Metal Cutting Processes MCQ Quiz - Objective Question with Answer for Metal Cutting Processes - Download Free PDF

Last updated on Jun 21, 2025

Latest Metal Cutting Processes MCQ Objective Questions

Metal Cutting Processes Question 1:

In a parting-off operation, how is the cutting tool fed into the workpiece?

  1. By moving the carriage along the lathe bed
  2. By rotating the cross-slide screw by hand
  3. By adjusting the tailstock manually
  4. By rotating the headstock spindle directly

Answer (Detailed Solution Below)

Option 2 : By rotating the cross-slide screw by hand

Metal Cutting Processes Question 1 Detailed Solution

Explanation:

Parting-Off Operation

  • Parting-off is a machining operation performed on a lathe where a thin cutting tool, known as a parting tool, is used to cut off a portion of the workpiece. This operation is commonly used to separate finished components from the raw material or to create grooves or recesses in the workpiece. The parting tool is fed perpendicular to the axis of the workpiece, slicing through it as the lathe spindle rotates.
  • In a parting-off operation, the cutting tool is mounted on the tool post, and the tool post is attached to the cross-slide of the lathe. The cross-slide allows the tool to move in a direction perpendicular to the axis of the workpiece. During the operation, the operator manually rotates the cross-slide screw to feed the parting tool into the workpiece. This manual control ensures precision and allows the operator to adjust the feed rate based on the material type, tool geometry, and cutting conditions.

Why Option 2 is Correct:

1. Mechanism of Cross-Slide Feed: The cross-slide screw is a threaded component connected to the cross-slide of the lathe. When the operator rotates the screw, the cross-slide moves along its dovetail ways, carrying the tool closer to or away from the workpiece. This movement is precisely controlled, which is essential for delicate operations like parting-off.

2. Perpendicular Feed: Parting-off requires the tool to move perpendicular to the workpiece's axis to achieve a clean separation. The cross-slide screw facilitates this perpendicular feed, making it the most appropriate choice for the operation.

3. Manual Control: Unlike automated feeds, manual control allows the operator to feel the resistance encountered during cutting. This tactile feedback helps the operator avoid excessive tool pressure, which can lead to tool breakage or deformation of the workpiece.

4. Practical Application: In most lathe operations, especially in small workshops or for custom jobs, manual control of the cross-slide is preferred for parting-off. It provides the flexibility to accommodate variations in material properties and tool wear.

Metal Cutting Processes Question 2:

Data of an orthogonal cutting is:

Chip thickness ratio = 0.4

Rake angle = 20°

What will be the shear plane angle?

[given, , ]

  1. cos1(0.711)
  2. tan1(0.435)
  3. tan1(0.711)
  4. tan1(0.211)

Answer (Detailed Solution Below)

Option 2 : tan1(0.435)

Metal Cutting Processes Question 2 Detailed Solution

Concept:

The shear plane angle \( \phi \) in orthogonal cutting is calculated using:

\( \tan \phi = \frac{r \cos \alpha}{1 - r \sin \alpha} \)

Given:

Chip thickness ratio \( r = 0.4 \), Rake angle \( \alpha = 20^\circ \)

\( \cos(20^\circ) = 0.94 \), \( \sin(20^\circ) = 0.34 \)

Calculation:

\( \tan \phi = \frac{0.4 \cdot 0.94}{1 - 0.4 \cdot 0.34} = \frac{0.376}{0.864} \approx 0.435 \)

\( \phi = \tan^{-1}(0.435) \)

 

Metal Cutting Processes Question 3:

______ is a process where small pieces from the tool rake face material are adhered to the flowing chip and thus removed from the surface.

  1. Adhesive wear
  2. Flank wear
  3. Crater wear
  4. Diffusion wear

Answer (Detailed Solution Below)

Option 3 : Crater wear

Metal Cutting Processes Question 3 Detailed Solution

Concept:

Crater wear is a specific type of tool wear that occurs on the rake face of a cutting tool. This wear is characterized by the formation of a concave depression or crater on the rake face, which is the surface of the tool that comes into contact with the chip as it flows away from the cutting edge. The process involves the removal of small pieces of tool material due to the high temperature and pressure conditions at the tool-chip interface.

Explanation:

When a cutting tool is used in machining operations, it experiences various types of wear due to the intense conditions at the cutting zone. One of the critical areas where wear occurs is the rake face of the tool. The rake face is the surface over which the chip flows after being sheared off from the workpiece. The wear on this face can significantly affect the tool's performance and the quality of the machined surface.

Crater wear occurs due to the combination of mechanical and thermal factors at the tool-chip interface. As the chip flows over the rake face, it exerts a high shear force and generates significant heat due to friction and plastic deformation. This high temperature can cause the tool material to soften and weaken, making it more susceptible to wear.

Several mechanisms contribute to crater wear:

  1. Adhesion: Small pieces of tool material may adhere to the flowing chip due to the high temperature and pressure. As the chip moves away, it pulls these adhered particles off the rake face, creating a crater.
  2. Abrasion: The hard particles in the workpiece material or the chip itself can abrade the tool surface, leading to material removal and crater formation.
  3. Diffusion: At elevated temperatures, atoms from the tool material may diffuse into the chip material, weakening the tool and causing wear.
  4. Oxidation: The high temperature can also lead to oxidation of the tool material, forming brittle oxides that can be easily worn away, contributing to crater formation.

Crater wear is particularly significant in high-speed machining operations where the cutting temperatures are very high. It can affect the tool's geometry, reducing its effectiveness and leading to poor surface finish and dimensional inaccuracies in the machined part.

To mitigate crater wear, several strategies can be employed:

  • Using coatings: Applying wear-resistant coatings on the tool can help reduce the wear rate and extend tool life.
  • Optimizing cutting parameters: Adjusting cutting speed, feed rate, and depth of cut to minimize the temperature and pressure at the tool-chip interface can help reduce crater wear.
  • Using appropriate tool materials: Selecting tool materials with high hot hardness and wear resistance can improve tool performance in high-temperature conditions.
  • Using cutting fluids: Applying cutting fluids can help reduce the temperature and friction at the tool-chip interface, thereby reducing wear.

Conclusion:

Crater wear is a critical issue in machining operations, particularly in high-speed cutting where temperatures are elevated. Understanding the mechanisms of crater wear and employing appropriate strategies to mitigate it can significantly enhance tool life and machining performance. In the context of the given options, crater wear is the correct answer for the process where small pieces from the tool rake face material are adhered to the flowing chip and thus removed from the surface.

Correct Option:

Option 3: Crater wear

Metal Cutting Processes Question 4:

Which of the following is NOT an advantage of increasing the side-cutting edge angle? 

  1. Increases tool life
  2. Diminishes chip thickness
  3. Dissipates heat quickly
  4. Provides better surface finish
  5. Poor surface finish

Answer (Detailed Solution Below)

Option 5 : Poor surface finish

Metal Cutting Processes Question 4 Detailed Solution

Explanation:

Side cutting edge angle

The following are the advantages of increasing this angle,

  • It increases tool life as, for the same depth of cut; the cutting force is distributed on a wider surface
  • It diminishes the chip thickness for the same amount of feed and permits greater cutting speed.
  • It dissipates heat quickly for having wider cutting edge.
    • The side cutting edge angle of the tool has practically no effect on the value of cutting force or power consumed for a given depth of cut and feed
    • Large side cutting edge angles are lightly to cause the tool to chatter

Metal Cutting Processes Question 5:

The angle between face of the tool and a line parallel with the base of the tool, measured in a perpendicular plane through the side cutting edge is

  1. Side relief angle
  2. Back rake angle
  3. Side rake angle
  4. End relief angle
  5. Clearance Angle

Answer (Detailed Solution Below)

Option 2 : Back rake angle

Metal Cutting Processes Question 5 Detailed Solution

Explanation:

Single point cutting tool

RRB JE ME 74 6 Qs 28thAug 2015 Shift2 Diag(Nita) Hindi i mages satya Q2


Face: Face is the surface of the tool on which the chip impinges when separated from work-piece

Side Cutting Edge Angle:

  • The angle between side cutting edge and side of the tool shank is called side cutting edge angle.
  • It is also called a lead angle or principle cutting angle.

End Cutting Edge Angle:

  • The angle between the end cutting edge and a line perpendicular to the shank of the tool is called end cutting edge angle.

Side Relief Angle:

  • The angle between the portion of the side flank immediately below the side cutting edge and the line perpendicular to the base of the tool measured at right angles to the side flank is known as side relief angle.
  • It is the angle that prevents interference, as the tool enters the work material.

End Relief Angle:

  • The end relief angle is the angle between the portion of the end flank immediately below the end cutting edge and the line perpendicular to the base of the tool, measured at right angles to the end flank.
  • It is the angle that allows the tool to cut without rubbing on the workpiece.

Back Rake Angle:

  • The angle between the face of the tool and a line parallel with the base of the tool, measured in a perpendicular plane through the side cutting edge is called the back rake angle. So, the correct answer is option 2.
  • It is the angle that measures the slope of the face of the tool from the nose toward the rear.
  • If the slope is downward toward the nose, it is a negative back rake angle. And if the slope is down from the nose, it is a positive back rake angle. If there is not any slope, the back rake angle is zero.

Side Rake Angle:

  • The angle between the face of the tool and a line parallel with the base of the tool, measured in a plane perpendicular to the base and side cutting edge is called the side rake angle.
  • It is the angle that measures the slope of the tool face from the cutting edge. If the slope is towards the cutting edge, it is a negative side rake angle.
  • If the slope is away from the cutting edge, it is a positive side rake angle.

Important Points

  • Normal Plane —it is a plane perpendicular to the principal cutting edge of the tool. 
  • Orthogonal Plane —it is a plane perpendicular to the reference plane and also perpendicular to the cutting plane.
  • Machine Longitudinal Plane —it is a plane perpendicular to the reference plane and along the direction of longitudinal feed.
  • Machine Transverse Plane —it is a plane perpendicular to the reference plane and along the direction of transverse feed.

Top Metal Cutting Processes MCQ Objective Questions

The feed rate of single point cutting tool is 3 mm/revolution and the workpiece is rotating at 600 r.p.m. Determine the total machining time to turn the cylindrical surface of length 300 mm of the workpiece.

  1. 5 sec
  2. 15 sec
  3. 20 sec
  4. 10 sec

Answer (Detailed Solution Below)

Option 4 : 10 sec

Metal Cutting Processes Question 6 Detailed Solution

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Concept:

Time for machining \(= \frac{L}{{f\times N}}\)

where L is job length (mm), f is feed (mm/rev), N is job speed (rpm)

Calculation:

Given:

f = 3 mm/rev, N = 600 rpm, L = 300 mm

Time for machining \(= \frac{L}{{f.N}} = \frac{ 300}{3×600} \) = 0.1666 minutes = 0.1666 × 60 = 10 sec

The total machining time to turn the cylindrical surface is 10 sec.

In an orthogonal cutting system the cutting edge is:

  1. In line with direction of tool travel
  2. Perpendicular to direction of tool-travel
  3. Perpendicular to shear plane
  4. Perpendicular to direction of depth of cut

Answer (Detailed Solution Below)

Option 2 : Perpendicular to direction of tool-travel

Metal Cutting Processes Question 7 Detailed Solution

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Explanation:

Orthogonal Cutting

Oblique Cutting

In orthogonal cutting, cutting edge is perpendicular to the direction of tool-travel.

In oblique cutting, the angle between cutting edge and direction of tool-travel is less than 90°.

Chip flow angle is zero.

Chip flow angle is more than zero.

Tool life is less.

Tool life is more.

Two components of forces.

  • Axial (Px)
  • Tangential (Pz)

Three components of forces.

  • Axial (Px)
  • Tangential (Pz)
  • Radial (Pxy)

Poor surface finish.

Good surface finish.

Used in slotting, parting, grooving, pipe cutting.

Used in turning, milling, drilling, grinding.

 An orthogonal cutting operation is being carried out under the following conditions : Cutting Speed = 2 m/sec, Depth of cut = 0.5 mm, Chip thickness = 0.6 mm. What is the chip velocity? 

  1. 2 m/sec
  2. 2.4 m/sec
  3. 1 m/sec
  4. 1.66 m/sec 

Answer (Detailed Solution Below)

Option 4 : 1.66 m/sec 

Metal Cutting Processes Question 8 Detailed Solution

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Concept:
We know that, 

\(r = \frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)

r = chip thickness ratio, t = chip thickness before cutting/(uncut chip thickness) (mm), tc = chip thickness after cutting (mm), V = cutting speed (m/s), Vc = chip velocity (m/s)

Calculation:

Given:

V = 2 m/s, Depth of cut = 0.5 mm

In orthogonal cutting,

t = f(feed) 

and b(width) = d(depth of cut)

but in the question, nothing is mentioned about the uncut chip thickness and feed so we have taken the uncut chip thickness equal to the depth of cut. (Official Question of UPPSC AE)

tc = 0.6 mm

\(\frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)

\(\frac{{0.5}}{{0.6}} = \frac{{{V_c}}}{2} \Rightarrow {V_c} = 1.66\;m/s\)

Which pair of following statement is correct for orthogonal cutting using a single-point cutting tool ?

P.  Reduction in friction angle increases cutting force

Q.  Reduction in friction angle decreases cutting force

R.  Reduction in friction angle increases chip thickness

S.  Reduction in friction angle decreases chip thickness

  1. P and R
  2. P and S
  3. Q and R
  4. Q and S

Answer (Detailed Solution Below)

Option 4 : Q and S

Metal Cutting Processes Question 9 Detailed Solution

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Explanation:

In orthogonal cutting,

(i) Reduction in friction angle decreases the cutting force

(ii) Reduction in friction angle decreases chip thickness

According to Merchant's theory,

\(ϕ = \frac{{\rm{\pi }}}{4} + \frac{{\rm{α }}}{2} - \frac{{\rm{β }}}{2}\) where, ϕ = shear angle, α = rake angle, β = friction angle

As β decreases. ϕ increases because α is constant.

  • If all other factors remain the same, a higher shear angle results in a smaller shear plane area. Since the shear strength is applied across this area, the shear force required to form the chip will decrease when the shear plane area is decreased.

Now \(\frac{t}{{{t_c}}} = \frac{{sinϕ }}{{\cos \left( {ϕ - α } \right)}}\)

\({t_c} = \frac{{\cos \left( {ϕ - α } \right)}}{{sinϕ }} \times t\)

  • As ϕ increases, cos (ϕ -α) will decrease and Sinϕ will increase. This chip thickness will decrease.
  • Hence from the above options Q and S, both are correct..

The chip thickness ratio 'r' in Orthogonal metal cutting operation is always:

  1. Greater than unity
  2. Equal to unity
  3. Less than unity
  4. Equal to depth of cut

Answer (Detailed Solution Below)

Option 3 : Less than unity

Metal Cutting Processes Question 10 Detailed Solution

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Explanation:

Chip thickness ratio / Cutting ratio (r):

It is the ratio of chip thickness before cut (t1) to the chip thickness after cut (t2).

\(r=\frac{Chip\;thickness\;before\;cut\;(t_1)}{Chip\;thickness\;after\;cut\;(t_2)}\Rightarrow \frac{uncut\;chip\;thickness}{chip\;thickness}\)

chip thickness after the cut (t2) is always greater than the chip thickness before the cut (t1), ∴ r is always < 1, i.e. the uncut chip thickness value is less than the chip thickness value.

Assuming volume to be constant:

t1b1L1 = t2b2L2

\(\frac{t_1}{t_2}=\frac{L_2}{L_1}\;\;\;\;(\because b_1=b_2)\)

as t2 > t1, ∴ L2 < L1 i.e. length after cut is less than the length before the cut.

Assuming discharge to be constant:

t1b1V = t2b2Vc

\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)

as t2 > t1, ∴ V > Vc i.e. cutting velocity is greater than the chip velocity.

The shear strain in  metal cutting is _______

  1. \(\tan \left( {\phi - \alpha } \right) + \cot \phi \)
  2. \(\sin \left( {\phi - \alpha } \right) + \cot \phi \)
  3. \(\cos \left( {\phi - \alpha } \right) + \cot \phi \)
  4. \(\cot \left( {\phi - \alpha } \right) + \cot \phi \)

Answer (Detailed Solution Below)

Option 1 : \(\tan \left( {\phi - \alpha } \right) + \cot \phi \)

Metal Cutting Processes Question 11 Detailed Solution

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Explanation:

F1 S.S Madhu 2.12.19 D 9

The above diagram is called "Merchant Circle Diagram", which represents the inter-relationship of the different force components in continuous chip formation under orthogonal cutting.

The term (Fs) represents the "Shear force" which is mainly responsible for the chip-separation from the parent body by shearing and is used to determine the yield strength of the work material. This shear force acts along the shear angle (ϕ).

Cutting strain (ϵ):

The magnitude of the average strain that develops along the shear plane due to machining action is called "cutting strain".

It is given by-

\(\epsilon=\cot ϕ \;+\;\tan \left( {ϕ - α } \right) \)

where ϕ = shear angle and α = rake angle.

At what rpm should a lathe be run to give a cutting speed of 20 m/min when turning a rod of diameter 40 mm?

  1. 100 rpm
  2. 160 rpm
  3. 200 rpm
  4. 250rpm

Answer (Detailed Solution Below)

Option 2 : 160 rpm

Metal Cutting Processes Question 12 Detailed Solution

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Concept:

Cutting speed is given by

V = πDN

Where V = cutting speed, D = diameter, N = speed in RPM

Calculation:

Given:

V = 20 m/min , D = 40 mm = 0.04 m

V = πDN

20 = π × 0.04 × N

N = 160 rpm

A 150 mm long, 12.5 mm diameter stainless steel rod is being reduced to a diameter of 12 mm by turning operation. The lathe spindle rotates at N = 400 rpm. Calculate the cutting speed.

  1. 16 m/min
  2. 20 m/min
  3. 15.7 m/min
  4. 31.4 m/min

Answer (Detailed Solution Below)

Option 3 : 15.7 m/min

Metal Cutting Processes Question 13 Detailed Solution

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Concept:

Cutting speed is given by

V = πDN

Where V = cutting speed, D = diameter, N = speed in RPM

Calculation:

Given:

N = 400 rpm, D = 12.5 mm = 0.0125 m

V = πDN

V = π × 0.0125 × 400

V = 15.7 m/min 

Extreme Pressure Additive (EPA) is mixed with cutting fluid for improving its power of

  1. cooling
  2. lubrication
  3. cleaning of cutting zone
  4. protection of machined surface

Answer (Detailed Solution Below)

Option 2 : lubrication

Metal Cutting Processes Question 14 Detailed Solution

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Explanation:

Purposes of application of cutting fluid in machining:

  • The main aim of the application of cutting fluid is to improve machinability through reduction of cutting forces and temperature, improvement by surface integrity, and enhancement of tool life.
  • Cooling of the job and the tool to reduce the detrimental effects of cutting temperature on the job and the tool.
  • Lubrication at the chip–tool interface and the tool flanks to reduce cutting forces and friction and thus the amount of heat generation.
  • Cleaning the machining zone by washing away the chip – particles and debris.
  • To form a solid lubricating layer under high pressure and temperature some extreme pressure additive (EPA) is deliberately added in a reasonable amount in the mineral oil or soluble oil. ​

Additional Information

​Selection of Cutting Fluid:

Machining Material

Cutting Fluid

Grey cast iron

Dry air

Steels

Soluble oil

Aluminum and its alloys

kerosene oil

Stainless steels

High-performance soluble oil

Neat oil with high concentration with chlorinated EP additive 

Copper and its alloys

Water-based fluids

The rake angle of a single-point cutting tool corresponds to ______ of a twist drill.

  1. Chiesel edge angle
  2. Point angle
  3. Helix angle
  4. Lip clearance angle

Answer (Detailed Solution Below)

Option 3 : Helix angle

Metal Cutting Processes Question 15 Detailed Solution

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Explanation:

  • Chip removal can be done in three ways:
    1. Single point cutting (e.g.Turning, shaping),
    2. Multi-point cutting (e.g.milling),
    3. Abrasive machining (e.g grinding).
  • Single point cutting tools have only one cutting edge, such as cutting tools used in lathe and shaper.
  • Multi-point Cutting tools have more than one cutting edge, such as milling cutters, twist drills.
  • In abrasive machining, the abrasive particles act as cutting edges.

Single point cutting tool nomenclature:

tool geometry

Multi-point cutting tool(Twist drill):

  • Flutes help in easy chip removal and oil is pumped which flows over the flute.
  • Helix:
    • To provide a variable rake angle to drill.
    • Helps in easy chip flow.
    • Maximum rake angle = Helix angle(at outer periphery)

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