Question 1 
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 1 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 2 
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 2 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 3 
In modern CNC machine tools, the backlash has been eliminated by
preloaded ballscrews  
rack and pinion  
ratchet and pinion  
slider crank mechanism 
Question 3 Explanation:
Question 4 
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 4 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 5 
Interpolator in a CNC machine
controls spindle speed  
coordinates axes movements  
operates tool changer
 
commands canned cycle

Question 5 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 6 
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 6 Explanation:
Question 7 
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 7 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 8 
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 8 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.
Question 9 
For the situation shown in the figure below the expression for H in terms of r, R and D is
H=D+\sqrt{r^{2}+R^{2}}  
H=(R+r)+(D+r)  
H=(R+r)+\sqrt{D^{2}R^{2}}  
H=(R+r)+\sqrt{2D(R+r)D^{2}} 
Question 9 Explanation:
\begin{aligned} A B &=\sqrt{(R+r)^{2}(DRr)^{2}} \\ H &=R+A B+r \\ &=R+r+\sqrt{(R+r+DRr)(R+rD+R+r)} \\ &=R+r+\sqrt{(R+r+DRr)(R+rD+R+r)} \\ &=R+r+\sqrt{D[2(R+r)D]} \\ &=R+r+\sqrt{2 D(R+r)D^{2}} \end{aligned}
Question 10 
The figure below represents a triangle PQR with initial coordinates of the vertices as P(1,3), Q(4,5) and R(5,3.5). The triangle is rotated in the XY plane about the vertex P by angle \theta in clockwise direction. If sin\theta = 0.6 and cos\theta = 0.8, the new coordinates of the vertex Q are
(4.6, 2.8)  
(3.2, 4.6)  
(7.9, 5.5)  
(5.5, 7.9) 
Question 10 Explanation:
If the triangle PQR rotates by \theta clockwise then abscissa value of Q must increase whereas ordinate value of Q must decrease and only satisfying option is (4.6,2.8) i.e. A
There are 10 questions to complete.