ 64 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=16\left[\begin{array}{c} 32 \\ 4 \end{array}\right](\because A X=\lambda X) \end{array}
\therefore One of eigen value of the given matrix A is given by \lambda=16
\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=\left[\begin{array}{l} 32 \\ 16 \\ 64 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=16\left[\begin{array}{c} 32 \\ 4 \end{array}\right](\because A X=\lambda X) \end{array}
\therefore One of eigen value of the given matrix A is given by \lambda=16
Question 2 |
The directional derivative of the function f(x,y)=x^2+y^2 along aline directed from (0,0) to (1,1), evaluated at the point x=1, y=1 is
\sqrt{2} | |
2 | |
2 \sqrt{2} | |
4 \sqrt{2} |
Question 2 Explanation:
Given function is f(x, y)=x^{2}+y^{2}
The direction vector a is given by
\overline{\mathrm{a}}=(1,1)-(0,0)=\overline{\mathrm{i}}+\overline{\mathrm{j}}
Let the given point be P=(x, y)=(1,1)
Now, the directional directive of f(x, y) in the direction of vector a at point is given by
\mathrm{D} . \mathrm{D}=(\nabla \mathrm{f}) \cdot \cdot \frac{\overline{\mathrm{a}}}{|\mathrm{a}|}
\Rightarrow D . D=(2 x \bar{i}+2 x \bar{j}) \cdot \frac{(\bar{i}+\bar{j})}{\sqrt{1+1}}
\therefore D . D=\frac{2+2}{\sqrt{2}}=2 \sqrt{2}
The direction vector a is given by
\overline{\mathrm{a}}=(1,1)-(0,0)=\overline{\mathrm{i}}+\overline{\mathrm{j}}
Let the given point be P=(x, y)=(1,1)
Now, the directional directive of f(x, y) in the direction of vector a at point is given by
\mathrm{D} . \mathrm{D}=(\nabla \mathrm{f}) \cdot \cdot \frac{\overline{\mathrm{a}}}{|\mathrm{a}|}
\Rightarrow D . D=(2 x \bar{i}+2 x \bar{j}) \cdot \frac{(\bar{i}+\bar{j})}{\sqrt{1+1}}
\therefore D . D=\frac{2+2}{\sqrt{2}}=2 \sqrt{2}
Question 3 |
The differential equation \frac{dy}{dx}+4y=5 is valid in the domain 0\leq x\leq 1 with y(0)=2.25. The solution of the differential equation is
y=e^{-4x}+5 | |
y=e^{-4x}+1.25 | |
y=e^{4x}+5 | |
y=e^{4x}+1.25 |
Question 3 Explanation:
\begin{array}{l} \text { Given } \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}=5,0 \leq \mathrm{x} \leq 1 \; \; \ldots(1) \\ \because \frac{d y}{d x}+P(x, y)=Q(x) \\ \text { With } \mathrm{y}(0)=2.25\;\;\; \ldots(2) \\ \text { Here, I.F }=\mathrm{e}^{\int 4 \mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}} \end{array}
The general solution of (1) is given by
y \cdot e^{4 x}=\int(5)\left(e^{4 x}\right) d x+c
\Rightarrow \mathrm{y} \cdot \mathrm{e}^{4 \mathrm{x}}=\frac{5}{4} \mathrm{e}^{4 \mathrm{x}}+\mathrm{c}\cdots(3)
Using (2) and (3)
(2.25)(1)=\left(\frac{5}{4}\right)(1)+c
c=1 \cdots(4)
The solution of (1) from (3) and (4) is
y \cdot e^{4 x}=\frac{5}{4} e^{4 x}+1
or y=\frac{5}{4}+e^{4 x}=e^{-4 x}+1.25
The general solution of (1) is given by
y \cdot e^{4 x}=\int(5)\left(e^{4 x}\right) d x+c
\Rightarrow \mathrm{y} \cdot \mathrm{e}^{4 \mathrm{x}}=\frac{5}{4} \mathrm{e}^{4 \mathrm{x}}+\mathrm{c}\cdots(3)
Using (2) and (3)
(2.25)(1)=\left(\frac{5}{4}\right)(1)+c
c=1 \cdots(4)
The solution of (1) from (3) and (4) is
y \cdot e^{4 x}=\frac{5}{4} e^{4 x}+1
or y=\frac{5}{4}+e^{4 x}=e^{-4 x}+1.25
Question 4 |
An analytic function f(z) of complex variable z=x+iy may be written as f(z)=u(x,y)+iv(x,y). Then u(x,y) and v(x,y) must satisfy
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} |
Question 4 Explanation:
Given that the complex function f(z)=u(x,y)+ i v(x,y) is an analytic function.
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
Question 5 |
A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are V_P \; and \; V_R, in the x and y directions, respectively. The magnitude of the angular velocity of the body is
2V_R | |
2V_P | |
V_R/\sqrt{3} | |
V_P/\sqrt{3} |
Question 5 Explanation:
\begin{array}{l} \Rightarrow \mathrm{V}_{\mathrm{R}}=(\mathrm{IR}) \omega \\ \Rightarrow \omega=\frac{\mathrm{V}_{\mathrm{R}}}{(\mathrm{IR})} \\ \Rightarrow \omega \times \frac{\mathrm{V}_{\mathrm{R}}}{\frac{1}{2}} \\ \Rightarrow \omega=2 \mathrm{V}_{\mathrm{R}} \end{array}
Question 6 |
Consider a linear elastic rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume plane stress conditions in the plane normal to the thickness. The Young's modulus E=200 MPa and Poisson's ratio v=0.3 are given. The principal strains in the plane of the sheet are
(0.35, -0.15) | |
(0.5, 0.0) | |
(0.5, -0.15) | |
(0.5, -0.5) |
Question 6 Explanation:
\begin{aligned} \sigma_{\mathrm{x}}&=100 \mathrm{MPa} \\ v&=\mu=0.3 \\ \sigma_{\mathrm{y}}&=0, \sigma_{\mathrm{z}}=0, \mathrm{E}=200 \mathrm{MPa} \\ &\text { Principal strain in x-direction } \\ &=\epsilon_{1}=\epsilon_{\mathrm{x}}=\frac{\sigma_{\mathrm{x}}}{\mathrm{E}}-\mu \frac{\sigma_{\mathrm{y}}}{\mathrm{E}} \\ &=\frac{100}{200}-0=0.5\\ &\text { Principal strain in y-direction } \\ &=\epsilon_{2}=\epsilon_{\mathrm{y}}=\frac{\sigma_{\mathrm{y}}}{\mathrm{E}}-\mu \frac{\sigma_{\mathrm{x}}}{\mathrm{E}}\\ &=0-(0.3)\left(\frac{100}{200}\right)=-0.15\\ &\therefore\left(\epsilon_{\mathrm{x}}, \epsilon_{\mathrm{y}}\right)=(0.5-0.15) \end{aligned}
Question 7 |
A spur gear has pitch circle diameter D and number of teeth T. The circular pitch of the gear is
\frac{\pi D}{T} | |
\frac{T}{D} | |
\frac{D}{T} | |
\frac{2 \pi D}{T} |
Question 7 Explanation:
Circular pitch : It is the distance between two similar points on adjacent teeth measured along pitch
circle circumference circular pitch
\left(P_{c}\right)=\frac{\text { Pitch circlecircum }}{\text { Number of teeth }}=\frac{\pi D}{T}
circle circumference circular pitch
\left(P_{c}\right)=\frac{\text { Pitch circlecircum }}{\text { Number of teeth }}=\frac{\pi D}{T}
Question 8 |
Endurance limit of a beam subjected to pure bending decreases with
decrease in the surface roughness and decrease in the size of the beam | |
increase in the surface roughness and decrease in the size of the beam | |
increase in the surface roughness and increase in the size of the beam | |
decrease in the surface roughness and increase in the size of the beam |
Question 8 Explanation:
Endurance limit decreases with increase in surface roughness and with increase in size of the
beam.
Question 9 |
A two-dimensional incompressible frictionless flow field is given by \vec{u}=x\hat{i}-y\hat{j}. If \rho is the density of the fluid, the expression for pressure gradient vector at any point in the flow field is given as
\rho (x\hat{i}+y\hat{j}) | |
-\rho (x\hat{i}+y\hat{j}) | |
\rho (x\hat{i}-y\hat{j}) | |
-\rho (x^2 \hat{i} + y^2 \hat{j}) |
Question 9 Explanation:
Given, 2-D incompressible frictionless fluid flow.
\overrightarrow{\mathrm{u}}=x \hat{\mathrm{i}}-y \hat{\mathrm{j}}
Thus, velocity components in x and y directions are :
\mathrm{u}=\mathrm{x} \text { and } \mathrm{v}=-\mathrm{y}
Navier-Stokes equation for incompressible, frictionless fluid flow reduces to
\rho \frac{\mathrm{DV}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}+\overrightarrow{\rho \mathrm{g}}
There are no components of body force in x and y direction. Hence,
\rho \frac{\mathrm{D} \overrightarrow{\mathrm{V}}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}
where, \nabla \overrightarrow{\mathrm{P}} is the pressure gradient vector
Hence,
\begin{aligned} \bigtriangledown \vec{P}&=\rho \left [ \left ( u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right )\hat{i}+\left ( u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\right ) \hat{j}\right ]\\ &=-\rho \left [ \{ x(1)+(-y)(0)\}\hat{i}+\{ x(0)+(-y)(-1)\} \hat{j}\right ]\\ &=-\rho (x\hat{i}+y\hat{j}) \end{aligned}
\overrightarrow{\mathrm{u}}=x \hat{\mathrm{i}}-y \hat{\mathrm{j}}
Thus, velocity components in x and y directions are :
\mathrm{u}=\mathrm{x} \text { and } \mathrm{v}=-\mathrm{y}
Navier-Stokes equation for incompressible, frictionless fluid flow reduces to
\rho \frac{\mathrm{DV}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}+\overrightarrow{\rho \mathrm{g}}
There are no components of body force in x and y direction. Hence,
\rho \frac{\mathrm{D} \overrightarrow{\mathrm{V}}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}
where, \nabla \overrightarrow{\mathrm{P}} is the pressure gradient vector
Hence,
\begin{aligned} \bigtriangledown \vec{P}&=\rho \left [ \left ( u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right )\hat{i}+\left ( u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\right ) \hat{j}\right ]\\ &=-\rho \left [ \{ x(1)+(-y)(0)\}\hat{i}+\{ x(0)+(-y)(-1)\} \hat{j}\right ]\\ &=-\rho (x\hat{i}+y\hat{j}) \end{aligned}
Question 10 |
Sphere-1 with a diameter of 0.1 m is completely enclosed by another sphere-2 of diameter 0.4 m. The view factor F_{12} is
0.0625 | |
0.25 | |
0.5 | |
1 |
Question 10 Explanation:
Given data:
\mathrm{d}_{1}=0.1 \mathrm{m}
\mathrm{d}_{2}=0.4 \mathrm{m}
\mathrm{F}_{1-1}=0 (from the geometry)
From the additive rule
\mathrm{F}_{1-1}+\mathrm{F}_{1-2}=1
\mathrm{F}_{1-2}=1
There are 10 questions to complete.