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}

Question 6 |

For an opaque surface, the absorptivity (\alpha), transmissivity (\tau) and reflectivity (\rho) are related by the equation:

\alpha +\rho =\tau | |

\rho +\alpha +\tau =0 | |

\alpha +\rho =1 | |

\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

\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

6.5 | |

8.9 | |

9.1 | |

27 |

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

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

1200 | |

900 | |

875 | |

720 |

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}

\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

zero | |

r\alpha | |

\sqrt{(r\alpha )^{2}+(r\omega ^{2})^{2}} | |

r\omega |

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

0 | |

1 | |

1.08 | |

2.02 |

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}

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.