GATE ME 2012

Question 1
In abrasive jet machining, as the distance between the nozzle tip and the work surface increases, the material removal rate
A
Increases continuously
B
Decreases continuously
C
Decreases, becomes stable and then increases
D
Increases, becomes stable and then decreases
Manufacturing Engineering   Machining and Machine Tool Operation
Question 1 Explanation: 
In abrasive jet machining the variation is as shown

MRR = Metal removal rate
NTD = Nozzle tip distance
Question 2
Match the following metal forming processes with their associated stresses in the workpiece.
A
1-S, 2-P, 3-Q, 4-R
B
1-S, 2-P, 3-R, 4-Q
C
1-P, 2-Q, 3-S, 4-R
D
1-P, 2-R, 3-Q, 4-S
Manufacturing Engineering   Forming Process
Question 2 Explanation: 
1. Coining - Compressive
2. Wire drawing - Tensile
3. Blanking - Shear
4. Deep drawing-Tensile and compressive
Question 3
In an interchangeable assemble, shafts of size 25.000^{_{-0.010}^{+0.040}} mm mate with holes of size 25.000^{_{-0.020}^{+0.030}} mm. The maximum interference (in microns) in the assembly is
A
40
B
30
C
20
D
10
Manufacturing Engineering   Metrology and Inspection
Question 3 Explanation: 
\begin{array}{l} \text { Size of shaft }=25.000^{\overset{+0.040}{-0.010}} \mathrm{mm} \\ \text { Size of hole }=25.000^{\overset{+0.030}{+0.020}} \mathrm{mm} \end{array}

Maximum interference
=0.040-0.020=0.020 \mathrm{mm}=20 \mu \mathrm{m}
Question 4
During normalizing process of steel, the specimen is heated
A
Between the upper and lower critical temperature and cooled in still air
B
Above the upper critical temperature and cooled in furnace
C
Above the upper critical temperature and cooled in still air
D
Between the upper and lower critical temperature and cooled in furnace
Manufacturing Engineering   Engineering Materials
Question 4 Explanation: 
During normalizing process of steel. The specimen in heated upto 50^{\circ} \mathrm{C} above the upper critical temperature and is then cooled is still air.
Question 5
Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m. The volumetric flow rate is 0.2 m^{3}/s . The head loss (in m) due to friction is (assume g = 9.81 m/s^{2})
A
116.18
B
0.116
C
18.22
D
232.36
Fluid Mechanics   Flow Through Pipes
Question 5 Explanation: 
Given data:
\begin{aligned} d=200 \mathrm{mm}&=0.2 \mathrm{m} \\ \therefore \qquad A=\frac{\pi}{4} d^{2}&=\frac{3.14}{4} \times(0.2)^{2}\\ &=0.0314 \mathrm{m}^{2} \\ f &=0.0225 \\ L &=500 \mathrm{m} \\ Q &=0.2 \mathrm{m}^{3} / \mathrm{s} \\ also \qquad Q &=A V\\ \therefore \qquad 0.2&=0.0314 \times V\\ or \qquad V&=6.369 \mathrm{m} / \mathrm{s} \end{aligned}
We know that the Darcy's formula, head loss due to friction,
\begin{aligned} h_{f} &=\frac{f L V^{2}}{2 g d} \\ &=\frac{0.0225 \times 500 \times(6.369)^{2}}{2 \times 9.81 \times 0.2} \\ &=116.29 \mathrm{m} \text { of oil } \end{aligned}
Question 6
For an opaque surface, the absorptivity (\alpha), transmissivity (\tau) and reflectivity (\rho) are related by the equation:
A
\alpha +\rho =\tau
B
\rho +\alpha +\tau =0
C
\alpha +\rho =1
D
\alpha +\rho =0
Heat Transfer   Radiation
Question 6 Explanation: 
For any surface
\alpha+\tau+\rho=1
But for opaque surface, transmissivity (\tau)=0
\therefore \quad \alpha+\rho=1
Question 7
Steam enters an adiabatic turbine operating at steady state with an enthalpy of 3251.0kJ/kg and leaves as a saturated mixture at 15 kPa with quality (dryness fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are h_f=225.94kJ/kg and h_g=2598.3 kJ/kg respectively. The mass flow rate of steam is 10kg/s. Kinetic and potential energy changes are negligible. The power output of the turbine in MW is
A
6.5
B
8.9
C
9.1
D
27
Thermodynamics   Power System
Question 7 Explanation: 
\text{Power}=m_r \times (h_1 -h_2)=10 \times (3251-(225.94+0.9 \times(2598.3-225.94)))=8900\; KJ/S=8.9\; MW
Question 8
The following are the data for two crossed helical gears used for speed reduction:
Gear I: pitch circle diameter in the plane of rotation 80mm and helix angle 30^{\circ}
Gear II: pitch circle diameter in the plane of rotation 120mm and helix angle 22.5^{\circ}
If the input speed is 1440 rpm. The output speed in rpm is
A
1200
B
900
C
875
D
720
Theory of Machine   Gear and Gear Train
Question 8 Explanation: 
For helical gear
\begin{aligned} \text { Velocity ratio } &=\frac{N_{2}}{N_{1}}=\frac{D_{1} \cos \phi}{D_{2} \cos \phi} \\ &=\frac{80 \cos 30^{\circ}}{120 \cos 22.5^{\circ}}\\ \therefore \quad N_{2} &=1440 \times 0.625 \\ &=900 \mathrm{rpm} \end{aligned}
Question 9
A solid disk of radius r rolls without slipping on a horizontal floor with angular velocity \omega and angular acceleration \alpha. The magnitude of the acceleration of the point of contact on a disc is
A
zero
B
r\alpha
C
\sqrt{(r\alpha )^{2}+(r\omega ^{2})^{2}}
D
r\omega
Engineering Mechanics   Plane Motion
Question 9 Explanation: 


Motion of body can be expressed as sum of pure rotational and pure translation.
For no slip condition.
Translational acceleration of centre of mass (a)
=\alpha r
Velocity of centre of mass (v)=\omega r
Pure rotational II Pure translation

Now superimposing the two motions.

Linear acceleration of bottom point
=\alpha r-\alpha r+r \omega^{2}
Velocity of bottom point =\omega r-\omega r=0
\therefore Net acceleration of bottom point =r \omega^{2}
Question 10
A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is
A
0
B
1
C
1.08
D
2.02
Strength of Materials   Thin Cylinder
Question 10 Explanation: 
Hoop stress =\frac{p d}{4 t}
p= internal pressure
d= diameter of shell
t= thickness of shell
\begin{array}{l} \text { Now } \sigma=\frac{p d^{\prime}}{4 t^{\prime}} \\ \begin{aligned} d^{\prime}=1.01 d &\; \text { and } t^{\prime}=0.99 t \\ \therefore \quad \sigma^{\prime} &=\frac{p \times 1.01 d}{4 \times 0.99 t}=1.0202 \sigma \\ \text { \% change } &=\frac{\sigma^{\prime}-\sigma}{\sigma} \times 100 \\ &=\frac{(1.0202-1) \sigma}{\sigma} \times 100=2.02 \% \end{aligned} \end{array}
There are 10 questions to complete.

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