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Maharashtra HSC Mechanical Maintenance (Bifocal) study materials
Maharashtra HSC Mechanical Maintenance (Bifocal) study resources

Maharashtra HSC Mechanical Maintenance (Bifocal) practice questions

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Question 1Engineering Drawing and Workshop Calculation

In first-angle projection, where is the top view placed relative to the front view?

  1. Above the front view
  2. To the right of the front view
  3. Below the front view Correct answer.
  4. To the left of the front view

Explanation

In first-angle projection the object is imagined between the observer and the plane, so the view seen from above is projected onto a plane below the object and appears below the front view. This is why first-angle layouts place the top view under the front view and the left side view on the right.

Question 2Engineering Drawing and Workshop Calculation

A workshop drawing carries the note 'Scale 1:5'. A measured length of 24 mm on the drawing represents what actual length on the component?

  1. 240 mm
  2. 120 mm Correct answer.
  3. 4.8 mm
  4. 29 mm

Explanation

A scale of 1:5 is a reducing scale, meaning one unit on the drawing equals five units on the real object. Multiplying the drawn length by the scale factor gives 24 × 5 = 120 mm. Dividing instead would wrongly enlarge the drawing.

Question 3Engineering Drawing and Workshop Calculation

Which line type is used in an engineering drawing to show hidden internal features such as a drilled hole that is not visible in the view?

  1. Continuous thick line
  2. Continuous thin wavy line
  3. Dashed thin line Correct answer.
  4. Chain thin line

Explanation

Hidden edges and outlines are drawn with short dashes of medium thickness so they can be distinguished from visible outlines. The dash pattern signals that the feature lies behind the visible surface. Continuous thick lines are reserved for visible outlines.

Question 4Engineering Drawing and Workshop Calculation

A hexagonal head bolt must be drawn in a view where the head appears as a regular hexagon. Which construction gives the correct shape?

  1. A circle divided into six equal parts by stepping the radius around the circumference Correct answer.
  2. Two concentric circles joined by radial lines
  3. A circle divided into six parts by stepping the diameter around the circumference
  4. A square with its corners joined to the centre

Explanation

For a regular hexagon inscribed in a circle, the side length equals the circumradius, so stepping the radius six times around the circumference locates all six vertices. This is the standard workshop construction for a bolt head across corners.

Question 5Engineering Drawing and Workshop Calculation

In a drawing, a dimension is written as 'Ø40 H7/g6'. What does the symbol Ø indicate?

  1. The feature is a radius
  2. The feature is a diameter Correct answer.
  3. The feature is a square section
  4. The feature is a spherical surface

Explanation

The symbol Ø precedes a dimension to show that the value is a diameter of a cylindrical feature. The letters H7 and g6 that follow are ISO tolerance designations for the hole and shaft respectively, but the diameter symbol itself only fixes the type of size.

Question 6Engineering Drawing and Workshop Calculation

A shaft measures 24.96 mm and its permitted limits are 25.00 mm maximum and 24.90 mm minimum. What is the actual deviation of the shaft from the basic size of 25 mm?

  1. −0.04 mm Correct answer.
  2. +0.06 mm
  3. −0.10 mm
  4. +0.04 mm

Explanation

Deviation is the algebraic difference between the actual size and the basic size, so 24.96 − 25.00 = −0.04 mm. The negative sign shows the shaft is undersize but still inside the permitted limits, so the part is acceptable.

Question 7Engineering Drawing and Workshop Calculation

A steel plate of length 1.2 m, width 0.5 m and thickness 8 mm is made of material with density 7850 kg/m³. What is its mass?

  1. 376.8 kg
  2. 0.03768 kg
  3. 37.68 kg Correct answer.
  4. 3.768 kg

Explanation

Volume equals 1.2 × 0.5 × 0.008 = 0.0048 m³, and mass equals volume times density, so 0.0048 × 7850 = 37.68 kg. Thickness must be converted to metres before multiplying, otherwise the units are inconsistent.

Question 8Engineering Drawing and Workshop Calculation

A taper has a large diameter of 60 mm, a small diameter of 50 mm and a length of 100 mm. What is its taper ratio expressed as 1 in x?

  1. 1 in 10 Correct answer.
  2. 1 in 2
  3. 1 in 5
  4. 1 in 20

Explanation

Taper is the difference in diameter divided by the length, so (60 − 50)/100 = 10/100 = 1/10, giving a taper of 1 in 10. Note that conicity, the half-angle measure, would use half this difference and give 1 in 20.

Question 9Engineering Drawing and Workshop Calculation

A drawing shows a metric thread marked M20 × 2.5. What does the value 2.5 represent?

  1. The thread length in millimetres
  2. The depth of the thread in millimetres
  3. The nominal diameter in millimetres
  4. The pitch of the thread in millimetres Correct answer.

Explanation

In the ISO metric thread designation the first number is the nominal diameter and the number after the multiplication sign is the pitch. For M20 × 2.5 the major diameter is 20 mm and the pitch is 2.5 mm, which is the coarse pitch for that size.

Question 10Engineering Drawing and Workshop Calculation

A spur gear has 40 teeth and a module of 2.5 mm. What is its pitch circle diameter?

  1. 100 mm Correct answer.
  2. 40 mm
  3. 250 mm
  4. 16 mm

Explanation

Pitch circle diameter equals module multiplied by the number of teeth, so 2.5 × 40 = 100 mm. The module is the ratio of pitch circle diameter to tooth number, which is why this single multiplication gives the answer.

Question 11Engineering Drawing and Workshop Calculation

A section view of a machined steel part is required. Which hatching convention is correct?

  1. Thick continuous lines parallel to the main outlines
  2. Thin continuous lines inclined at 45° to the main outlines, evenly spaced Correct answer.
  3. Dashed lines at 60° with wide spacing
  4. Chain lines crossing at right angles

Explanation

Cut surfaces are hatched with thin continuous lines, usually at 45° to the principal outlines of the part, and spaced evenly for clarity. Adjacent parts in an assembly are hatched at different angles or spacings so the boundary between them is clear.

Question 12Engineering Drawing and Workshop Calculation

In a drawing, a hole is dimensioned '4 × Ø10 THRU'. What does this note tell the workshop?

  1. Four 10 mm counterbores of through depth
  2. Four holes of 10 mm diameter drilled through the part Correct answer.
  3. Four holes of 10 mm depth drilled part way
  4. A single 40 mm hole drilled through the part

Explanation

When several identical features are repeated, the quantity precedes the size, so 4 × Ø10 means four holes each 10 mm in diameter. The word THRU shows the holes pass completely through the material rather than to a stated depth.

Question 13Engineering Drawing and Workshop Calculation

A right regular hexagonal prism of side 25 mm and axis length 60 mm rests on one of its base edges on the horizontal plane. Which view is drawn first when using the change-of-position method?

  1. The view showing the true shape of the hexagon in the vertical plane
  2. The view with the axis perpendicular to the horizontal plane
  3. The view with the axis inclined at 45° to both planes
  4. The view with the axis parallel to the vertical plane and the base edge perpendicular to it Correct answer.

Explanation

In the change-of-position method the solid is first drawn in the simple position with its axis parallel to the vertical plane, then tilted so that the required condition, here resting on a base edge, is satisfied in the second stage. This staged approach keeps the true lengths and shapes accessible at each step.

Question 14Engineering Drawing and Workshop Calculation

A plain turning job of diameter 50 mm is machined at a cutting speed of 30 m/min. Using π = 3.14, what spindle speed should be set?

  1. 95 rev/min
  2. 382 rev/min
  3. 191 rev/min Correct answer.
  4. 600 rev/min

Explanation

Spindle speed equals cutting speed divided by the circumference, N = 1000V/(πD) = 1000 × 30/(3.14 × 50), which gives about 191 rev/min. The diameter must be in millimetres because of the 1000 factor converting metres to millimetres.

Question 15Engineering Drawing and Workshop Calculation

A hole and shaft are designated 40 H8/f7. What type of fit does this combination produce?

  1. Transition fit that may give clearance or interference
  2. Clearance fit with the hole always larger than the shaft Correct answer.
  3. Line-to-line fit with zero clearance in all cases
  4. Interference fit with the shaft always larger than the hole

Explanation

The hole tolerance H8 starts at the basic size and goes plus, while the shaft tolerance f7 lies wholly below the basic size. Because the shaft maximum is smaller than the hole minimum, clearance always exists and the assembly is a running clearance fit.

Question 16Engineering Drawing and Workshop Calculation

On a drawing, a surface is marked with the symbol Ra 1.6. What does this specify?

  1. The total tolerance on the surface is 1.6 mm
  2. The surface hardness is 1.6 on the Rockwell scale
  3. The maximum permissible arithmetic mean roughness is 1.6 micrometres Correct answer.
  4. The minimum material removal is 1.6 mm

Explanation

Ra denotes the arithmetic mean roughness value of the surface profile, expressed in micrometres, so Ra 1.6 limits the average roughness to 1.6 µm. This is a surface texture requirement and is independent of dimensional tolerances.

Question 17Engineering Drawing and Workshop Calculation

A drawing shows a welded lap joint with the symbol for a fillet weld, a triangle with the size 6 placed to the left of the symbol and a length of 50 to the right. What does this mean?

  1. A 6 mm butt weld of 50 mm length
  2. A 6 mm fillet weld of 50 mm length Correct answer.
  3. A 50 mm fillet weld of 6 mm length
  4. A 6 mm fillet weld applied all round

Explanation

In the standard welding symbol the size of the weld, here the 6 mm leg of the fillet, is written to the left of the weld symbol and the length to the right. So the note calls for a 6 mm fillet weld running for 50 mm.

Question 18Engineering Drawing and Workshop Calculation

A conical funnel is to be developed from a flat sheet. Which method is appropriate for finding the true lengths needed for the development?

  1. Triangulation method
  2. Radial line method Correct answer.
  3. Parallel line method
  4. Offset method

Explanation

The radial line method is used for solids whose surfaces converge to an apex, such as cones and pyramids, because all generators radiate from that apex. The development is then an arc of radius equal to the slant height, with the arc length equal to the base circumference.

Question 19Engineering Drawing and Workshop Calculation

Two views of a component show a line that appears as a point in the front view. What does this indicate about the edge it represents?

  1. The edge is inclined to both reference planes
  2. The edge is parallel to the frontal plane and appears true length
  3. The edge lies in the frontal plane at an angle
  4. The edge is perpendicular to the frontal plane and appears as a point Correct answer.

Explanation

A line projects as a point only when the line of sight is along the line itself, which happens when the edge is perpendicular to the plane of projection. In the adjacent view the same edge will appear at true length, confirming its orientation.

Question 20Engineering Drawing and Workshop Calculation

A drawing uses a scale of 1:2.5 and the drawing sheet measures 420 mm by 297 mm. What actual length does a drawn line of 168 mm represent, and is this a reducing or enlarging scale?

  1. 420 mm on a reducing scale Correct answer.
  2. 67.2 mm on a reducing scale
  3. 168 mm on a full scale
  4. 420 mm on an enlarging scale

Explanation

A ratio of 1:2.5 means one unit on the drawing stands for 2.5 units in reality, so 168 × 2.5 = 420 mm. Because the representative fraction is less than one, the drawing is smaller than the object and the scale is reducing.

Question 21Engineering Materials and Heat Treatment

In a workshop, a mild steel component is to be case hardened by carburising. Which property change is the main purpose of this treatment?

  1. Increasing the hardness of the core while keeping the surface soft
  2. Relieving internal stresses after welding throughout the section
  3. Increasing the hardness of the surface while keeping the core tough Correct answer.
  4. Reducing the carbon content of the surface to improve weldability

Explanation

Carburising raises the carbon content of the outer layer of low-carbon steel, so that after quenching the surface becomes hard and wear resistant while the low-carbon core stays tough and shock resistant. This combination of hard skin and ductile core is the reason gears and shafts are carburised. The carbon is added to the surface only, not removed, and stress relief is a separate treatment.

Question 22Engineering Materials and Heat Treatment

A steel specimen is heated to 30 degrees C above its upper critical temperature and then cooled in still air. Which heat treatment has been carried out?

  1. Annealing
  2. Austempering
  3. Normalising Correct answer.
  4. Tempering

Explanation

Normalising means heating steel about 30 to 50 degrees C above the upper critical temperature, holding for a uniform structure, and cooling in still air. It refines the grain structure and gives a more uniform, moderately harder structure than annealing. Tempering is done below the critical range after hardening, and austempering uses a molten salt bath.

Question 23Engineering Materials and Heat Treatment

Which of the following is the correct approximate carbon range for medium carbon steel used for shafts and axles?

  1. 1.20 to 1.60 percent carbon
  2. 0.30 to 0.60 percent carbon Correct answer.
  3. 0.05 to 0.15 percent carbon
  4. 2.50 to 3.50 percent carbon

Explanation

Medium carbon steels contain roughly 0.30 to 0.60 percent carbon, which gives a balance of strength and toughness suitable for shafts, axles and connecting rods after suitable heat treatment. Low carbon steels are below about 0.25 percent, and high carbon steels run from about 0.6 to 1.4 percent. Carbon above about 2 percent defines cast iron, not steel.

Question 24Engineering Materials and Heat Treatment

During a hardness test a hardened steel part is pressed with a 10 mm hardened steel ball under a 3000 kg load, and the diameter of the impression is measured. Which hardness scale is being used?

  1. Shore hardness
  2. Brinell hardness Correct answer.
  3. Rockwell C hardness
  4. Vickers hardness

Explanation

The Brinell test uses a hardened steel or carbide ball, commonly 10 mm diameter, with standard loads such as 3000 kg, and the hardness is calculated from the measured diameter of the spherical impression. Vickers uses a diamond pyramid and Rockwell C uses a diamond cone with depth measurement. Shore hardness is used mainly for rubbers and plastics.

Question 25Engineering Materials and Heat Treatment

A machined mild steel bracket is to be given a stress-relief treatment after welding. Which treatment is appropriate?

  1. Cold working the weld zone with a hammer
  2. Heating above the upper critical temperature and quenching in water
  3. Heating to about 600 to 650 degrees C, holding, then slow cooling Correct answer.
  4. Heating to 950 degrees C and cooling in still air

Explanation

Stress relief is a sub-critical treatment at roughly 600 to 650 degrees C, held long enough to allow residual stresses to relax, followed by slow cooling. The steel is not taken above the critical range, so the structure and hardness are largely retained. Quenching would introduce new stresses and hardening instead of relieving them.

Question 26Engineering Materials and Heat Treatment

Which iron is characterised by carbon present mainly as free graphite flakes, giving good damping capacity and machinability?

  1. Grey cast iron Correct answer.
  2. Wrought iron
  3. Mild steel
  4. White cast iron

Explanation

In grey cast iron most of the carbon is present as free graphite in flake form, which gives the characteristic grey fracture. The graphite flakes provide excellent damping capacity, good machinability and useful compressive strength, though tensile strength and ductility are low. White cast iron has carbon combined as cementite, and wrought iron is a low-carbon ferrous product with slag inclusions.

Question 27Engineering Materials and Heat Treatment

A steel component has been hardened and is now too brittle for service. Which treatment should follow hardening to improve toughness?

  1. Tempering Correct answer.
  2. Normalising
  3. Carburising
  4. Cyaniding

Explanation

Hardening produces martensite, which is very hard but brittle and contains high internal stress. Tempering reheats the part to a temperature below the critical range, allowing the martensite to transform partly and toughness to rise at some sacrifice of hardness. Carburising and cyaniding are surface-hardening treatments applied before hardening, and normalising is a separate softening and refining treatment.

Question 28Engineering Materials and Heat Treatment

Which quenching medium generally produces the slowest cooling rate and therefore the least severe hardening of steel?

  1. Brine solution
  2. Still air Correct answer.
  3. Cold water
  4. Agitated water

Explanation

Cooling severity depends on the medium and its agitation: brine is the most severe, followed by water, then oil, and still air is the mildest. Slow cooling in still air gives lower hardness and less risk of cracking, which is why it is used for normalising and for steels that do not need a fully martensitic structure. Water and brine cool fast enough to harden but also increase distortion and cracking risk.

Question 29Engineering Materials and Heat Treatment

In the iron-carbon equilibrium diagram, the eutectoid point occurs at approximately what carbon content and temperature?

  1. 0.02 percent carbon and 1147 degrees C
  2. 0.8 percent carbon and 723 degrees C Correct answer.
  3. 2.1 percent carbon and 723 degrees C
  4. 0.8 percent carbon and 1147 degrees C

Explanation

The eutectoid reaction in plain carbon steel occurs at about 0.8 percent carbon and 723 degrees C, where austenite transforms directly into pearlite, a lamellar mixture of ferrite and cementite. Steels below this carbon are hypoeutectoid and contain proeutectoid ferrite, while those above are hypereutectoid and contain proeutectoid cementite. The 1147 degrees C line is the eutectic temperature of the cast iron region.

Question 30Engineering Materials and Heat Treatment

A steel with 0.4 percent carbon is slowly cooled from the austenitic region. Which microstructural constituent forms first at the austenite grain boundaries before pearlite?

  1. Proeutectoid ferrite Correct answer.
  2. Martensite
  3. Ledeburite
  4. Proeutectoid cementite

Explanation

A 0.4 percent carbon steel is hypoeutectoid, so as it cools below the A3 line ferrite begins to precipitate along the austenite grain boundaries. The remaining austenite becomes richer in carbon until it reaches the eutectoid composition and transforms to pearlite. Proeutectoid cementite forms in hypereutectoid steels, ledeburite in cast irons, and martensite only on rapid quenching.

Question 31Engineering Materials and Heat Treatment

Which surface-hardening process involves heating steel in contact with a carbon-rich solid compound such as charcoal, without melting the surface?

  1. Nitriding
  2. Flame hardening
  3. Induction hardening
  4. Pack carburising Correct answer.

Explanation

Pack carburising places the component in a box with charcoal or other carbonaceous material and heats it so that carbon diffuses into the austenitic surface. Induction and flame hardening harden the surface of an already hardenable steel by rapid heating without adding carbon, and nitriding introduces nitrogen rather than carbon. Diffusion of carbon from a solid medium is the distinguishing feature of pack carburising.

Question 32Engineering Materials and Heat Treatment

A plain carbon steel is to be made fully martensitic throughout a thick section. Which property of the steel most directly determines whether this is possible?

  1. Weldability
  2. Ductility
  3. Hardenability Correct answer.
  4. Electrical conductivity

Explanation

Hardenability is the ability of a steel to form martensite at depth under a given cooling rate, and it depends mainly on alloy content and grain size rather than on carbon content alone. A steel of low hardenability hardens only a shallow skin, while a high-hardenability steel can be through-hardened in thicker sections. Ductility, weldability and conductivity do not govern the depth of martensite formation.

Question 33Engineering Materials and Heat Treatment

Which non-ferrous metal is commonly used for bearing shells and bushes because of its low friction and good embeddability?

  1. Zinc
  2. Aluminium
  3. White metal (babbitt) Correct answer.
  4. Copper

Explanation

White metal, or babbitt, is a tin or lead based alloy with antimony and copper that offers low friction, good conformability and the ability to embed dirt particles, which makes it suitable for bearing linings. Copper and aluminium are used as structural metals and zinc mainly for coatings and die castings. Bearing alloys are chosen for their tribological behaviour rather than structural strength.

Question 34Engineering Materials and Heat Treatment

A steel shaft fails by fatigue after prolonged service under reversed bending. Which feature of the fracture surface is most characteristic of fatigue failure?

  1. Large cup-and-cone necking
  2. Uniform dimples across the whole surface
  3. Smooth beach marks with a final rough overload zone Correct answer.
  4. Intergranular cracking with no distinct zones

Explanation

Fatigue failure begins at a stress raiser and propagates progressively, producing smooth beach marks or clamshell patterns where the crack front rested, ending in a rough final fracture zone when the remaining section can no longer carry the load. Dimples and cup-and-cone necking are features of ductile overload, not fatigue. Recognising beach marks helps maintenance engineers trace the origin of failure.

Question 35Engineering Materials and Heat Treatment

Which of the following best describes the structure of pearlite in slowly cooled carbon steel?

  1. A uniform solid solution of carbon in austenite
  2. Alternate lamellae of ferrite and cementite Correct answer.
  3. A single phase of carbon dissolved in ferrite
  4. Free graphite nodules in a ferrite matrix

Explanation

Pearlite is a two-phase lamellar aggregate in which thin plates of ferrite alternate with thin plates of cementite, formed when austenite of eutectoid composition cools slowly. The lamellar spacing affects hardness and strength: finer pearlite is harder. A single-phase solid solution describes ferrite or austenite, and graphite nodules describe malleable or nodular cast iron.

Question 36Engineering Materials and Heat Treatment

A component is to be hardened by heating in a molten salt bath containing sodium cyanide. Which process is being described?

  1. Nitriding
  2. Pack carburising
  3. Liquid cyaniding Correct answer.
  4. Austempering

Explanation

Liquid cyaniding uses a molten cyanide-bearing salt bath, typically around 800 to 870 degrees C, so that both carbon and nitrogen diffuse into the surface before quenching. It gives a thin, hard case quickly but involves toxic salts requiring strict safety controls. Pack carburising uses a solid carbon medium, nitriding uses nitrogen-bearing gas, and austempering is a quenching technique rather than a surface treatment.

Question 37Engineering Materials and Heat Treatment

A steel contains 1.2 percent carbon and is slowly cooled from the austenitic range. Which constituent forms at the prior austenite grain boundaries?

  1. Pearlite nodules
  2. Proeutectoid ferrite
  3. Proeutectoid cementite network Correct answer.
  4. Graphite flakes

Explanation

At 1.2 percent carbon the steel is hypereutectoid, so on slow cooling below the Acm line cementite precipitates along the austenite grain boundaries as a network before the remaining austenite transforms to pearlite. This brittle grain-boundary network is one reason high carbon steels need careful heat treatment. Proeutectoid ferrite appears only in hypoeutectoid steels, and graphite flakes are characteristic of grey cast iron.

Question 38Engineering Materials and Heat Treatment

Two steels have the same carbon content but one contains nickel and chromium. In the Jominy end-quench test, what result is expected for the alloyed steel?

  1. Hardness at the quenched end is lower than for the plain steel
  2. Hardness is uniform along the entire specimen length
  3. Hardness drops to the low value within a shorter distance
  4. Hardness remains high to a greater distance from the quenched end Correct answer.

Explanation

The Jominy test measures hardenability by quenching one end of a standard bar and plotting hardness against distance from that end. Alloying elements such as nickel, chromium and molybdenum slow the transformation of austenite, so martensite forms further from the quenched end and hardness stays high over a greater distance. The plain carbon steel of the same carbon content loses hardness sooner.

Question 39Engineering Materials and Heat Treatment

A hardened tool steel shows a network of fine cracks on the surface after quenching in water. Which measure is most appropriate to prevent this in future?

  1. Increase the section thickness of the part
  2. Quench in oil and temper immediately after quenching Correct answer.
  3. Quench in brine instead of water
  4. Increase the austenitising temperature before quenching

Explanation

Quench cracks arise from the high thermal and transformation stresses produced by very severe cooling, especially in high carbon and alloy steels. Using a slower oil quench and tempering promptly after quenching reduces the stress before it can cause cracking. Brine and higher austenitising temperatures would both increase severity, and increasing section thickness raises the risk rather than reducing it.

Question 40Engineering Materials and Heat Treatment

An annealed 0.8 percent carbon steel and a hardened and tempered 0.8 percent carbon steel are compared. Which statement correctly describes the difference in their structures?

  1. Both contain identical coarse pearlite because carbon content is the same
  2. The annealed steel contains lamellar pearlite, while the hardened and tempered steel contains fine carbides in a ferrite matrix Correct answer.
  3. The annealed steel contains martensite, while the tempered steel contains coarse pearlite
  4. The annealed steel contains free graphite, while the tempered steel contains cementite network

Explanation

Annealing a eutectoid steel gives coarse lamellar pearlite, whereas hardening followed by tempering produces tempered martensite, in which fine carbide particles are dispersed in a ferrite matrix. The fine dispersion gives higher strength and toughness at the same carbon content. Carbon content alone does not fix the structure, because the cooling and tempering history determine the arrangement of the carbides.

Question 41Measuring Instruments and Metrology

A fitter in a workshop must measure the outside diameter of a finished shaft to the nearest 0.01 mm using a standard workshop instrument. Which instrument is most suitable for this routine measurement?

  1. Outside micrometer Correct answer.
  2. Try square
  3. Steel rule
  4. Vernier height gauge

Explanation

An outside micrometer has a spindle-and-anvil measuring faces and a thimble graduated to read directly to 0.01 mm, which matches the required resolution for shaft diameters. A steel rule reads only to about 0.5 mm, and the other tools are not designed for external diameter measurement to that precision.

Question 42Measuring Instruments and Metrology

On a metric vernier calliper, the main scale is graduated in 1 mm divisions and the vernier has 50 divisions. What is the least count of the instrument?

  1. 0.5 mm
  2. 0.05 mm
  3. 0.1 mm
  4. 0.02 mm Correct answer.

Explanation

Least count equals the value of one main scale division divided by the number of vernier divisions, so 1 mm divided by 50 gives 0.02 mm. This is the standard least count of a 50-division metric vernier calliper used in workshops.

Question 43Measuring Instruments and Metrology

What is the reading of a metric outside micrometer whose barrel shows 7.5 mm and whose thimble, with 50 divisions, is on the 22nd division?

  1. 7.22 mm
  2. 7.94 mm
  3. 7.72 mm Correct answer.
  4. 7.50 mm

Explanation

Each thimble division on a 0.5 mm pitch micrometer equals 0.01 mm, so 22 divisions give 0.22 mm. Adding this to the barrel reading of 7.5 mm gives a total of 7.72 mm.

Question 44Measuring Instruments and Metrology

Which instrument works on the principle of a calibrated screw converting rotary motion into linear displacement?

  1. Dial indicator
  2. Micrometer Correct answer.
  3. Bevel protractor
  4. Surface plate

Explanation

In a micrometer the spindle moves axially through a nut when the thimble is rotated, so each revolution produces a known linear advance equal to the screw pitch. The thimble scale then subdivides that pitch to give fine readings.

Question 45Measuring Instruments and Metrology

A dial indicator has a spindle travel of 10 mm and its large dial is graduated into 100 divisions, one full pointer revolution covering 1 mm. What does each small dial division represent?

  1. 0.01 mm
  2. 10 mm
  3. 0.1 mm
  4. 1 mm Correct answer.

Explanation

The small dial of a dial indicator counts full revolutions of the large pointer, and one revolution corresponds to 1 mm of spindle travel. Therefore each small dial division represents 1 mm, which is used to record total travel beyond one revolution.

Question 46Measuring Instruments and Metrology

Which instrument is used to measure the angle of a machined V-groove with a resolution of 5 minutes of arc?

  1. Universal bevel protractor Correct answer.
  2. Combination set square
  3. Vernier calliper
  4. Radius gauge

Explanation

A universal bevel protractor carries a graduated main arc and a vernier scale that reads to 5 minutes, and its adjustable blade allows it to be set into angular features such as V-grooves. The other instruments measure linear dimensions, squareness, or radii, not angles to 5 minutes.

Question 47Measuring Instruments and Metrology

A vernier calliper main scale reads 24 mm and the 13th vernier division coincides with a main scale line. The vernier has 50 divisions and the least count is 0.02 mm. What is the total reading?

  1. 24.65 mm
  2. 24.02 mm
  3. 24.26 mm Correct answer.
  4. 24.13 mm

Explanation

The vernier contribution is the coinciding division number multiplied by the least count, that is 13 times 0.02 mm, which equals 0.26 mm. Adding this to the main scale reading of 24 mm gives 24.26 mm.

Question 48Measuring Instruments and Metrology

A micrometer is checked with its spindle closed on the anvil and the thimble zero is found to be 3 divisions below the barrel zero line. How should a subsequent reading be corrected?

  1. Subtract 0.03 mm from the observed reading
  2. Add 0.03 mm to the observed reading Correct answer.
  3. Add 0.06 mm to the observed reading
  4. No correction is needed

Explanation

When the thimble zero sits below the barrel datum line at closure, the instrument reads less than the true size, so the zero error is negative and must be added to every observed reading. With 0.01 mm per division, 3 divisions equal 0.03 mm of correction.

Question 49Measuring Instruments and Metrology

Which instrument is best suited for measuring the depth of a blind bore to a resolution of 0.02 mm?

  1. Outside micrometer
  2. Vernier depth gauge Correct answer.
  3. Feeler gauge
  4. Divider

Explanation

A vernier depth gauge has a graduated beam and a base that seats on the workpiece face, with the beam end probing the bottom of the bore, giving direct depth readings to 0.02 mm. The other tools cannot measure internal depth to this resolution.

Question 50Measuring Instruments and Metrology

A sine bar with a roller centre distance of 100 mm is set up with slip gauges of 30 mm under one roller. What angle is established, taking sine of the angle as 0.3?

  1. 30.00 degrees
  2. 17.46 degrees Correct answer.
  3. 18.00 degrees
  4. 16.70 degrees

Explanation

For a sine bar the sine of the set angle equals the slip gauge height divided by the roller centre distance, giving 30 divided by 100, or 0.3. The angle whose sine is 0.3 is approximately 17.46 degrees.