Question 1 
Which one of the following CANNOT impart linear
motion in a CNC machine?
Linear motor  
Ball screw  
Lead screw  
Chain and sprocket 
Question 1 Explanation:
Chain and Sprocket mechanism is not used in CNC
Machines.
Linear Motors Recently linear motors are being increasingly considered for use in high performance CNC machine tools. The linear motor consists of a series of magnets attached to the machine base and a set of electrical coils potted around a steel laminate core attached to the moving slide.
The fact that there are no mechanical parts in contact means that there is no wear periodic maintenance required. Linear motors are not limited in travel like ball screws. Larger bare required to achieve high velocity, with a longer travel to prevent undue vibration.
This larger ball screw results in a higher inertia. This means a larger motor with more torque is required (introduction inertia) and the responsiveness and bandwidth of the system is reduced, resulting in poor servo performance.
Machines built with linear motors and alldigital drive systems can produce parts with higher accuracy and tighter tolerances at higher feeds and speeds. Also, they reduce significantly the nonmachining time with high acceleration and deceleration rates.
Linear Motors Recently linear motors are being increasingly considered for use in high performance CNC machine tools. The linear motor consists of a series of magnets attached to the machine base and a set of electrical coils potted around a steel laminate core attached to the moving slide.
The fact that there are no mechanical parts in contact means that there is no wear periodic maintenance required. Linear motors are not limited in travel like ball screws. Larger bare required to achieve high velocity, with a longer travel to prevent undue vibration.
This larger ball screw results in a higher inertia. This means a larger motor with more torque is required (introduction inertia) and the responsiveness and bandwidth of the system is reduced, resulting in poor servo performance.
Machines built with linear motors and alldigital drive systems can produce parts with higher accuracy and tighter tolerances at higher feeds and speeds. Also, they reduce significantly the nonmachining time with high acceleration and deceleration rates.
Question 2 
Match the additive manufacturing technique in
Column I with its corresponding input material in
Column II.
Additive manufacturing technique (Column I)
P. Fused deposition modelling
Q. Laminated object manufacturing
R. Selective laser sintering
Input material (Column II)
1. Photo sensitive liquid resin
2. Heat fusible powder
3. Filament of polymer
4. Sheet of thermoplastic or green compacted metal sheet
Additive manufacturing technique (Column I)
P. Fused deposition modelling
Q. Laminated object manufacturing
R. Selective laser sintering
Input material (Column II)
1. Photo sensitive liquid resin
2. Heat fusible powder
3. Filament of polymer
4. Sheet of thermoplastic or green compacted metal sheet
P3, Q4, R2  
P1, Q2, R4  
P2, Q3, R1  
P4, Q1, R4 
Question 2 Explanation:
Fuseddeposition modeling consists of a computercontrolled extruder, through which a polymer
filament is deposited to produce a part slice by slice.
Laminatedobject manufacturing uses a laser beam or vinyl cutter to first cut the slices on paper or plastic sheets (laminations); then it applies an adhesive layer, if necessary and finally stacks the sheets to produce the part.
Selective laser sintering uses a highpowered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.
Laminatedobject manufacturing uses a laser beam or vinyl cutter to first cut the slices on paper or plastic sheets (laminations); then it applies an adhesive layer, if necessary and finally stacks the sheets to produce the part.
Selective laser sintering uses a highpowered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.
Question 3 
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 3 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 positionand velocity control loops, to generate the required machining path. Typical interpolators are capable of generating linear and circular paths.
Question 4 
The XY table of a NC machine tool is to move from P(1,1) to Q(51,1); all coordinates are in mm. The pitch of the NC drive leadscrew is 1 mm. If the backlash between the leadscrew and the nut is 1.8^{\circ}, then the total backlash of the table on moving from P to Q is _______mm (round off to two decimal places).
0.15  
0.25  
0.55  
0.85 
Question 4 Explanation:
P(1, 1) to P(51, 1) is 50 mm distance. As pitch of the leadscrew is 1 mm. It has to move 50 rotation and in each rotation, backlash is 1.8 degree. Therefore total backlash
= 1.8 x 50 degree.
In one rotation i.e. 360^{\circ} rotation  1 mm movement take place
In 1.8 \times 50^{\circ}=\frac{1.8 \times 50}{360} \mathrm{~mm}=0.25 \mathrm{~mm}
Question 5 
In modern CNC machine tools, the backlash has been eliminated by
preloaded ballscrews  
rack and pinion  
ratchet and pinion  
slider crank mechanism 
Question 5 Explanation:
Question 6 
A point 'P' on a CNC controlled XYstage is moved to another point 'Q' using the coordinate
system shown in the figure below and rapid positioning command (G00).
A pair of stepping motors with maximum speed of 800 rpm, controlling both the X and Y motion of the stage, are directly coupled to a pair of lead screw, each with a uniform pitch of 0.5 mm. The time needed to position the point 'P' to the point 'Q' is _______ minutes. (round off to 2 decimal places).
A pair of stepping motors with maximum speed of 800 rpm, controlling both the X and Y motion of the stage, are directly coupled to a pair of lead screw, each with a uniform pitch of 0.5 mm. The time needed to position the point 'P' to the point 'Q' is _______ minutes. (round off to 2 decimal places).
0.75  
1.5  
2  
1.2 
Question 6 Explanation:
\begin{aligned} N&=800 \mathrm{rpm}, P=0.5 \mathrm{mm} / \mathrm{rev} \\ V &=N \times P\\ &=\mathrm{rev} / \mathrm{min} \times \mathrm{mm} / \mathrm{rev}\\&=400 \mathrm{mm} / \mathrm{min} \\ \Delta t_{x} &=\frac{600}{400}=1.5 \mathrm{min} \\ \Delta t_{y} &=\frac{300}{400}=0.75 \mathrm{min} \end{aligned}
There are two stepper motor so both will work till 0.75 min then y axis motor will stop then only x axis motor will run for 0.75 more, so total time will be 1.5 min.
There are two stepper motor so both will work till 0.75 min then y axis motor will stop then only x axis motor will run for 0.75 more, so total time will be 1.5 min.
Question 7 
Interpolator in a CNC machine
controls spindle speed  
coordinates axes movements  
operates tool changer
 
commands canned cycle

Question 7 Explanation:
As interpolator provides two functions:
1. It computes individual axis velocities to drive the tool along the programmed path at given feed rate.
2. It generates intermediate coordinate positions along the programmed path.
1. It computes individual axis velocities to drive the tool along the programmed path at given feed rate.
2. It generates intermediate coordinate positions along the programmed path.
Question 8 
Circular arc on a part profile is being machined on a vertical CNC milling machine. CNC part program using metric units with absolute dimensions is listed below:

N60 G01 X 30 Y 55 Z 5 F
50 N70 G02 X 50 Y 35 R 20
N80 G01 Z 5

The coordinates of the centre of the circular arc are :

N60 G01 X 30 Y 55 Z 5 F
50 N70 G02 X 50 Y 35 R 20
N80 G01 Z 5

The coordinates of the centre of the circular arc are :
(30, 55)  
(50, 55)  
(50, 35)  
(30, 35) 
Question 8 Explanation:
Question 9 
A point P(1,3,5) is translated by 2\hat{i}+3\hat{j}4\hat{k} and then rotated counter clockwise by 90^{\circ} about the zaxis. The new position of the point is
(6, 3, 9)  
(6, 3, 9)  
(6, 3, 9)  
(6, 3, 9) 
Question 9 Explanation:
Initial point is (1, 3, 5)
After tranlated by 2i+3j4k, it becomes
=(1+2, 3+3, 54)= (3, 6, 9)
Rotation by 90 CCW about z axis (NOTE: : when rotated about zaxis the z component will remain same.)
Hence the point (3, 6, 9) changes to (6, 3, 9).
After tranlated by 2i+3j4k, it becomes
=(1+2, 3+3, 54)= (3, 6, 9)
Rotation by 90 CCW about z axis (NOTE: : when rotated about zaxis the z component will remain same.)
Hence the point (3, 6, 9) changes to (6, 3, 9).
Question 10 
Match the following part programming codes with their respective functions
P II, Q I, R IV, S III  
P IV, Q II, R III, S I  
P IV, Q III, R II, S I  
P III, Q IV, R II, S I 
Question 10 Explanation:
G01 is used for linear interpolation, G03 is used for circular interpolation counter clockwise, M03 for spindle rotation clockwise and M05 for spindle top.
There are 10 questions to complete.