Question 1 |
In abrasive jet machining, as the distance between the nozzle tip and the work surface increases, the material removal rate
Increases continuously | |
Decreases continuously | |
Decreases, becomes stable and then increases | |
Increases, becomes stable and then decreases |
Question 1 Explanation:
In abrasive jet machining the variation is as shown
MRR = Metal removal rate
NTD = Nozzle tip distance
MRR = Metal removal rate
NTD = Nozzle tip distance
Question 2 |
Match the following metal forming processes with their associated stresses in the workpiece.
1-S, 2-P, 3-Q, 4-R | |
1-S, 2-P, 3-R, 4-Q | |
1-P, 2-Q, 3-S, 4-R | |
1-P, 2-R, 3-Q, 4-S |
Question 2 Explanation:
1. Coining - Compressive
2. Wire drawing - Tensile
3. Blanking - Shear
4. Deep drawing-Tensile and 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
40 | |
30 | |
20 | |
10 |
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}
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
Between the upper and lower critical temperature and cooled in still air | |
Above the upper critical temperature and cooled in furnace | |
Above the upper critical temperature and cooled in still air | |
Between the upper and lower critical temperature and cooled in furnace |
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})
116.18 | |
0.116 | |
18.22 | |
232.36 |
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}
\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}
There are 5 questions to complete.