Question 1:For a 3-axes CNC table, the slide along the vertical axis of the table driven by a DC servo motor via lead screw nut mechanism. The lead screw has pitch of 5 mm. This lead screw is fitted with a relative (incremental) circular encoder. The basic length unit (BLU) of the slide along the vertical axis of the table is 0.005 mm. When the table moves along the vertical axis by 9 mm. The corresponding number of pulses generated by the encoder is
Question 2:In a CNC feed drive, a stepper motor with step angle of 1.8 degree drives a lead screw with pitch of 2 mm. The basic length unit (BLU) for this drive is
Question 3:The interpolator in a CNC machine controls
Question 4:Match the following: Group-I P. G08 Q. G41 R. G01 S. G02 Group-II 1. Linear interpolation 2. Acceleration 3. Circular interpolation 4. Cutter radius compensation P Q R S
Question 5:For machining the circular arc shown in the figure below on a NC machining, the interpolation parameters (I, K) and the incremental movements in the direction of the X and Z axes are programmed as
Question 6:Match the following
Group I P. G09 Q. G41 R. G01 S. G03 Group II 1. Linear interpolation 2. Retardation 3. Circular interpolation 4. Cutter radius compensation
P Q R S
Question 7:A DC servomotor is directly driving an NC table. The pitch of the lead screw of the table is 5 mm. The motor rotates at 100 rpm for an applied voltage of 10 V. If the voltage speed characteristic of the motor is linear, the applied voltage for a table of 3 m/min is equal to
Question 8:Feed motion can be provided with stepper motors in CNC
Question 9:A axis control in a typical CNC machine provides
Question 10:The most common interpolation methods in continuous path NC machining are
Question 11:If the z-axis an d x-axis o f CNC lath e are provided with straight line controls, it is possible to carry out
Question 12:Interpolator in a CNC machine
Question 13: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 F50N70 G02 X 50 Y 35 R 20N80 G01 Z 5............................................ The coordinates of the centre of the circular arc are:
Question 14:In a CNC milling operation, the tool has to machine the circular arc from point (20, 20) to (10, 10) at sequence number 5 of the CNC part program. If the center of the arc is at (20,10) and the machine has incremental mode of defining position coordinates, the correct tool path command is
Question 15:A drill is positioned at point P and its has to proceed to point Q. The coordinates of point Q in the incremental system of defining position of a point in CNC part program will be
Question 16:The function of interpolator in a CNC machine controller is to
Question 17:For the CNC part programming, match Group A with Group B : Group A P. Circular interpolation, counter clock wise Q. dwell R. circular interpolation, clock wise S. Point to point countering
Group B I. G02 II. G03 III. G04 IV. G00
Question 18:For machining a rectangular island represented by coordinates P(0,0), Q(100,0), fl( 100,50) and S(0,50) on a casting using CNC milling machine, an end mill with a diameter of 16 mm is used. The trajectory of the cutter centre to machine the island PQRS is
Question 19:A CNC vertical milling machine has to cut a straight slot of 10 mm width and 2 mm depth by a cutter of 10 mm diameter between points (0, 0) and (100, 100) on-the XY plane (dimensions in mm). The feed rate used for milling is 50 mm/min. Milling time for the slot (in seconds) is
Question 20:In a CNC program block, N002 G02 G91 X40 Z40..., G02 and G91 refer to
Question 21:Match the following: NC Code Definition P. M05 1. Absolute coordinate system Q. Q01 2. Dwell R. Q04 3. Spindle stop S. G90 4. Linear interpolation
Question 22:A customer insists on a modification to change the BLU of the CNC drive to 10 microns without changing the table speed. The modification can be accomplished by
Question 23:The Basic Length Unit (BLU), i.e., the table movement corresponding to 1 pulse of the pulse generator, is
Question 24:Which type of motor is NOT used in axis or spindle drives of CNC machine tools?
step angle (α) = 360/200 = 1.8ο (α) = 360/200=1.8ο x = PA/360,A = αnp np is number of pluse np = 1 for BLU
x= 5 microns.
Question 25:NC contouring is an example of
Question 26:The tool of an NC machine has to move along a circular arc from (5,5) to (10,10) while performing an operation. The centre of the arc is at (10,5). Which one of the following NC tool path commands performs the above mentioned operation?
Question 27:Which among the NC operations given below are continuous path operations? Arc Welding (AW) Drilling (D) Laser Cutting of Sheet (LC) Milling (M) Punching in Sheet Metal (P) Spot Welding (SW)
Question 28:During the execution of a CNC part program block NO20 GO2 X45.0 Y25.0 R5.0the type of tool motion will be
Question 29:Machining of complex shapes on CNC machine requires
Question 30:In finish machining of an island on a casting with CNC milling machine, an end mill with 10 mm diameter is employed, The corner points of the Island are represented by (0, 0), (0, 30), (60, 30) and (50, 0). By applying cutter radius compensation, the trajectory of the cutter will be
Given = Dia of end milling cutter = 10 mm
Question 31:In a point to point control NC machine, the slides positioned by an integrally mounted stepper motor drive. If the specification of the motor is 1 deg/pulse, and the pitch Of the lead screw is 3.6 mm.What is the expected positioning accuracy?
For 1 revolution of lead screw side will move by 3.6 mm 1 revolution = 360° = 360 pulses 360 pulses will cause motion of 3.6 mm. 1 pulse will result in 3.6/360° mm of motion i.e. 0.01 mm or 10 µm.
Question 32:With reference to NC machine, which of the following statement is wrong?
Question 33:In PTP type of NC system
In a point to point type of NC system, only end position of tool is controlled. Neither path nor velocity of the tool is important.
Question 34: In NC part programming spindle speed of 730 rpm will be coded by the magic -3 rule as
Past Year Questions: Numerical Control Notes | EduRev notes for Mechanical Engineering is made by best teachers who have written some of the best books of
Mechanical Engineering. It has gotten 45 views and also has 4.9 rating. You can download Free Past Year Questions: Numerical Control Notes | EduRev pdf from EduRev by
using search above. You can also find Past Year Questions: Numerical Control Notes | EduRev ppt and other Mechanical Engineering slides as well. If you want Past Year Questions: Numerical Control Notes | EduRev
Tests & Videos, you can search for the same too. Mechanical Engineering Past Year Questions: Numerical Control Notes | EduRev Summary and Exercise are very important for
perfect preparation. You can see some Past Year Questions: Numerical Control Notes | EduRev sample questions with examples at the bottom of this page. Complete
Past Year Questions: Numerical Control Notes | EduRev chapter (including extra questions, long questions, short questions, mcq) can be found on EduRev, you can check
out Mechanical Engineering lecture & lessons summary in the same course for Mechanical Engineering Syllabus. EduRev is like a wikipedia
just for education and the Past Year Questions: Numerical Control Notes | EduRev images and diagram are even better than Byjus! Do check out the sample questions
of Past Year Questions: Numerical Control Notes | EduRev for Mechanical Engineering, the answers and examples explain the meaning of chapter in the best manner. This is
your solution of Past Year Questions: Numerical Control Notes | EduRev search giving you solved answers for the same. To Study Past Year Questions: Numerical Control Notes | EduRev for Mechanical Engineering
this is your one stop solution.