Question 1 |

A 76.2 mm gauge block is used under one end of a 254 mm sine bar with roll diameter of 25.4 mm. The height of gauge blocks required at the other end of the sine bar to measure an angle of 30^{\circ} is __________mm (round off to two decimal places).

203.2 | |

145.6 | |

214.3 | |

542.6 |

Question 1 Explanation:

\begin{aligned} \sin 30^{\circ} &=\frac{h_{1}}{254} \\ \text{or}\qquad h_{1} &=254 \sin 30^{\circ}=127 \mathrm{~mm} \\ H_{2} &=76.2+127=203.2 \mathrm{~mm} \end{aligned}\\ Case II:

\begin{aligned} \sin 30^{\circ} &=\frac{76.2-h_{1}}{254} \\ \text{or}\qquad h_{1} &=76.2-254 \sin 30^{\circ}\\ &=-50.8 \mathrm{~mm} \end{aligned}

Which is not feasible

Therefore final answer 203.2 mm.

Question 2 |

The thickness, width and length of a metal slab are 50 mm, 250 mm and 3600 mm, respectively. A rolling operation on this slab reduces the thickness by 10% and increases the width by 3%. The length of the rolled slab is ________mm (round off to one decimal place).

3254.2 | |

2453.6 | |

4521.3 | |

3883.5 |

Question 2 Explanation:

\begin{array}{l} h_{1}=50 \mathrm{~mm} ; \quad h_{2}=0.9 h_{1} ; \\ b_{1}=250 \mathrm{~mm} ; \quad b_{2}=1.03 b_{1} ; \\ L_{1}=3600 \mathrm{~mm} ; \quad L_{2}=? \end{array}

Volume remains const. in theory of plasticity

\begin{aligned} h_{1} b_{1} L_{1} &=h_{2} b_{2} L_{2} \\ &=0.9 h_{1} \times 1.03 b_{1} \times L_{2} \\ L_{2} &=\frac{3600}{0.9 \times 1.03}=3883.5 \mathrm{~mm}=3883.5 \mathrm{~mm} \end{aligned}

Volume remains const. in theory of plasticity

\begin{aligned} h_{1} b_{1} L_{1} &=h_{2} b_{2} L_{2} \\ &=0.9 h_{1} \times 1.03 b_{1} \times L_{2} \\ L_{2} &=\frac{3600}{0.9 \times 1.03}=3883.5 \mathrm{~mm}=3883.5 \mathrm{~mm} \end{aligned}

Question 3 |

In a pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has been computed to be 400 N. The cutting velocity, V_c=100 \; m/min, depth of cut, t=2.2 \; mm, feed, s_0=0.1 \; mm/revolution and chip velocity, V_f=20 \; m/min, the shear strength, \tau _s of the material will be _______MPa (round off to two decimal places).

458.25 | |

392.23 | |

254.36 | |

214.36 |

Question 3 Explanation:

\begin{aligned} \alpha &=0^{\circ} \\ F_{s} &=400 \mathrm{~N} \\ \text { Cutting velocity }(\mathrm{V}) &=100 \mathrm{~m} / \min \left(V_{c}\right) \\ d &=2.0 \mathrm{~mm}(t) \\ f &=0.1 \mathrm{~mm} / \mathrm{rev}\left(S_{o}\right) \\ \text { Chip velocity }\left(V_{c}\right) &=20 \mathrm{~m} / \min \quad\left(V_{f}\right) \\ r &=\frac{t}{t_{c}}=\frac{l_{c}}{l}=\frac{V_{c}}{V} \quad\left(V_{c}=\text { chip velocity; } V=\right.\text { cutting velocity) }\\ &=\frac{20}{100}=0.2\\ \tan \phi &=\frac{r \cos \alpha}{1-r \sin \alpha}=\frac{0.2 \cos 0^{\circ}}{1-0.2 \sin 0^{\circ}}=0.2 \\ \phi &=11.31^{\circ} \\ & \frac{F_{S}}{A_{S}}=\frac{F_{S}}{\left(\frac{b t}{\sin \phi}\right)}=\frac{F_{S} \sin \phi}{f d} \\ t &=f \sin 90^{\circ} \\ b &=d \sin 90^{\circ} \\ b t &=f d \\ &=\frac{400 \sin 11.31^{\circ}}{0.1 \times 2.0} \mathrm{~N} / \mathrm{mm}^{2} \\ &=392.23 \mathrm{MPa} \end{aligned}

Question 4 |

A surface grinding operation has been performed on a Cast Iron plate having dimensions 300 mm (length) x 10 mm (width) x 50 mm (height). The grinding was performed using an alumina wheel having a wheel diameter of 150 mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table speed is 5 m/min, depth of cut per pass is 50 \mu m and the number of grinding passes is 20. The average tangential and average normal force for each pass is found to be 40 N and 60 N respectively. The value of the specific grinding energy under the aforesaid grinding conditions is __________J/mm^3 (round off to one decimal place).

24.8 | |

32.6 | |

38.4 | |

48.2 |

Question 4 Explanation:

\begin{aligned} \text { Power } &=F_{c} \cdot V \\ &=40 \mathrm{~N} \times 40 \mathrm{~m} / \mathrm{s}=1600 \mathrm{~W} \\ \mathrm{MRR} &=10 \times 0.050 \times \frac{5000}{60} \mathrm{~mm}^{3} / \mathrm{s} \\ &=41.667 \mathrm{~mm}^{3} / \mathrm{s} \end{aligned}

e=\frac{\text { Power }}{\text { MRR }}=\frac{1600 \mathrm{~J} / \mathrm{s}}{41.667 \mathrm{~mm}^{3} / \mathrm{s}}=38.40 \mathrm{~J} / \mathrm{mm}^{3}

e=\frac{\text { Power }}{\text { MRR }}=\frac{1600 \mathrm{~J} / \mathrm{s}}{41.667 \mathrm{~mm}^{3} / \mathrm{s}}=38.40 \mathrm{~J} / \mathrm{mm}^{3}

Question 5 |

A spot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5 mm and a thickness of 1 mm. The welding time was 0.1 s. The melting energy for the steel is 20 J/mm^3. Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding operation is _________kW (round off to two decimal places).

30.5 | |

39.25 | |

45.75 | |

55.75 |

Question 5 Explanation:

Heat transfer efficiency =10\%

Energy required to melt =20 \mathrm{~J} / \mathrm{mm}^{3}

Diameter of nugget =5 \mathrm{~mm}

Thickness =1 \mathrm{~mm}

Time =0.1 \mathrm{~second}

Heat required to melt, Q=20 \mathrm{~J} / \mathrm{mm}^{3} \times \frac{\pi}{4}(5)^{2} \times 1=392.5 \mathrm{~J}

\begin{aligned} \text { Power } &=\frac{\text { Heat required to melt }}{\text { Time }} \\ P &=\frac{392.5}{0.1} \mathrm{~J} / \mathrm{s} \\ P &=3925 \mathrm{~J} / \mathrm{s}=3925 \mathrm{~W} \\ \text { Actual power supplied } &=\frac{3925}{\eta_{\text {th }}} \\ \eta_{\text {th }} &=10 \% \\ P &=\frac{3925}{0.1}=39250 \mathrm{~W}=39.25 \mathrm{~kW} \end{aligned}

Energy required to melt =20 \mathrm{~J} / \mathrm{mm}^{3}

Diameter of nugget =5 \mathrm{~mm}

Thickness =1 \mathrm{~mm}

Time =0.1 \mathrm{~second}

Heat required to melt, Q=20 \mathrm{~J} / \mathrm{mm}^{3} \times \frac{\pi}{4}(5)^{2} \times 1=392.5 \mathrm{~J}

\begin{aligned} \text { Power } &=\frac{\text { Heat required to melt }}{\text { Time }} \\ P &=\frac{392.5}{0.1} \mathrm{~J} / \mathrm{s} \\ P &=3925 \mathrm{~J} / \mathrm{s}=3925 \mathrm{~W} \\ \text { Actual power supplied } &=\frac{3925}{\eta_{\text {th }}} \\ \eta_{\text {th }} &=10 \% \\ P &=\frac{3925}{0.1}=39250 \mathrm{~W}=39.25 \mathrm{~kW} \end{aligned}

Question 6 |

A cast product of a particular material has dimensions 75 mm x 125 mm x 20 mm. The total solidification time for the cast product is found to be 2.0 minutes as calculated using Chvorinov's rule having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a cylinder having diameter = 50 mm and height = 50 mm, the total solidification time will be __________minutes (round off to two decimal places).

1.25 | |

2.83 | |

2.48 | |

3.12 |

Question 6 Explanation:

Casting dimensions =75 \mathrm{~mm} \times 125 \mathrm{~mm} \times 20 \mathrm{~mm}

Total solidification time \left(t_{s}\right)_{\mathrm{cu}}=2.0 \mathrm{~min}.

Cylindrical casting =\mathrm{H}=\mathrm{D}=50 \mathrm{~min}

Total solidification Time \left(t_{s}\right)_{c y}=? Chvorinov's rule,

\begin{aligned} t_{s} &\propto\left(\frac{V}{A}\right)^{n} \quad n=2 \\ \frac{\left(t_{s}\right)_{c u}}{\left(t_{s}\right)_{c y}}&=\frac{\left(\frac{V}{A}\right)_{c u}^{2}}{\left(\frac{V}{A}\right)_{c y}^{2}}\\ \left(\frac{V}{A}\right)_{c y} &=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2} h+\pi d h} \\ h &=d \\ \left(\frac{V}{A}\right)_{c y} &=\frac{d}{6} \\ \frac{\left(t_{s}\right)_{c U}}{\left(t_{s}\right)_{c_{y}}} &=\frac{\left[\frac{75 \times 125 \times 20}{2(75 \times 125+125 \times 20+20 \times 75)}\right]^{2}}{\left(\frac{50}{6}\right)^{2}} \\ \frac{2}{\left(t_{s}\right)_{c y}} &=\frac{49.131}{69.44}=0.7075 \\ \left(t_{s}\right)_{c y} &=2.83 \mathrm{~min} . \end{aligned}

Total solidification time \left(t_{s}\right)_{\mathrm{cu}}=2.0 \mathrm{~min}.

Cylindrical casting =\mathrm{H}=\mathrm{D}=50 \mathrm{~min}

Total solidification Time \left(t_{s}\right)_{c y}=? Chvorinov's rule,

\begin{aligned} t_{s} &\propto\left(\frac{V}{A}\right)^{n} \quad n=2 \\ \frac{\left(t_{s}\right)_{c u}}{\left(t_{s}\right)_{c y}}&=\frac{\left(\frac{V}{A}\right)_{c u}^{2}}{\left(\frac{V}{A}\right)_{c y}^{2}}\\ \left(\frac{V}{A}\right)_{c y} &=\frac{\frac{\pi}{4} d^{2} h}{2 \frac{\pi}{4} d^{2} h+\pi d h} \\ h &=d \\ \left(\frac{V}{A}\right)_{c y} &=\frac{d}{6} \\ \frac{\left(t_{s}\right)_{c U}}{\left(t_{s}\right)_{c_{y}}} &=\frac{\left[\frac{75 \times 125 \times 20}{2(75 \times 125+125 \times 20+20 \times 75)}\right]^{2}}{\left(\frac{50}{6}\right)^{2}} \\ \frac{2}{\left(t_{s}\right)_{c y}} &=\frac{49.131}{69.44}=0.7075 \\ \left(t_{s}\right)_{c y} &=2.83 \mathrm{~min} . \end{aligned}

Question 7 |

The allowance provided in between a hole and a shaft is calculated from the difference between

lower limit of the shaft and the upper limit of the hole | |

upper limit of the shaft and the upper limit of the hole | |

upper limit of the shaft and the lower limit of the hole | |

lower limit of the shaft and the lower limit of the hole |

Question 7 Explanation:

It is minimum clearance or maximum interference. It is the intentional difference between the basic dimensions of the mating parts. The allowance may be positive or negative.

Question 8 |

The machining process that involves ablation is

Abrasive Jet Machining | |

Chemical Machining | |

Electrochemical Machining | |

Laser Beam Machining |

Question 8 Explanation:

Laser beam machining (LBM) is a nonconventional machining process, which broadly refers to the process of material removal, accomplished through the interactions between the laser and target materials. The processes can include laser drilling, cutting, grooving, writing, scribing, ablation, welding, cladding, milling, and so on. LBM is a thermal process, and unlike conventional mechanical processes, LBM removes material without mechanical engagement. In general, the workpiece is heated to melting or boiling point and removed by melt ejection, vaporization, or ablation.

Question 9 |

In a CNC machine tool, the function of an interpolator is to generate

signal for the lubrication pump during machining | |

error signal for tool radius compensation during machining | |

NC code from the part drawing during post processing | |

reference signal prescribing the shape of the part to be machined |

Question 9 Explanation:

In contouring systems the machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify the coordinates of the initial and final points of each segment, and the feed rate. The operation of producing the required shape based on this information is termed interpolation and the corresponding unit is the "interpolator". The interpolator coordinates the motion along the machine axes, which are separately driven, by providing reference positions instant by instant for the position-and velocity control loops, to generate the required machining path. Typical interpolators are capable of generating linear and circular paths.

Question 10 |

The size distribution of the powder particles used in Powder Metallurgy process can be determined by

Laser scattering | |

Laser reflection | |

Laser absorption | |

Laser penetration |

Question 10 Explanation:

Particle Size, Shape, and Distribution:

Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:

1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.

2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.

3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.

4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.

5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:

1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.

2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.

3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.

4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.

5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

There are 10 questions to complete.