GATE Mechanical Engineering 2019 SET-1

Question 1
Consider the matrix
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
A
0
B
1
C
2
D
3
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\text { Given: } A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
Question 2
A parabola x=y^2 \; with \; 0\leq x\leq 1 is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360^{\circ} around the x-axis is
A
\frac{\pi}{4}
B
\frac{\pi}{2}
C
\pi
D
2 \pi
Engineering Mathematics   Calculus
Question 2 Explanation: 
\text { Given: } y^{2}=x, 0 \leq x \leq 1
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
Question 3
For the equation \frac{dy}{dx} + 7x^2 y=0, if y(0)=3/7, then the value of y(1)is
A
\frac{7}{3}e^{-7/3}
B
\frac{7}{3}e^{-3/7}
C
\frac{3}{7}e^{-7/3}
D
\frac{3}{7}e^{-3/7}
Engineering Mathematics   Differential Equations
Question 3 Explanation: 
Given \frac{d y}{d x}+7 x^{2} y=0 \ldots(1)
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
Question 4
The lengths of a large stock of titanium rods follow a normal distribution with a mean (\mu) of 440 mm and a standard deviation (\sigma) of 1mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
A
81.85%
B
68.40%
C
99.75%
D
86.64%
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 



\begin{array}{l} \mathrm{Z}=\frac{x-\mu}{\sigma} \\ \mathrm{Z}(\mathrm{x}=438)=\frac{438-440}{1}=-2 \\ \mathrm{P}(\mathrm{Z}=-2)=2.28 \% \\ \mathrm{Z}(\mathrm{x}=441)=\frac{441-440}{1}=1 \\ \mathrm{P}(\mathrm{Z}=1)=84.13 \% \\ \end{array}
The percentage of rods whose lengths lie between 438 mm and 441 mm =
\begin{array}{c} =\mathrm{P}(\mathrm{Z}=1)-\mathrm{P}(\mathrm{Z}=-2) \\ =84.13 \%-2.28 \%=81.85 \% \end{array}
Question 5
A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity \omega. At time t=0, the vertical position of the followeris y(0)=0, and the system is in the configuration shown below.

The vertical position of the follower face, y(t) is given by
A
e \sin \omega t
B
e(1+ \cos 2\omega t)
C
e(1- \cos \omega t)
D
e \sin 2 \omega t
Theory of Machine   Cams
Question 5 Explanation: 


\begin{aligned} \mathrm{AA}_{1}=\mathrm{y} &=\mathrm{e}(1-\cos \theta) \\ &=\mathrm{e}(1-\cos \omega \mathrm{t}) \end{aligned}
Question 6
The natural frequencies corresponding to the spring-mass systems I and II are \omega _I \; and \; \omega _{II}, respectively. The ratio \frac{\omega _I}{\omega _{II}} is
A
\frac{1}{4}
B
\frac{1}{2}
C
2
D
4
Theory of Machine   Vibration
Question 6 Explanation: 
System: I
k_{e q}=\frac{k \cdot k}{k+k}=\frac{k}{2}
\omega_{\mathrm{n}_{1}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}
System - II
\mathrm{k}_{\mathrm{eq}}=2 \mathrm{k} \quad \omega_{\mathrm{n}_{2}}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}
\frac{\omega_{\mathrm{n}_{1}}}{\omega_{\mathrm{n}_{2}}}=\frac{\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}}}{\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}} \times \frac{\mathrm{m}}{2 \mathrm{k}}}=\frac{1}{2}
Question 7
A spur gear with 20^{\circ} full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is
A
0.36 kN
B
0.73 kN
C
1.39 kN
D
2.78kN
Theory of Machine   Gear and Gear Train
Question 7 Explanation: 
\begin{array}{l} \phi=20^{\circ}, P=20 k W, \omega=200 \mathrm{rad} / \mathrm{s}, d=100 \mathrm{mm}=0.1 \mathrm{m} \\ \text { Torque }=\text { Power } / \omega \\ \mathrm{T}=\frac{20000}{200}=100 \mathrm{Nm} \\ \text { Now, } \mathrm{T}=\mathrm{F}_{\mathrm{T}} \times \frac{\mathrm{d}}{2} \\ \Rightarrow 100=\mathrm{F}_{\mathrm{T}} \times \frac{0.1}{2} \\ \Rightarrow \mathrm{F}_{\mathrm{T}}=2000 \mathrm{N}\\ \frac{F_{R}}{F_{T}}=\tan \phi \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=2000 \times \tan 20^{\circ} \\ \Rightarrow \mathrm{F}_{\mathrm{R}}=727.94 \mathrm{N}=0.73 \mathrm{kN} \end{array}
Question 8
During a non-flow thermodynamic process(1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q_{1-2}=W_{1-2}) when the process is
A
Isentropic
B
Polytropic
C
Isothermal
D
Adiabatic
Thermodynamics   Thermodynamic System and Processes
Question 8 Explanation: 
Given, \mathrm{Q}_{1-2}=\mathrm{W}_{1-2}
\therefore \Delta \mathrm{U}_{1-2}=0
\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0
\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}
So, the process is isothermal.
Question 9
For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross-section, the Nusselt number at constant wall heat flux (Nu_q) and that at constant wall temperature (Nu_T) are related as
A
Nu_q \gt Nu_T
B
Nu_q \lt Nu_T
C
Nu_q = Nu_T
D
Nu_q = ( Nu_T)^2
Heat Transfer   Heat Transfer in Flow Over Plates and Pipes
Question 9 Explanation: 
For laminar flow through circular tube:
\mathrm{Nu}_{\mathrm{q}}=4.36 (For constant heat flux)
\mathrm{Nu}_{\mathrm{T}}=3.66 (For constant wall temperature)
\mathrm{Nu}_{\mathrm{q}} \gt \mathrm{Nu}_{\mathrm{T}}
Question 10
As per common design practice, the three types of hydraulic turbines, in descending order of flow rate, are
A
Kaplan, Francis, Pelton
B
Pelton, Francis, Kaplan
C
Francis, Kaplan, Pelton
D
Pelton, Kaplan, Francis
Fluid Mechanics   Turbines and Pumps
Question 10 Explanation: 
Kaplan turbine has highest flow area hence it can handle highest discharge. On the other hand, Pelton turbine has lowest flow area hence it works on low discharge.
\therefore \mathrm{Q}_{\mathrm{Kaplan}} \gt \mathrm{Q}_{\mathrm{Francis}}>\mathrm{Q}_{\mathrm{Pclton}}
There are 10 questions to complete.

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