# 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 $\alpha +\rho =0$
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. 