Question 1 |
The limit
p=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2\pi ^2}{x-\pi+2 \sin x } \right )
has a finite value for a real \alpha . The value of \alpha and the corresponding limit p are
p=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2\pi ^2}{x-\pi+2 \sin x } \right )
has a finite value for a real \alpha . The value of \alpha and the corresponding limit p are
\alpha =-3\pi, \text{ and }p= \pi | |
\alpha =-2\pi, \text{ and }p= 2\pi | |
\alpha =\pi, \text{ and }p= \pi | |
\alpha =2\pi, \text{ and }p= 3\pi |
Question 1 Explanation:
\begin{aligned}
p&=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\
p&=\left ( \frac{\pi ^2+\alpha \pi +2 \pi ^2}{\pi-\pi+2 \sin \pi } \right ) \\
&= \frac{2 \pi ^2+\alpha \pi}{0}\\
\therefore \;\; \alpha &= -3 \pi\\
p&=\lim_{x \to \pi}\left ( \frac{x^2- 3 \pi x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\
&=\lim_{x \to \pi}\left ( \frac{2x- 3 \pi }{1+2 \cos \pi} \right ) \\
&= \frac{2 \pi-3 \pi}{1-2}=\frac{-\pi}{-1}=\pi\\
\therefore \; \alpha &=-3 \pi \text{ and }p= \pi
\end{aligned}
Question 2 |
Solution of \triangledown^2T=0 in a square domain (0 \lt x \lt 1
and 0 \lt y \lt 1) with boundary conditions:
T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is
T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is
T(x,y)=x-xy+y | |
T(x,y)=x+y | |
T(x,y)=-x+y | |
T(x,y)=x+xy+y |
Question 2 Explanation:
T(x, 0) = x \Rightarrow
option (c) is not correct.
T(0, y) = y \Rightarrow all options satisfied.
T(x, 1) = 1 + x; \Rightarrow only option (b) is satisfied.
T(1, y) = 1 + y is \Rightarrow only option (b) is satisfied.
T(0, y) = y \Rightarrow all options satisfied.
T(x, 1) = 1 + x; \Rightarrow only option (b) is satisfied.
T(1, y) = 1 + y is \Rightarrow only option (b) is satisfied.
Question 3 |
Given a function \varphi =\frac{1}{2}(x^2+y^2+z^2) in threedimensional Cartesian space, the value of the
surface integral
\oiint_{S}{\hat{n}.\triangledown \varphi dS}
where S is the surface of a sphere of unit radius and \hat{n} is the outward unit normal vector on S, is
\oiint_{S}{\hat{n}.\triangledown \varphi dS}
where S is the surface of a sphere of unit radius and \hat{n} is the outward unit normal vector on S, is
4 \pi | |
3 \pi | |
4 \pi/3 | |
0 |
Question 3 Explanation:
\begin{aligned}
\varphi &=\frac{1}{2}(x^2+y^2+z^2)\\
\triangledown \varphi &=(x\hat{i}+y\hat{j}+z\hat{k})=\bar{F}\\
\oiint_{S}(\triangledown \varphi\cdot \bar{n})dS&=\int \int _v\int Div\; \bar{F} dv\\
&=\int \int \int 3dv\\
&=3v\\
&=3\left ( \frac{4}{3} \pi \right )=4\pi
\end{aligned}
Question 4 |
The Fourier series expansion of x^3
in the interval -1\leq x\leq 1 with periodic continuation has
only sine terms | |
only cosine terms | |
both sine and cosine terms | |
only sine terms and a non-zero constant |
Question 4 Explanation:
f(x)=x^3, \;\; -1 \leq x \leq 1
It is an odd function
Fourier series contains only sine terms.
It is an odd function
Fourier series contains only sine terms.
Question 5 |
If A=\begin{bmatrix}
10 &2k+5 \\
3k-3 & k+5
\end{bmatrix} is a symmetric matrix, the
value of k is ___________.
8 | |
5 | |
-0.4 | |
\frac{1+\sqrt{1561}}{12} |
Question 5 Explanation:
A=\begin{bmatrix}
10 & 2k+5\\
3k-3 & k+5
\end{bmatrix}
(2k + 5) = (3k - 3)
k=8
(2k + 5) = (3k - 3)
k=8
Question 6 |
A uniform light slender beam AB of section modulus
EI is pinned by a frictionless joint A to the ground
and supported by a light inextensible cable CB to
hang a weight W as shown. If the maximum value
of W to avoid buckling of the beam AB is obtained
as \beta \pi ^2 EI, where \pi is the ratio of circumference to
diameter of a circle, then the value of \beta is
0.0924\; m^{-2} | |
0.0713\; m^{-2} | |
0.1261\; m^{-2} | |
0.1417\; m^{-2} |
Question 6 Explanation:
Draw FBD of AB
\Sigma M_A=0
W \times 2.5=T \sin 30^{\circ} \times 2.5
T=2W
Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W
Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}
\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})
W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI
W0.0924 \pi^2 EI=\beta \pi ^2 EI
\beta =0.0924 m^{-2}
\Sigma M_A=0
W \times 2.5=T \sin 30^{\circ} \times 2.5
T=2W
Compressive load acting on AB =T \cos 30^{\circ}=2W \times \frac{\sqrt{3}}{2}=\sqrt{3}W
Buckling happens when \sqrt{3}W=P_{cr}=\frac{\pi ^2 EI}{L_e^2}
\sqrt{3}W=\frac{\pi ^2 EI}{L^2} \;\;(\because L_e=L \text{as both ends hinged})
W=\frac{1 \times \pi ^2 EI}{\sqrt{3} \times (2.5)^2}=0.0924 \pi^2 EI
W0.0924 \pi^2 EI=\beta \pi ^2 EI
\beta =0.0924 m^{-2}
Question 7 |
The figure shows a schematic of a simple Watt
governor mechanism with the spindle O_1O_2 rotating
at an angular velocity \omega about a vertical axis. The
balls at P and S have equal mass. Assume that there
is no friction anywhere and all other components are
massless and rigid. The vertical distance between
the horizontal plane of rotation of the balls and the
pivot O_1
is denoted by h . The value of h=400 mm
at a certain \omega . If \omega is doubled, the value of h will be
_________ mm.
50 | |
100 | |
150 | |
200 |
Question 7 Explanation:
h_1
= 400 mm, h_2
= ?
\omega _1=\omega \;\;\;\omega _2=2\omega
For Watt governor,
\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}
\omega _1=\omega \;\;\;\omega _2=2\omega
For Watt governor,
\begin{aligned} h&=\frac{g}{\omega ^2}\\ h\propto \frac{1}{\omega ^2}\\ \Rightarrow h_1\omega _1^2&=h_2\omega _2^2\\ 400 \times \omega ^2&=h_2 \times (2\omega )^2\\ h_2&=100mm \end{aligned}
Question 8 |
A square threaded screw is used to lift a load W
by applying a force F. Efficiency of square threaded
screw is expressed as
The ratio of work done by W per revolution to work done by F per revolution | |
W/F | |
F/W | |
The ratio of work done by F per revolution to work done by W per revolution |
Question 8 Explanation:
\text{Screw efficiency}=\frac{\text{Work done by the applied force/rev}}{\text{Work done in lifting the load/rev}}
Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}
Efficiency depends on helix angle and friction angle.
Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}
Efficiency depends on helix angle and friction angle.
Question 9 |
A CNC worktable is driven in a linear direction by
a lead screw connected directly to a stepper motor.
The pitch of the lead screw is 5 mm. The stepper
motor completes one full revolution upon receiving
600 pulses. If the worktable speed is 5 m/minute
and there is no missed pulse, then the pulse rate
being received by the stepper motor is
20 KHz | |
10 kHz | |
3 kHz | |
15 kHz |
Question 9 Explanation:
No. of steps required for one full revolution of
stepper motor shaft or lead screws n_S= 600
Pitch (p) = 5 mm
Linear table speed V_{table}= 5 m/min = 5000 mm/min
RPM of lead Screw (N_S)= \frac{V_{table}}{p}=1000 rpm
We have equation of frequency of pulse generator
\begin{aligned} f_p&= N_s \times n_S\\ f_p&= 1000 \times 600=600,000 pulses/min\\ f_p&=\frac{600000}{60}pulses/sec\\ f_p&=10000 pulses/sec \text{ or }Hz\\ f_p&=10kHz \end{aligned}
Pitch (p) = 5 mm
Linear table speed V_{table}= 5 m/min = 5000 mm/min
RPM of lead Screw (N_S)= \frac{V_{table}}{p}=1000 rpm
We have equation of frequency of pulse generator
\begin{aligned} f_p&= N_s \times n_S\\ f_p&= 1000 \times 600=600,000 pulses/min\\ f_p&=\frac{600000}{60}pulses/sec\\ f_p&=10000 pulses/sec \text{ or }Hz\\ f_p&=10kHz \end{aligned}
Question 10 |
The type of fit between a mating shaft of diameter 25.0^{\begin{matrix}
+0.010\\
-0.010
\end{matrix}}mm and a hole of diameter 25.015^{\begin{matrix}
+0.015\\
-0.015
\end{matrix}}mm is __________.
Clearance | |
Transition | |
Interference | |
Linear |
Question 10 Explanation:
If,
D_{hole} = 25.00 mm,
D_{shaft} = 25.01 mm ( Interference fit.)
D_{hole} = 25.03 mm,
D_{shaft} = 24.99 mm
(Clearance fit)
Some of the assemblies provide clearance fit and some provides interference fit.
Hence, It is transition fit
There are 10 questions to complete.