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GATE & IES — MECHANICAL ENGINEERING
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Machining & Machine Tools

Machining Fundamentals

Principles and applications of machining fundamentals for Mechanical Engineering and GATE & IES preparation.

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60 Active Questions
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Question 1 of 60
Original Exam: GATE (2003)
In orthogonal cutting, the velocity relationship between cutting velocity (V), chip velocity (Vc), and shear velocity (Vs) is:
✓ Correct Answer: A
Explanation:
The velocity hodograph (velocity diagram) in orthogonal cutting relates the three velocities using the sine rule: V/sin(90°-φ+α) = Vc/sin φ = Vs/sin(90°-α), where V is cutting velocity, Vc is chip velocity, Vs is shear velocity, φ is shear plane angle, and α is rake angle. From this: Vc = V × sin φ / cos(φ-α) and Vs = V × cos α / cos(φ-α). The chip thickness ratio r = t1/t2 = Vc/V = sin φ / cos(φ-α).
Difficulty: intermediate Type: conceptual
Question 2 of 60
Original Exam: IES (2004)
The cutting speed in turning is given by V = πDN/1000 (m/min). For a workpiece of diameter 100 mm rotating at 500 rpm, the cutting speed is:
✓ Correct Answer: A
Explanation:
Cutting speed V = πDN/1000 = π × 100 × 500 / 1000 = π × 50 = 157.08 ≈ 157 m/min. Where D is workpiece diameter in mm and N is spindle speed in rpm. The factor 1000 converts mm to m. This formula is fundamental to all turning operations. The cutting speed determines tool life, surface finish, and productivity. Recommended cutting speeds: HSS tools 20-50 m/min for steel, carbide tools 100-300 m/min for steel.
Difficulty: basic Type: numerical
Question 3 of 60
Original Exam: GATE (2005)
The power consumed in machining is given by:
✓ Correct Answer: A
Explanation:
Power consumed in machining P = Fc × V, where Fc is the cutting force (principal force, in the direction of cutting velocity) and V is the cutting velocity. The thrust force (Ft) does not contribute to power in orthogonal cutting (perpendicular to cutting velocity). P = Fc × V = u × MRR, where u is specific cutting energy. Torque in turning: T = Fc × D/2. Power: P = T × ω = Fc × (D/2) × (2πN/60) = Fc × V. Units: P in watts if Fc in N and V in m/s.
Difficulty: basic Type: conceptual
Question 4 of 60
Original Exam: IES (2006)
The tool wear mechanism that is dominant at high cutting speeds and temperatures is:
✓ Correct Answer: A
Explanation:
At high cutting speeds and temperatures (above 700°C for steel), diffusion wear dominates. Tool atoms diffuse into the chip material and vice versa, weakening the tool material. Diffusion wear causes crater wear on the rake face (where temperatures are highest). Abrasion wear (mechanical scratching) dominates at low speeds. Adhesion wear (BUE formation and tearing) dominates at medium speeds. Oxidation wear occurs at very high temperatures. Crater wear is characterised by KT (crater depth) and KB (crater width).
Difficulty: intermediate Type: conceptual
Question 5 of 60
Original Exam: GATE (2007)
The uncut chip thickness (feed per tooth) in milling is given by:
✓ Correct Answer: A
Explanation:
In milling, the uncut chip thickness varies with the angular position of the tooth: tc = fz × sin θ, where fz is feed per tooth (mm/tooth) and θ is the instantaneous angle of engagement (0 at entry, φ at exit for up-milling). The chip thickness varies from 0 at entry to maximum at exit (up-milling) or from maximum at entry to 0 at exit (down-milling). Average chip thickness: tc_avg = fz × sin(φ/2) for symmetric engagement. This variation causes variable cutting forces in milling.
Difficulty: intermediate Type: conceptual
Question 6 of 60
Original Exam: IES (2008)
The nose radius of a cutting tool affects the surface roughness. The theoretical surface roughness (Rmax) in turning is:
✓ Correct Answer: A
Explanation:
Theoretical maximum surface roughness in turning: Rmax = f²/(8r), where f is feed (mm/rev) and r is nose radius (mm). This is derived geometrically from the scallop height left by the tool nose. Arithmetic mean roughness Ra ≈ Rmax/4 = f²/(32r). To reduce surface roughness: (1) Decrease feed, (2) Increase nose radius. Actual Ra is higher than theoretical due to BUE, vibration, and tool wear. Example: f = 0.2 mm/rev, r = 0.8 mm → Rmax = 0.04/(8×0.8) = 0.00625 mm = 6.25 μm.
Difficulty: intermediate Type: numerical
Question 7 of 60
Original Exam: GATE (2009)
The material removal rate (MRR) in drilling is given by:
✓ Correct Answer: A
Explanation:
MRR in drilling = (π/4) × D² × f × N, where D is drill diameter (mm), f is feed (mm/rev), N is spindle speed (rpm). This equals the volume of material removed per minute = cross-sectional area × feed rate = (πD²/4) × (f × N). Alternatively: MRR = (πD²/4) × fr, where fr = f × N is the feed rate (mm/min). Cutting speed in drilling: V = πDN/1000 (m/min). Machining time: tm = L/(f × N), where L is hole depth.
Difficulty: basic Type: numerical
Question 8 of 60
Original Exam: IES (2010)
The relationship between the shear plane angle (φ), rake angle (α), and chip thickness ratio (r) is:
✓ Correct Answer: A
Explanation:
From the geometry of the chip formation: chip thickness ratio r = t1/t2 = sin φ / cos(φ-α). Rearranging: r cos(φ-α) = sin φ → r(cos φ cos α + sin φ sin α) = sin φ → r cos α cos φ = sin φ(1 - r sin α) → tan φ = r cos α / (1 - r sin α). This is the fundamental relationship in orthogonal cutting. Higher r (thinner chip) corresponds to larger φ (steeper shear plane), which means more efficient cutting.
Difficulty: intermediate Type: numerical
Question 9 of 60
Original Exam: GATE (2011)
The cutting temperature in machining is most significantly affected by:
✓ Correct Answer: A
Explanation:
Cutting temperature θ ∝ V^a × f^b × d^c, where the exponents are typically a ≈ 0.4, b ≈ 0.2, c ≈ 0.1. Cutting speed has the greatest influence on temperature (highest exponent). This is because: (1) Higher speed → more heat generated per unit time, (2) Less time for heat dissipation. Temperature affects: tool wear rate (exponentially), surface integrity, dimensional accuracy. Cutting temperature can be measured by: tool-work thermocouple, radiation pyrometer, embedded thermocouple, metallographic methods.
Difficulty: intermediate Type: conceptual
Question 10 of 60
Original Exam: IES (2012)
The tool life equation with extended Taylor's formula VT^n × f^a × d^b = C gives the tool life T as:
✓ Correct Answer: A
Explanation:
From VT^n × f^a × d^b = C: T^n = C/(V × f^a × d^b) → T = [C/(V × f^a × d^b)]^(1/n) = (C/(V × f^(a/n) × d^(b/n)))^(1/n). Wait, more precisely: T = [C/(V × f^a × d^b)]^(1/n). This shows that tool life decreases with increasing V, f, and d. The exponent 1/n for V is largest (n is smallest for V), confirming that cutting speed has the greatest effect on tool life. Typical n values: HSS 0.1-0.2, carbide 0.2-0.4, ceramic 0.4-0.6.
Difficulty: intermediate Type: numerical
Question 11 of 60
Original Exam: GATE (2013)
The optimum cutting speed for minimum cost (Vopt_cost) in machining is:
✓ Correct Answer: A
Explanation:
Optimum tool life for minimum cost: T_opt_cost = (1/n - 1)(Ct/Cm + tc), where n is Taylor's exponent, Ct is tool cost per edge, Cm is machining cost rate ($/min), tc is tool change time (min). Optimum cutting speed: Vopt_cost = C/[T_opt_cost]^n. For maximum production rate: T_opt_prod = (1/n - 1) × tc (no tool cost term). Vopt_prod > Vopt_cost (maximum production speed is higher than minimum cost speed). The optimum speed balances machining time cost against tool cost.
Type: conceptual
Question 12 of 60
Original Exam: IES (2014)
The flank wear (VB) in a cutting tool is measured on:
✓ Correct Answer: A
Explanation:
Flank wear (VB) is measured on the flank face (clearance face) of the tool, as the width of the wear land parallel to the machined surface. ISO standard tool life criterion: VB = 0.3 mm (uniform wear) or VBmax = 0.6 mm (non-uniform wear). Flank wear causes: (1) Increased cutting forces, (2) Poor surface finish, (3) Dimensional inaccuracy. Crater wear (KT) occurs on the rake face. Flank wear is caused by: abrasion, adhesion, diffusion. Flank wear is the most common tool failure mode in finish machining.
Difficulty: basic Type: conceptual
Question 13 of 60
Original Exam: GATE (2015)
The specific cutting energy (u) in machining is related to the cutting force (Fc), cutting speed (V), and MRR by:
✓ Correct Answer: A
Explanation:
Specific cutting energy u = Power/MRR = (Fc × V)/(V × f × d) = Fc/(f × d) = Fc/(b × t1), where f is feed, d is depth of cut, b is width of cut, t1 is uncut chip thickness. Units: N/mm² = J/mm³. u is a material property used to estimate cutting forces: Fc = u × f × d. Typical values: aluminium 0.4-1.0 J/mm³, cast iron 1.1-5.4 J/mm³, steel 2.0-4.0 J/mm³, titanium 3.0-6.0 J/mm³. u increases at small chip thickness (size effect).
Difficulty: intermediate Type: conceptual
Question 14 of 60
Original Exam: IES (2016)
The machining time for a turning operation is given by:
✓ Correct Answer: A
Explanation:
Machining time in turning: tm = L/(f × N), where L is the length of cut (mm), f is feed (mm/rev), N is spindle speed (rpm). Since V = πDN/1000, N = 1000V/(πD). Therefore: tm = L × πD/(1000 × V × f). For a complete pass: L = workpiece length + approach + overtravel. Number of passes = (total depth of cut)/(depth per pass). Total machining time = tm × number of passes. This formula is fundamental for production planning and cost estimation.
Difficulty: basic Type: numerical
Question 15 of 60
Original Exam: GATE (2017)
The angle between the tool face (rake face) and the shear plane is called:
✓ Correct Answer: A
Explanation:
The friction angle β = arctan(μ) = arctan(F/N), where F is friction force and N is normal force on the rake face, and μ is the coefficient of friction. The friction angle is NOT the angle between the rake face and shear plane. The angle between the rake face and shear plane = φ + β - α (from Merchant's circle geometry). Merchant's equation: 2φ + β - α = 90° → φ = 45° + α/2 - β/2. The friction angle determines the shear plane angle and hence the chip thickness and cutting forces.
Difficulty: intermediate Type: conceptual
Question 16 of 60
Original Exam: IES (2003)
The clearance angle (relief angle) in a cutting tool is provided to:
✓ Correct Answer: B
Explanation:
The clearance angle (relief angle) is the angle between the tool flank and the machined surface. It prevents rubbing (interference) of the tool flank against the machined surface, which would cause excessive friction, heat, and tool wear. Typical clearance angles: 5-15°. Too small: rubbing and excessive wear. Too large: weakens the cutting edge (less support). End clearance prevents rubbing at the end of the tool. Side clearance prevents rubbing on the side. Clearance angle does not directly affect chip formation (rake angle does).
Difficulty: basic Type: conceptual
Question 17 of 60
Original Exam: GATE (2004)
In the Merchant's circle diagram, the resultant force R is related to the cutting force (Fc) and thrust force (Ft) by:
✓ Correct Answer: B
Explanation:
The resultant force R = √(Fc² + Ft²), where Fc is the cutting force (tangential, in direction of cutting velocity) and Ft is the thrust force (feed force, perpendicular to cutting velocity in orthogonal cutting). R is the vector sum of Fc and Ft. From Merchant's circle: R = F/sin β = N/cos β = Fs/cos(φ+β-α) = Fn/sin(φ+β-α), where F is friction force, N is normal force on rake face, Fs is shear force, Fn is normal force on shear plane, β is friction angle, φ is shear plane angle, α is rake angle.
Difficulty: basic Type: conceptual
Question 18 of 60
Original Exam: IES (2005)
The chip compression ratio (chip thickness ratio) r in orthogonal cutting is always:
✓ Correct Answer: B
Explanation:
Chip thickness ratio r = t1/t2, where t1 is uncut chip thickness and t2 is chip thickness. Since the chip is always thicker than the uncut chip (t2 > t1) due to the chip compression effect, r < 1 always. The chip compression ratio (reciprocal) = t2/t1 = 1/r > 1. Higher r (closer to 1) means thinner chip, larger shear plane angle, lower cutting forces, and more efficient cutting. r is related to shear plane angle: r = sin φ / cos(φ-α). Typical r values: 0.2-0.8.
Difficulty: basic Type: conceptual
Question 19 of 60
Original Exam: GATE (2006)
The process of grinding uses abrasive grains as cutting tools. The effective rake angle of abrasive grains is typically:
✓ Correct Answer: B
Explanation:
Abrasive grains in grinding have random orientations, and most grains present a highly negative rake angle (typically -30° to -60°) to the workpiece. This is because the grains are irregular in shape and are bonded in the wheel with random orientations. The highly negative rake angle causes: (1) High specific cutting energy (10-100× higher than turning), (2) High temperatures, (3) Ploughing and rubbing (not just cutting), (4) Small chip size. The negative rake angle is why grinding requires high power despite small material removal rates.
Difficulty: intermediate Type: conceptual
Question 20 of 60
Original Exam: IES (2007)
The feed in turning is defined as the distance the tool advances per revolution of the workpiece. The feed rate (fr) is:
✓ Correct Answer: B
Explanation:
Feed rate fr = f × N (mm/min), where f is feed (mm/rev) and N is spindle speed (rpm). The feed rate gives the linear velocity of the tool along the workpiece axis. Machining time: tm = L/fr = L/(f × N). MRR = V × f × d = (πDN/1000) × f × d. In CNC machining, feed rate is programmed directly (mm/min). Feed affects: surface roughness (Ra ∝ f²), cutting forces (Fc ∝ f^0.75), and tool life (T ∝ f^(-a/n)).
Difficulty: basic Type: conceptual
Question 21 of 60
Original Exam: GATE (2008)
The depth of cut in turning affects the cutting force approximately as:
✓ Correct Answer: B
Explanation:
Cutting force Fc = Ks × f^x × d^y, where Ks is specific cutting force, f is feed, d is depth of cut. Typically x ≈ 0.75 and y ≈ 1.0. The depth of cut has a nearly linear effect on cutting force (y ≈ 1.0) because increasing d proportionally increases the chip width (b = d for orthogonal cutting). Feed has a less-than-linear effect (x ≈ 0.75) due to the size effect (specific cutting energy increases at small chip thickness). This is why depth of cut is preferred over feed for increasing MRR (less effect on tool life).
Difficulty: intermediate Type: conceptual
Question 22 of 60
Original Exam: IES (2009)
The process of lapping is used to achieve:
✓ Correct Answer: B
Explanation:
Lapping uses loose abrasive (Al2O3, SiC, diamond) in a vehicle (oil, water) between a lap (cast iron, copper) and the workpiece. Lapping achieves: (1) Very high surface finish (Ra 0.025-0.1 μm), (2) Excellent flatness and dimensional accuracy (±0.001-0.005 mm), (3) Correction of minor geometric errors. MRR is very low (0.001-0.01 mm/min). Used for: gauge blocks, optical flats, valve seats, precision bearings. Lapping is a finishing process after grinding. The lap conforms to the workpiece shape.
Difficulty: intermediate Type: conceptual
Question 23 of 60
Original Exam: GATE (2010)
The cutting force in turning can be estimated using the formula Fc = Ks × b × t1, where Ks is the specific cutting force. For Ks = 2000 N/mm², b = 2 mm (width of cut), t1 = 0.2 mm (uncut chip thickness), the cutting force is:
✓ Correct Answer: B
Explanation:
Cutting force Fc = Ks × b × t1 = 2000 × 2 × 0.2 = 800 N. Where Ks is specific cutting force (N/mm²), b is width of cut (mm) = depth of cut in turning, t1 is uncut chip thickness (mm) = feed in turning. This formula is equivalent to Fc = u × f × d (using specific cutting energy u = Ks). Power = Fc × V = 800 × V. For V = 100 m/min = 1.67 m/s: P = 800 × 1.67 = 1333 W ≈ 1.33 kW.
Difficulty: basic Type: numerical
Question 24 of 60
Original Exam: IES (2011)
The process of superfinishing (microhoning) differs from honing in that:
✓ Correct Answer: B
Explanation:
Superfinishing (microhoning) uses: (1) Very light pressure (0.07-0.3 MPa vs 0.5-3 MPa for honing), (2) Short-stroke oscillating motion (1-6 mm amplitude, 500-3000 strokes/min), (3) Fine abrasive stones (Al2O3 or SiC, 400-600 grit). Superfinishing removes only the surface asperities (peaks) left by previous operations, achieving Ra 0.01-0.1 μm. Stock removal: 0.002-0.02 mm. Used for: bearing races, crankshaft journals, camshafts. The process stops automatically when the surface becomes smooth (contact area increases, pressure drops).
Difficulty: intermediate Type: conceptual
Question 25 of 60
Original Exam: GATE (2012)
The grinding ratio (G-ratio) is defined as:
✓ Correct Answer: B
Explanation:
Grinding ratio G = Volume of workpiece removed / Volume of wheel wear. Wait — actually G = Volume of workpiece removed / Volume of wheel wear. A high G-ratio means the wheel removes more material per unit of wheel wear (efficient grinding). Typical G-ratios: conventional abrasives (Al2O3, SiC): G = 20-80, CBN wheels: G = 1000-60,000, diamond wheels: G = 1000-100,000. G-ratio depends on: workpiece material, wheel specification, grinding conditions. High G-ratio is desired for economic grinding.
Difficulty: intermediate Type: conceptual
Question 26 of 60
Original Exam: IES (2013)
The process of electrochemical machining (ECM) follows Faraday's law. The material removal rate in ECM is:
✓ Correct Answer: B
Explanation:
ECM material removal rate from Faraday's law: m = ZIt/F (mass removed), where Z is electrochemical equivalent (g/A·s), I is current (A), t is time (s), F = 96,485 C/mol (Faraday's constant). MRR (volume/time) = ZI/(ρF), where ρ is density. For iron: Z = 56/(2×96485) = 2.9×10⁻⁴ g/A·s. ECM current: 1000-40,000 A. Voltage: 5-30 V. Gap: 0.1-0.5 mm. MRR is proportional to current I (not I²). ECM has no tool wear (tool is cathode, only workpiece dissolves).
Difficulty: intermediate Type: numerical
Question 27 of 60
Original Exam: GATE (2014)
The process of thread cutting on a lathe uses a single-point tool. The lead screw pitch (p_ls) and workpiece thread pitch (p_w) are related by the gear ratio as:
✓ Correct Answer: B
Explanation:
In thread cutting on a lathe: Gear ratio = N_spindle/N_lead_screw = p_w/p_ls, where p_w is the workpiece thread pitch and p_ls is the lead screw pitch. This ensures that for every revolution of the spindle, the tool advances by exactly one thread pitch. Example: to cut M20×2.5 thread (p_w = 2.5 mm) with lead screw pitch p_ls = 5 mm: Gear ratio = 2.5/5 = 1:2 (spindle rotates twice for each lead screw revolution). Change gears are used to achieve the required ratio.
Difficulty: intermediate Type: conceptual
Question 28 of 60
Original Exam: IES (2015)
The process of gear hobbing uses a hob (worm-shaped cutter) to generate gear teeth. The hob and gear blank rotate in a fixed ratio given by:
✓ Correct Answer: B
Explanation:
In gear hobbing, the hob and gear blank rotate in a timed relationship: N_hob/N_blank = Z_blank/Z_hob_starts, where Z_blank is the number of teeth on the gear being cut and Z_hob_starts is the number of starts (threads) on the hob. For a single-start hob: N_hob/N_blank = Z_blank (hob rotates Z_blank times for each revolution of the blank). This ensures each tooth space is generated correctly. Hobbing is a continuous generating process — all teeth are cut simultaneously as the hob and blank rotate.
Difficulty: intermediate Type: conceptual
Question 29 of 60
Original Exam: GATE (2016)
The process of jig boring is used for:
✓ Correct Answer: B
Explanation:
Jig boring machines are precision machine tools used for: (1) Boring holes with very accurate location (positional accuracy ±0.002-0.005 mm), (2) Producing jigs, fixtures, dies, and moulds, (3) Measuring and locating hole positions precisely. Features: (1) High-precision lead screws or optical scales for positioning, (2) Rigid construction to minimise vibration, (3) Temperature-controlled environment. Jig boring replaced the use of jigs for drilling — the machine itself provides the accuracy. Modern CNC machining centres have largely replaced dedicated jig borers.
Difficulty: intermediate Type: conceptual
Question 30 of 60
Original Exam: IES (2017)
The process of reaming is used to:
✓ Correct Answer: B
Explanation:
Reaming is a finishing operation that improves: (1) Dimensional accuracy (tolerance H7, ±0.01-0.02 mm), (2) Surface finish (Ra 0.4-1.6 μm), (3) Roundness and straightness of an existing drilled hole. Stock removal: 0.1-0.5 mm on diameter. Reamers have multiple cutting edges (6-16 flutes) and remove very little material. Types: hand reamer, machine reamer, adjustable reamer, shell reamer. Reaming cannot correct hole location errors (only improves size and finish). Drilling → reaming is a common sequence for precision holes.
Difficulty: basic Type: conceptual
Question 31 of 60
Original Exam: GATE (2001)
The type of chip formed during machining of brittle materials (like cast iron) is:
✓ Correct Answer: C
Explanation:
Discontinuous (segmented) chips are formed when machining brittle materials (cast iron, brass, bronze, hard plastics) because: (1) Brittle materials fracture rather than deform plastically, (2) The chip breaks into small segments as it forms. Discontinuous chips are also formed at: very low cutting speeds, large feed, small rake angle, or when machining ductile materials with poor cutting conditions. Continuous chips form with ductile materials at high speeds. Continuous chips with BUE form at medium speeds with ductile materials.
Difficulty: basic Type: conceptual
Question 32 of 60
Original Exam: IES (2002)
The process of up-milling (conventional milling) differs from down-milling (climb milling) in that:
✓ Correct Answer: C
Explanation:
Up-milling (conventional): cutter rotates against the feed direction. Chip thickness increases from 0 (at entry) to maximum (at exit). The cutter tends to lift the workpiece (upward force). Advantages: no backlash problems, suitable for rough surfaces. Down-milling (climb): cutter rotates with the feed direction. Chip thickness decreases from maximum (at entry) to 0 (at exit). The cutter pushes the workpiece down (better clamping). Advantages: better surface finish, longer tool life, lower cutting forces. Down-milling requires backlash-free machine.
Difficulty: intermediate Type: conceptual
Question 33 of 60
Original Exam: GATE (2003)
The process of cylindrical grinding uses a grinding wheel to finish cylindrical surfaces. The infeed (plunge) grinding differs from traverse grinding in that:
✓ Correct Answer: C
Explanation:
Infeed (plunge) grinding: the grinding wheel is fed radially (perpendicular to workpiece axis) into the workpiece without axial traverse. Used for: (1) Short workpieces where wheel width covers the entire length, (2) Form grinding (wheel profile is reproduced on workpiece), (3) Grinding of shoulders and grooves. Traverse grinding: the wheel traverses axially along the workpiece while rotating. Used for: long workpieces, better surface finish (overlapping passes). Infeed grinding is faster but requires a wider wheel and produces a characteristic pattern.
Difficulty: intermediate Type: conceptual
Question 34 of 60
Original Exam: IES (2004)
The process of tapping is used to produce:
✓ Correct Answer: C
Explanation:
Tapping produces internal threads in a pre-drilled hole using a tap (a hardened, threaded tool with cutting edges). The tap is rotated into the hole, cutting threads as it advances. Types: hand tap (taper, plug, bottoming), machine tap, spiral flute tap (for blind holes), spiral point tap (for through holes). Tap drill size = thread major diameter - thread pitch (approximately). For M10×1.5: tap drill = 10 - 1.5 = 8.5 mm. External threads are produced by: die (manual), thread cutting on lathe, thread rolling, thread milling.
Difficulty: basic Type: conceptual
Question 35 of 60
Original Exam: GATE (2005)
The process of gear shaping uses a reciprocating gear-shaped cutter. The gear shaper can produce:
✓ Correct Answer: C
Explanation:
Gear shaping (Fellows gear shaper) uses a gear-shaped cutter that reciprocates axially while rotating. It can produce: (1) External spur gears, (2) Internal spur gears (unique advantage over hobbing), (3) Helical gears (with helical guide), (4) Cluster gears, (5) Gears with shoulders. Gear shaping cannot produce worm gears or bevel gears. Hobbing is faster but cannot produce internal gears. Gear shaping is preferred for: internal gears, cluster gears, and gears with obstructions near the gear face.
Difficulty: intermediate Type: conceptual
Question 36 of 60
Original Exam: IES (2006)
The process of surface grinding uses a grinding wheel to produce flat surfaces. The wheel speed in surface grinding is typically:
✓ Correct Answer: C
Explanation:
In surface grinding: Wheel peripheral speed: 20-35 m/s (1200-2100 m/min) — this is the cutting speed. Workpiece speed (table speed): 0.1-0.5 m/s (6-30 m/min) — much lower. The high wheel speed ensures efficient material removal. Infeed (depth of cut): 0.005-0.05 mm per pass. Cross-feed: 1/3 to 3/4 of wheel width per pass. The wheel speed is limited by: wheel strength (bursting speed), machine power, and vibration. High-speed grinding (HEDG) uses wheel speeds up to 200 m/s.
Difficulty: basic Type: conceptual
Question 37 of 60
Original Exam: GATE (2007)
The process of EDM (Electric Discharge Machining) can machine:
✓ Correct Answer: C
Explanation:
EDM can machine any electrically conductive material regardless of hardness, including: hardened steel, carbides (WC-Co), titanium alloys, Inconel, copper, aluminium. The material removal is by thermal erosion (spark erosion), not mechanical cutting, so hardness is irrelevant. EDM cannot machine non-conductive materials (ceramics, polymers, glass) — for these, USM or LBM is used. EDM is particularly valuable for: hardened dies and moulds, complex shapes, deep narrow slots, and materials that are difficult to machine conventionally.
Difficulty: basic Type: conceptual
Question 38 of 60
Original Exam: IES (2008)
The process of chemical machining (CHM) removes material by:
✓ Correct Answer: C
Explanation:
Chemical Machining (CHM) removes material by controlled chemical etching: (1) Clean workpiece surface, (2) Apply maskant (rubber, neoprene, polyvinyl chloride) to protect areas not to be etched, (3) Scribe and peel maskant from areas to be etched, (4) Immerse in etchant (FeCl3 for aluminium, HNO3+HF for titanium, NaOH for aluminium), (5) Remove maskant and clean. Applications: aircraft skin panels (weight reduction), printed circuit boards, decorative etching. Advantages: no mechanical stresses, uniform material removal, complex shapes. Limitation: slow, hazardous chemicals.
Difficulty: intermediate Type: conceptual
Question 39 of 60
Original Exam: GATE (2009)
The process of centerless grinding differs from cylindrical grinding in that:
✓ Correct Answer: C
Explanation:
Centerless grinding: the workpiece is supported by a work rest blade (below) and a regulating wheel (smaller, rubber-bonded, tilted at 1-5°). The grinding wheel removes material while the regulating wheel controls workpiece rotation and feed rate. Advantages over cylindrical grinding: (1) No need for centres or chucks, (2) Continuous operation (through-feed), (3) Higher production rate, (4) Suitable for long, slender workpieces. Disadvantages: cannot grind eccentric parts, cannot grind blind holes, setup is complex. Used for: pins, rollers, shafts, needles.
Difficulty: intermediate Type: conceptual
Question 40 of 60
Original Exam: IES (2010)
The process of boring on a lathe differs from drilling in that:
✓ Correct Answer: C
Explanation:
Boring on a lathe: (1) Enlarges an existing hole (drilled or cast), (2) Uses a single-point boring tool, (3) Improves dimensional accuracy, roundness, and straightness, (4) Can correct hole location errors (unlike reaming). Boring can produce holes of any diameter (limited only by machine capacity). Drilling produces holes in solid material using a twist drill (multi-point tool). Boring is preferred for: large holes, accurate holes, holes with tight tolerances. Boring bar must be rigid to avoid chatter and deflection.
Difficulty: basic Type: conceptual
Question 41 of 60
Original Exam: GATE (2011)
The process of gear lapping is used to:
✓ Correct Answer: C
Explanation:
Gear lapping is a finishing process used after heat treatment to: (1) Improve surface finish of gear tooth flanks (Ra 0.1-0.4 μm), (2) Correct minor errors in tooth profile and spacing, (3) Improve gear noise and vibration characteristics. Process: the gear is run in mesh with a cast iron lap gear with abrasive compound. Gear lapping removes very little material (0.005-0.025 mm). It cannot correct large errors. Gear grinding is used for larger corrections. Gear shaving is used before heat treatment for finishing.
Difficulty: intermediate Type: conceptual
Question 42 of 60
Original Exam: IES (2012)
The process of thread rolling produces threads by:
✓ Correct Answer: C
Explanation:
Thread rolling is a cold forming process: hardened dies (flat dies or cylindrical dies) press against the workpiece, plastically deforming the material to form threads. Advantages over thread cutting: (1) Stronger threads (work hardening, compressive residual stresses), (2) Better surface finish (Ra 0.1-0.4 μm), (3) No material waste (material displaced, not removed), (4) Higher production rate, (5) Longer fatigue life. Limitations: only for ductile materials, requires precise blank diameter. Used for: bolts, screws, studs. Thread rolling is the preferred method for high-volume production.
Difficulty: intermediate Type: conceptual
Question 43 of 60
Original Exam: GATE (2013)
The process of electrical discharge grinding (EDG) is used for:
✓ Correct Answer: C
Explanation:
Electrical Discharge Grinding (EDG) combines EDM with a rotating wheel (graphite or metal-bonded) as the electrode. The wheel rotates but does not contact the workpiece — material is removed by spark erosion. Used for: (1) Grinding of carbides and hardened steels, (2) Form grinding of complex profiles, (3) Grinding of thin, fragile parts (no mechanical forces). Advantages: no grinding forces, no wheel loading, can grind very hard materials. Disadvantages: recast layer, lower surface finish than conventional grinding. EDG is different from ECG (electrochemical grinding).
Difficulty: intermediate Type: conceptual
Question 44 of 60
Original Exam: IES (2014)
The process of water jet machining (WJM) uses:
✓ Correct Answer: C
Explanation:
Water Jet Machining (WJM) uses a high-velocity, high-pressure water jet (pressure: 200-400 MPa, velocity: 600-900 m/s, nozzle diameter: 0.1-0.4 mm) to cut soft, non-metallic materials: rubber, foam, leather, textiles, food products, paper. No heat generation (cold cutting). Abrasive Water Jet Machining (AWJM) adds abrasive particles (garnet, Al2O3) to cut hard materials (metals, ceramics, composites). WJM advantages: no HAZ, no dust, can cut complex shapes, environmentally friendly. Limitation: cannot cut very hard materials without abrasive.
Difficulty: basic Type: conceptual
Question 45 of 60
Original Exam: GATE (2015)
The process of plasma arc machining (PAM) uses:
✓ Correct Answer: C
Explanation:
Plasma Arc Machining (PAM) uses a plasma jet (ionised gas at 8,000-28,000°C) to melt and remove material. The plasma is created by passing gas (Ar, N2, H2, or mixtures) through an electric arc. PAM can cut: any electrically conductive material (steel, stainless steel, aluminium, titanium, copper). Advantages: high MRR, can cut thick plates (up to 150 mm), faster than laser for thick sections. Disadvantages: large HAZ, poor surface finish, limited to conductive materials, produces fumes. Used for: structural steel cutting, shipbuilding, heavy fabrication.
Difficulty: intermediate Type: conceptual
Question 46 of 60
Original Exam: IES (2003)
The process of orthogonal cutting assumes:
✓ Correct Answer: D
Explanation:
Orthogonal cutting assumptions: (1) The cutting edge is straight and perpendicular to the cutting velocity direction, (2) The chip flows in the direction perpendicular to the cutting edge (no side flow), (3) The width of cut is much larger than the depth of cut (plane strain condition), (4) The tool is perfectly sharp (no nose radius), (5) Uniform chip thickness. Orthogonal cutting is a 2D simplification used for analysis. Oblique cutting (3D) occurs when the cutting edge is inclined to the cutting velocity (inclination angle ≠ 0°).
Difficulty: basic Type: conceptual
Question 47 of 60
Original Exam: GATE (2004)
The process of turning on a lathe produces a surface by:
✓ Correct Answer: D
Explanation:
Turning on a lathe: (1) Workpiece rotates (primary cutting motion), (2) Single-point cutting tool feeds axially (feed motion), (3) Tool is set at a depth of cut (depth of cut motion). The combination produces a helical cutting path. Operations on a lathe: turning (external cylindrical), facing (flat end surface), boring (internal cylindrical), threading, knurling, parting, taper turning. The lathe is the most versatile machine tool. Cutting speed V = πDN/1000, feed f (mm/rev), depth of cut d (mm).
Difficulty: basic Type: conceptual
Question 48 of 60
Original Exam: IES (2005)
The process of milling produces surfaces by:
✓ Correct Answer: D
Explanation:
Milling: (1) Multi-point rotating cutter (milling cutter) provides the primary cutting motion, (2) Workpiece feeds against the rotating cutter (feed motion). Each tooth of the cutter removes a small chip. Types: peripheral milling (horizontal axis), face milling (vertical axis), end milling. Milling can produce: flat surfaces, slots, pockets, contours, gears, threads. The milling cutter has multiple teeth, giving higher MRR than turning. Cutting speed V = πDN/1000 (D = cutter diameter). Feed per tooth fz = fr/(N × z), where z is number of teeth.
Difficulty: basic Type: conceptual
Question 49 of 60
Original Exam: GATE (2006)
The process of grinding is classified as a precision finishing process because:
✓ Correct Answer: D
Explanation:
Grinding is a precision finishing process because: (1) Tight dimensional tolerances: IT5-IT7 (±0.005-0.025 mm), (2) Fine surface finish: Ra 0.1-1.6 μm (vs. turning: Ra 0.8-6.3 μm), (3) Can machine hardened materials (after heat treatment), (4) Corrects geometric errors (roundness, flatness, straightness). Grinding uses abrasive wheels with thousands of cutting edges. Specific energy is 10-100× higher than turning. Used as a finishing operation after turning, milling, or heat treatment. Types: cylindrical, surface, centreless, internal, tool and cutter grinding.
Difficulty: basic Type: conceptual
Question 50 of 60
Original Exam: IES (2007)
The process of knurling on a lathe produces:
✓ Correct Answer: D
Explanation:
Knurling is a forming (not cutting) operation on a lathe that produces a textured pattern (diamond or straight) on the workpiece surface by pressing hardened knurling wheels against the rotating workpiece. The wheels plastically deform the surface, raising ridges. Knurling increases the diameter slightly (no material removed). Used for: handles, knobs, tool grips, and surfaces requiring improved grip. Knurling wheels are available in different pitches (fine, medium, coarse). The process requires high radial force and a rigid lathe.
Difficulty: basic Type: conceptual
Question 51 of 60
Original Exam: GATE (2008)
The process of gear hobbing is a generating process because:
✓ Correct Answer: D
Explanation:
Gear hobbing is a generating process: the involute tooth profile is generated by the envelope of successive cutting positions of the hob teeth as the hob and blank rotate in a timed relationship. The hob teeth are straight (rack-shaped in the normal plane), but the involute profile is generated by the relative motion. This is different from form cutting (gear milling) where the cutter profile matches the tooth space. Generating processes (hobbing, shaping, shaving) produce more accurate gears than form cutting because the profile is generated mathematically.
Difficulty: intermediate Type: conceptual
Question 52 of 60
Original Exam: IES (2009)
The process of abrasive water jet machining (AWJM) is used for:
✓ Correct Answer: D
Explanation:
Abrasive Water Jet Machining (AWJM) adds abrasive particles (garnet, Al2O3, SiC, 80-200 mesh) to the water jet, enabling cutting of hard materials: metals, ceramics, composites, glass, stone. Advantages: (1) No heat generation (cold cutting — no HAZ, no thermal distortion), (2) No mechanical stresses, (3) Can cut any material, (4) Complex shapes possible, (5) Environmentally friendly (no toxic fumes). Limitations: slow for thick sections, abrasive cost, noise, wet process. Used for: aerospace composites, armour plate, stone cutting, food processing.
Difficulty: intermediate Type: conceptual
Question 53 of 60
Original Exam: GATE (2010)
The process of ion beam machining (IBM) is used for:
✓ Correct Answer: D
Explanation:
Ion Beam Machining (IBM) / Focused Ion Beam (FIB) uses a focused beam of ions (Ga+, Ar+) accelerated to high energy (5-30 keV) to remove material atom by atom by sputtering. Applications: (1) Semiconductor device modification and repair, (2) TEM sample preparation, (3) Nano-scale machining, (4) Surface analysis (combined with SEM). MRR is extremely low (nm³/s). Achieves atomic-level precision. Used in: microelectronics, MEMS, nanotechnology. FIB can also deposit material (ion beam deposition). Not suitable for large-scale machining.
Type: conceptual
Question 54 of 60
Original Exam: IES (2011)
The process of magnetic abrasive finishing (MAF) uses:
✓ Correct Answer: D
Explanation:
Magnetic Abrasive Finishing (MAF) uses magnetic abrasive particles (iron particles bonded with abrasive like Al2O3 or SiC) that form a flexible 'brush' under a magnetic field. The magnetic field controls the pressure and arrangement of abrasive particles. The workpiece rotates and the magnetic brush finishes the surface. Achieves: Ra 0.01-0.1 μm, removes burrs, finishes complex surfaces. Used for: finishing of cylindrical surfaces, internal surfaces, flat surfaces, and complex shapes. Advantages: flexible tool, no tool wear, can finish complex geometries.
Type: conceptual
Question 55 of 60
Original Exam: GATE (2012)
The process of electrochemical honing (ECH) combines:
✓ Correct Answer: D
Explanation:
Electrochemical Honing (ECH) combines: (1) ECM (anodic dissolution) — removes 90% of material rapidly, (2) Conventional honing (abrasive stones) — removes the oxide film and provides dimensional accuracy and surface finish. The honing tool is the cathode; workpiece is the anode; electrolyte flows through the tool. Advantages over conventional honing: (1) Higher MRR (3-5× faster), (2) No glazing of abrasive stones, (3) No thermal damage, (4) Better surface finish. Used for: engine cylinder bores, hydraulic cylinders, gear teeth finishing.
Difficulty: intermediate Type: conceptual
Question 56 of 60
Original Exam: IES (2013)
The process of vibration-assisted machining (VAM) improves machining by:
✓ Correct Answer: D
Explanation:
Vibration-Assisted Machining (VAM) superimposes high-frequency vibration (ultrasonic: 20-40 kHz, amplitude 1-20 μm) on the cutting tool. Benefits: (1) Reduced cutting forces (30-50%), (2) Improved surface finish, (3) Reduced tool wear, (4) Better chip breaking, (5) Enables machining of brittle materials (glass, ceramics) and difficult-to-cut materials (titanium, CFRP). Mechanism: intermittent cutting (tool separates from workpiece each cycle), reducing friction and heat. Used in: precision turning, drilling of CFRP, micro-machining.
Type: conceptual
Question 57 of 60
Original Exam: GATE (2014)
The process of hard turning (turning of hardened steel, HRC > 45) using CBN tools is used as an alternative to grinding because:
✓ Correct Answer: D
Explanation:
Hard turning (HRC 45-65) using CBN (Cubic Boron Nitride) or ceramic tools can replace cylindrical grinding for certain applications: (1) Higher MRR than grinding (10-100× faster), (2) Single setup (no separate grinding machine), (3) Flexible (CNC turning centre), (4) Lower tooling cost than CBN grinding wheels, (5) Comparable accuracy (IT6-IT7) and surface finish (Ra 0.4-1.6 μm). Limitations: (1) Cannot match grinding accuracy for IT5 or better, (2) White layer formation (thermal damage), (3) Tool cost is high. Best for: medium-volume production of hardened components.
Difficulty: intermediate Type: conceptual
Question 58 of 60
Original Exam: IES (2015)
The process of micro-machining differs from conventional machining in that:
✓ Correct Answer: D
Explanation:
Micro-machining (chip thickness < 1 μm) differs from conventional machining: (1) Size effect — specific cutting energy increases dramatically at small chip thickness (chip thickness < cutting edge radius), (2) Ploughing dominates over cutting (material is pushed aside rather than cut), (3) Minimum chip thickness effect — below a critical chip thickness, no chip forms, (4) Grain size effects become significant, (5) Surface integrity is critical. Micro-machining requires: ultra-sharp tools (edge radius < 1 μm), high-precision machines, vibration isolation. Used for: MEMS, optical components, biomedical devices.
Type: conceptual
Question 59 of 60
Original Exam: GATE (2016)
The process of cryogenic machining uses liquid nitrogen (LN2) as a coolant to:
✓ Correct Answer: D
Explanation:
Cryogenic machining uses liquid nitrogen (LN2, -196°C) or liquid CO2 as coolant, applied directly to the cutting zone. Benefits: (1) Dramatically reduces cutting temperature, (2) Reduces tool wear (especially for titanium, Inconel, CFRP), (3) Improves surface integrity (compressive residual stresses), (4) Eliminates cutting fluid disposal problems (LN2 evaporates), (5) Improves chip breakability. Particularly effective for: titanium alloys (Ti-6Al-4V), nickel superalloys (Inconel 718), CFRP composites. Limitations: cost of LN2, special delivery system required.
Difficulty: intermediate Type: conceptual
Question 60 of 60
Original Exam: IES (2016)
The process of minimum quantity lubrication (MQL) in machining uses:
✓ Correct Answer: D
Explanation:
Minimum Quantity Lubrication (MQL) / Near-Dry Machining uses: (1) Very small amount of lubricant (50-500 mL/h vs. 10-100 L/min for flood cooling), (2) Mixed with compressed air (aerosol), (3) Directed precisely at the cutting zone. Benefits: (1) Reduces friction and tool wear (lubrication effect), (2) Minimal coolant cost and disposal, (3) Cleaner chips (easier recycling), (4) Better for environment and operator health. Limitations: less cooling than flood coolant. MQL is a compromise between dry machining (no lubrication) and flood cooling (excessive coolant). Used for: aluminium, cast iron, steel machining.
Difficulty: intermediate Type: conceptual