CBSE Class 12 Engineering Graphics Question Paper 2026 Solved Code 68

SECTION A

Q. 1 to Q. 14 : Answer the following multiple choice questions. Print the correct choice on your drawing sheet. [14 x 1 = 14 Marks]

1. Pictorial drawings, used to communicate the structure of objects to others are called _________ [1 Mark]
(A) Multiple plane drawings
(B) Three plane drawings
(C) Two plane drawings
(D) One plane drawings

Answer: (D) One plane drawings

Teacher's Note:
a) A pictorial drawing shows length, width and height together in a single view on one plane.
b) Orthographic views, in contrast, are multiple plane drawings.

2. A non-isometric line is drawn by [1 Mark]
(A) using angle of inclination.
(B) using co-ordinate method.
(C) using a line parallel to \( 45^{\circ} \) to horizontal.
(D) using a line perpendicular to horizontal.

Answer: (B) using co-ordinate method.

Teacher's Note:
a) Non-isometric lines are not parallel to the isometric axes, so their true length cannot be measured directly.
b) Locate both end points along isometric lines (co-ordinates) and then join them.

3. Isometric projection is a type of _________. [1 Mark]
(A) Axonometric projection
(B) Orthographic projection
(C) Oblique projection
(D) Perspective projection

Answer: (A) Axonometric projection

Teacher's Note:
a) Axonometric projections show all three axes in one view; isometric is the case where all three axes are equally inclined.
b) Do not confuse it with oblique or perspective projection.

4. Select the correct statements for the given figure. [1 Mark]
(i) A hexagonal pyramid is placed with the axis perpendicular to V.P. and parallel to H.P.
(ii) The solid is kept with a pair of base edges parallel to V.P.
(iii) The solid is an example of solid of revolution.
(iv) The solid has total no. of seven surfaces.
(A) (i) and (iv) only
(B) (i) and (ii) only
(C) (ii) and (iv) only
(D) (iii) and (iv) only

[Figure: Isometric projection of a hexagonal pyramid standing on its base on H.P., axis vertical, drawn inside a box of construction lines. Base edge 40, height 70. Isometric axes at \( 30^{\circ} \) to the horizontal on both sides; direction of viewing arrow F shown at the front. Label: ISOMETRIC PROJECTION.]

Answer: (C) (ii) and (iv) only

Teacher's Note:
a) The axis is vertical (perpendicular to H.P.), so statement (i) is wrong.
b) A hexagonal pyramid has 6 triangular faces + 1 hexagonal base = 7 surfaces.
c) A pyramid has flat faces, so it is a polyhedron, not a solid of revolution.

5. The side view of an object is shown on which plane ? [1 Mark]
(A) Vertical plane
(B) Profile plane
(C) Parallel plane
(D) Horizontal plane

Answer: (B) Profile plane

Teacher's Note:
a) Front view is on V.P., top view is on H.P. and side view is on the profile plane (P.P.).
b) The profile plane is perpendicular to both H.P. and V.P.

6. Match the List - I with List - II for the given isometric projection of combination of solids : [1 Mark]
List - I | List - II
(1) Pentagonal surface | (i) parallel to H.P. and perpendicular to V.P.
(2) Axis of upper solid | (ii) perpendicular to H.P. and V.P. both
(3) One rectangular surface | (iii) parallel to V.P. and H.P. both
(4) Axis of lower solid | (iv) perpendicular to H.P. and parallel to V.P.
(1) (2) (3) (4)
(A) (ii) (iv) (i) (iii)
(B) (ii) (i) (iii) (iv)
(C) (iii) (ii) (iv) (i)
(D) (iv) (ii) (i) (iii)

[Figure: Isometric projection of a combination of solids: a pentagonal prism (base edge 34, length 80) lying on H.P. on one rectangular face with its axis horizontal and parallel to V.P., its pentagonal ends facing the sides; a cone (base diameter 44, height 60) stands centrally on the top rectangular face of the prism with axis vertical through centre O. Isometric axes at \( 30^{\circ} \) and direction of viewing arrow F shown.]

Answer: (A) 1-(ii), 2-(iv), 3-(i), 4-(iii)

Teacher's Note:
a) The prism's pentagonal end faces are profile planes, so they are perpendicular to both H.P. and V.P.
b) The cone's axis is vertical: perpendicular to H.P. and parallel to V.P.
c) The prism's axis lies along its length, parallel to both H.P. and V.P.

7. Match the List - I with List - II : [1 Mark]
List - I (Fastener) | List - II (Tool)
(1) Screw | (i) Allen key
(2) Hexagonal socket head screw | (ii) Spanner
(3) Cotter | (iii) Screw driver
(4) Nut-bolt | (iv) Hammer
(1) (2) (3) (4)
(A) (iii) (ii) (iv) (i)
(B) (iv) (i) (iii) (ii)
(C) (iii) (i) (iv) (ii)
(D) (iv) (iii) (i) (ii)

Answer: (C) 1-(iii), 2-(i), 3-(iv), 4-(ii)

Teacher's Note:
a) A hexagonal socket head screw is turned with an Allen key (hexagonal key).
b) A cotter is driven into its slot with a hammer; a nut is tightened with a spanner.

8. The thickness of flat head rivet for diameter 'd' is : [1 Mark]
(A) 0.2 d
(B) 0.25 d
(C) 0.3 d
(D) 0.35 d

Answer: (B) 0.25 d

Teacher's Note:
a) Flat head rivet: head diameter = 2d and head thickness = 0.25d.
b) Learn the standard proportions of each rivet head as a table.

9. Chamfer is conical round off to the prism of bolt head at : [1 Mark]
(A) an angle of \( 15^{\circ} \) on the outer end face.
(B) an angle of \( 30^{\circ} \) on the inner end face.
(C) an angle of \( 15^{\circ} \) on both end face.
(D) an angle of \( 30^{\circ} \) on the outer end face.

Answer: (D) an angle of \( 30^{\circ} \) on the outer end face.

Teacher's Note:
a) The chamfer is \( 30^{\circ} \) and is given on the outer (top) face to remove sharp corners.
b) The chamfer produces the curved arcs seen in the front view of a hexagonal bolt head.

10. Select the correct option regarding plain washer. [1 Mark]
(i) A circular plate having a square hole in its centre.
(ii) The standard inner diameter is 'd + 1'.
(iii) Its outer diameter is '2d'.
(iv) It is placed below the nut.
(A) (i) and (ii) only
(B) (ii) and (iii) only
(C) (ii) and (iv) only
(D) (iii) and (iv) only

Answer: (C) (ii) and (iv) only

Teacher's Note:
a) A plain washer has a circular hole of diameter d + 1, not a square hole.
b) Its outer diameter is 2d + 3 and thickness is 0.15d, so statement (iii) is wrong.

11. Identify the combination of solids which is an example of combination of polyhedron. [1 Mark]
(A)
(B)
(C)
(D)

[Figure: Four isometric drawings as options. (A) A vertical cylinder (diameter 42, height 62) standing on the flat face of a hemisphere (diameter 80). (B) A pentagonal prism (edge 25, height 50) standing on a hexagonal prism (edge 45, thickness 25) lying on its base. (C) A pyramid (base edge 32, height 62) standing on a vertical cylinder (diameter 84, height 44). (D) A vertical cylinder (diameter 42, height 50) standing on a hexagonal prism (edge 30, length 102) lying with its axis horizontal.]

Answer: (B) Pentagonal prism placed on a hexagonal prism

Teacher's Note:
a) Polyhedra are solids bounded only by flat faces, such as prisms and pyramids.
b) Options (A), (C) and (D) contain a cylinder or hemisphere, which are solids of revolution.

12. The isometric projection of a sphere of radius 'R' is drawn using radius equal to : [1 Mark]
(A) Isometric R
(B) 0.5 R
(C) 2R
(D) True R

Answer: (D) True R

Teacher's Note:
a) The isometric projection of a sphere is a circle of true radius R.
b) Only the location of its centre is found using isometric length.

13. The holes for the bolts in the sole plate of bearings are elongated towards the width, to facilitate : [1 Mark]
(A) any misalignment while fitting.
(B) larger bolts than recommended.
(C) axial upliftment of bolt.
(D) to reduce the cost.

Answer: (A) any misalignment while fitting.

Teacher's Note:
a) Elongated (slotted) holes allow small adjustment of the bearing position during fitting.
b) This is why the bolt holes appear as slots in the top view of a bearing base.

14. Two square cross sectional rods are joined [1 Mark]
(A) by Sleeve and cotter joint
(B) by Gib and cotter joint
(C) by Tie rod joint
(D) by Flange joint

Answer: (B) by Gib and cotter joint

Teacher's Note:
a) A gib and cotter joint connects two square rods, for example in a connecting rod end.
b) Sleeve and cotter and socket and spigot joints are used for round rods.

SECTION B

Q. 15 to Q. 18 : Observe the given figure of combination of solids and answer the questions given below : [4 x 1 = 4 Marks]

[Figure: Isometric projection of a combination of solids: a triangular prism (base edge 86, height 40) standing on its triangular base on H.P. with axis vertical; a hexagonal pyramid (base edge 32, height 66) stands centrally on the top triangular face with its axis vertical. Direction of viewing arrow F shown.]

15. How many triangular surfaces are there ? [1 Mark]
(A) 5
(B) 6
(C) 7
(D) 8

Answer: (D) 8

Teacher's Note:
a) The hexagonal pyramid has 6 triangular slant faces.
b) The triangular prism has 2 triangular end faces (top and bottom), so the total is 6 + 2 = 8.

16. The total height of the combination of solids is : [1 Mark]
(A) 66 mm
(B) 40 mm
(C) 106 mm
(D) 118 mm

Answer: (C) 106 mm

Teacher's Note:
a) Total height = height of prism + height of pyramid = 40 + 66 = 106 mm.
b) Do not add the base edge (32 or 86) to the height.

17. The axis of lower solid is : [1 Mark]
(A) horizontal
(B) vertical
(C) inclined to horizontal at \( 60^{\circ} \)
(D) inclined to vertical at \( 60^{\circ} \)

Answer: (B) vertical

Teacher's Note:
a) The triangular prism rests on its triangular base, so its axis is vertical.
b) The 40 mm height is measured along the vertical isometric axis.

18. The upper solid is : [1 Mark]
(A) a pentagonal prism having axis perpendicular to V.P.
(B) a triangular prism having axis parallel to H.P.
(C) a hexagonal pyramid having axis perpendicular to V.P.
(D) a hexagonal pyramid having axis perpendicular to H.P.

Answer: (D) a hexagonal pyramid having axis perpendicular to H.P.

Teacher's Note:
a) The upper solid has a six-sided base and slant faces meeting at an apex, so it is a hexagonal pyramid.
b) It stands upright, so its axis is vertical, that is perpendicular to H.P.

Q. 19 to Q. 22 : Read the following paragraph and answer the questions given below : [4 x 1 = 4 Marks]

Rahul was repairing his bicycle for unknown noises. He found some fasteners were loose. In his tool box, he found various fasteners. Help him in resolving his queries :

19. Rahul found a bolt of 20 \( \times \) 2, where 2 represents : [1 Mark]
(A) Thickness of bolt
(B) Number of threads per mm
(C) Double start thread
(D) Distance between two corresponding points on the adjacent threads.

Answer: (D) Distance between two corresponding points on the adjacent threads.

Teacher's Note:
a) In the designation 20 \( \times \) 2, 20 is the nominal (major) diameter and 2 is the pitch in mm.
b) Pitch is the distance between corresponding points on adjacent threads.

20. Which one is not a temporary fastener ? [1 Mark]
(A) Bolt
(B) Rivet
(C) Screw
(D) Stud

Answer: (B) Rivet

Teacher's Note:
a) A rivet is a permanent fastener; it can be removed only by breaking it.
b) Bolts, screws and studs can be opened and refitted, so they are temporary fasteners.

21. M in M8, M10, M12 on bolts stands for [1 Mark]
(A) Metric thread
(B) Mean diameter
(C) Major diameter
(D) Minor diameter

Answer: (A) Metric thread

Teacher's Note:
a) M denotes an ISO metric thread; the number after it is the nominal diameter in mm.
b) M10 means a metric thread of 10 mm nominal diameter.

22. Distance moved by a nut in the axial direction in one rotation is called [1 Mark]
(A) Pitch
(B) Angle of thread
(C) Lead
(D) Flank

Answer: (C) Lead

Teacher's Note:
a) Lead is the axial advance in one complete turn.
b) Lead = number of starts \( \times \) pitch; for a single start thread, lead equals pitch.

23. (a) Fig. 1 shows the details of the parts of a Bush Bearing. Assemble all these parts correctly and then draw to scale 1 : 1 in its following views :
(i) Front view right half in section [13 Marks]
(ii) Top view [8 Marks]
Print the title and scale used. Draw the projection symbol. Give 6 important dimensions. [6 Marks]

[Figure: Fig. 1 - BUSH BEARING (Note : Figure not to scale. Use dimensions given). BASE (body) - Front view: sole plate 180 long and 17 thick with a recess 4 deep on the underside (R3 corners), 12 wide feet at each end, R5 rounds at the outer top corners and R5 fillets where the boss joins the plate; circular boss of \( \phi \) 60 outside with bore \( \phi \) 40, centre 50 above the bottom; oil hole at the top of the boss \( \phi \) 5 with a counterbore 3 deep; bolt holes 20 wide shown hidden, 100 CRS. Top view of base: plate 180 x 50, boss block 60 long, hidden rectangle of the recess 12 in from the edges, two elongated bolt holes 15 wide at 100 CRS, oil hole \( \phi \) 10 and \( \phi \) 5 at the centre. BUSH - Front view: two concentric circles \( \phi \) 30 (inside) and \( \phi \) 40 (outside). Top view: rectangle 60 long, bore shown by hidden lines, oil hole \( \phi \) 5 at mid-length (30 from one end).]

Answer:
Assembly: the bush (\( \phi \) 40 outside, \( \phi \) 30 inside, 60 long) is pressed into the \( \phi \) 40 bore of the body so that its \( \phi \) 5 oil hole lines up with the oil hole of the body. Title: ASSEMBLY OF BUSH BEARING, SCALE - 1:1.
(i) Front view, right half in section (13 marks):
1. Draw the right half of the body in section (2), the sole recess 4 deep on the underside (1) and the left half of the body in outside view (2), with the \( \phi \) 60 arc of the boss (1/2). Base length 180, thickness 17, centre height 50.
2. Draw the two circles representing the bush: \( \phi \) 40 and \( \phi \) 30 (2).
3. Draw the oil hole in the body, \( \phi \) 10 up to 3 deep then \( \phi \) 5 (1), and the \( \phi \) 5 oil hole in the bush (1/2).
4. Draw the bolt hole in the right half of the body with its axis (1 1/2) and the axis of the bolt hole in the left half (1/2), holes at 100 CRS.
5. Hatch the sectioned right half of the body (1) and the right half of the bush (1), with hatching of the bush in the opposite direction.
6. Show R5 rounds and fillets on the outer boundary of the body (1/2 mark is deducted if missed).
(ii) Top view (8 marks):
1. Draw the outline of the body 180 x 50 with the two visible 60 mm long vertical lines of the boss (2).
2. Draw four hidden vertical lines representing the bush (\( \phi \) 30 and \( \phi \) 40) (2).
3. Draw the hidden rectangle for the sole recess (1).
4. Draw two circles for the oil hole, \( \phi \) 10 and \( \phi \) 5 (1).
5. Draw the two elongated bolt holes (1) with their axis lines at 100 CRS (1/2).
6. Show the cutting plane X-X' on the top view (1/2).
Details (6 marks): print the title (1), scale used 1:1 (1), draw the first angle projection symbol (1) and give six important dimensions such as 180, 100 CRS, 50, 17, \( \phi \) 60, \( \phi \) 40, \( \phi \) 30, 60 (3).

Teacher's Note:
a) Marks split: front view 13, top view 8, title, scale, symbol and dimensions 6 = 27.
b) Common mistakes: hatching the left (unsectioned) half, drawing the bush and body hatching in the same direction, and forgetting the oil hole alignment.
c) Keep the top view exactly below the front view and use hidden lines for the bush and recess.

OR

(b) Fig.-2 shows the assembly of a Gib and Cotter Joint. Disassemble the parts correctly and then draw to scale 1 : 1, its following views of the following components. Keep their position same with respect to both H.P. and V.P.
(i) STRAP
(a) Front view lower half in section [8 Marks]
(b) Top view [6 Marks]
(ii) GIB
(a) Front view [4 Marks]
(b) Top view [3 Marks]
Print the titles of both and the scale used. Draw the projection symbol. Give 6 important dimensions. [6 Marks]

[Figure: Fig. 2 - GIB AND COTTER JOINT, FRONT VIEW (Note : Figure not to Scale. Use dimensions given). Two SQ 30 rods joined horizontally: the eye end (rod) on the left fits inside the strap/fork end on the right. Strap arms 12 thick above and below the SQ 30 rod; the strap ends in a SQ 30 rod on the right with R5 fillets; strap end 38 beyond the slot. Vertical cotter 10 thick, 120 long, taper 1 in 30, projecting 3 at the rounded ends. Gib 10 thick with heads 12 deep, head length 12 at the top and 25 at the bottom. Slot in the rod 40 wide (25 + 15) with 3 clearance; rod end 25 beyond the slot. Notes: (i) All fillets and Rounds R5 (ii) Slots 10 thick. Labels: EYE END, Strap/Fork End, Cotter 10 Thick, 10 Thick GIB.]

Answer:
Titles: DISASSEMBLY OF GIB AND COTTER JOINT - (i) STRAP, (ii) GIB. SCALE - 1:1.
(i) STRAP, (a) Front view lower half in section (8 marks):
1. Draw the boundary of the strap: two arms 12 thick enclosing the SQ 30 space (outer 54), and the solid SQ 30 end on the right (4), with R5 rounds and fillets (1/2) and a conventional break at the rod end (1/2).
2. Draw the slot for the gib and cotter in the lower half, 40 + 3 = 43 wide, keeping the 3 mm clearance (1 1/2 + 1/2).
3. Hatch the sectioned lower half of the strap (1).
(i) STRAP, (b) Top view (6 marks):
1. Draw the boundary of the strap, SQ 30 wide (1 1/2), with the conventional break at the end (1/2).
2. Draw the slot of the gib and cotter with clearance, 10 wide (2).
3. Draw two vertical lines at the slot, one visible and one hidden (2).
(ii) GIB, (a) Front view (4 marks):
1. Draw the boundary of the gib, 54 high between the heads, heads 12 deep, 10 thick (2 1/2), including the taper line of 1 in 30 (1) and R5 rounds (1/2).
(ii) GIB, (b) Top view (3 marks):
1. Draw the boundary of the gib, 10 thick (2).
2. Draw the vertical line showing the taper (1/2).
3. Draw the hidden vertical line (1/2).
Details (6 marks): print the titles of both parts (1), scale used (1), first angle projection symbol (1) and six important dimensions such as SQ 30, 12, 43, 38, 54, taper 1 in 30 (3).

Teacher's Note:
a) Marks split: strap 8 + 6, gib 4 + 3, details 6 = 27.
b) Common mistakes: forgetting the 3 mm clearance in the strap slot, showing the gib taper on the wrong side and hatching the upper half instead of the lower half.
c) Draw each part in the same position as in the assembly, as the question demands.

SECTION C

24. (a) Construct an isometric scale. [4 Marks]

Answer:
1. Draw a horizontal line PQ. From P draw one line at \( 45^{\circ} \) (true length / scale 1:1) and another at \( 30^{\circ} \) (isometric length) (1).
2. On the \( 45^{\circ} \) line mark divisions of 10 mm (0 to 80), and divide the first part into 1 mm divisions (1).
3. From each division drop vertical lines (perpendicular to PQ) to cut the \( 30^{\circ} \) line; these points give the isometric lengths, completing the isometric scale (1).
4. Print 'TRUE LENGTH / SCALE 1:1' and 'ISOMETRIC LENGTH / ISOMETRIC SCALE', and mark the angles \( 30^{\circ} \) and \( 45^{\circ} \) (1).

Teacher's Note:
a) Isometric length = 0.815 \( \times \) true length; the scale gives this without calculation.
b) Common mistake: projecting parallel to PQ instead of perpendicular to it.
c) One mark each for angles, divisions, projection and labelling.

(b) Draw the isometric projection of an upright cone (diameter 60 mm, axis 90 mm). It is placed on H.P. on its base. Indicate the direction of viewing. Give all the dimensions. [9 Marks]

Answer:
1. Draw the isometric ellipse of the \( \phi \) 60 base on the horizontal plane using isometric lengths (four-centre method in an isometric square), with its centre lines at \( 30^{\circ} \) (3 1/2 + 1/2).
2. From the centre draw the vertical axis of isometric length of 90 mm (about 73) to get the apex; draw the two generators from the apex tangent to the ellipse on both sides (3).
3. Mark the axis (1/2) and show the direction of viewing arrow (1/2).
4. Dimension the drawing with true values: \( \phi \) 60 and height 90, and mark the \( 30^{\circ} \) angles (1).

Teacher's Note:
a) Marks split: ellipse and centre lines 4, generators 3, axis and direction 1, dimensions 1.
b) Always write true dimensions on an isometric projection, though it is drawn with isometric lengths.
c) 1 mark is deducted for incorrect position; the cone must stand on its base with the axis vertical.

25. (a) Draw to scale 1 : 1, the standard profile of KNUCKLE THREAD taking enlarged pitch as 50 mm. Give standard dimensions. [8 Marks]

Answer:
Given P = 50 mm, so depth d = 0.5P = 25 mm and radius R = 0.25P = 12.5 mm.
1. Draw two horizontal lines 25 mm apart (crest and root lines) and one horizontal centre line midway; mark distances equal to half the pitch (25 mm) along it to get the centres of the arcs (2).
2. Draw semi-circular arcs of radius 12.5 mm alternately above and below, forming the crests and roots of the threads, at least two threads (3).
3. Hatch the thread profile and show a conventional break (1).
4. Give standard dimensions: P = 50, d = 0.5P = 25, R = 0.25P = 12.5 (2).

Teacher's Note:
a) Knuckle thread is made of semi-circles only, so the profile has no flat portions.
b) If the profile is sketched freehand instead of drawn to scale 1:1, 2 marks are deducted in all.

OR

(b) Draw to scale 1 : 1, the front view and top view of a HEXAGONAL NUT of diameter 25 mm. Keep its axis vertical. Give standard dimensions. [8 Marks]

Answer:
Given d = 25 mm: thickness of nut = d = 25 mm, width across flats = 1.5d + 3 = 40.5 mm, chamfer angle \( 30^{\circ} \), minor (hole) diameter 0.8d = 20 mm.
1. Front view (across flats or across corners): draw the hexagonal nut 25 thick, showing the chamfer arcs on the top face drawn at \( 30^{\circ} \) (with radius R = d for the across flats view), with the axis vertical (3).
2. Top view: draw a regular hexagon of 40.5 mm across flats, the chamfer circle of diameter 40.5 touching the flats, the \( \phi \) 20 (0.8d) hole circle and the \( \phi \) 25 thread circle as a three-quarter circle (3).
3. Give standard dimensions: d, 1.5d + 3, 0.8d, \( 30^{\circ} \) chamfer (2).

Teacher's Note:
a) Marks split: front view 3, top view 3, dimensions 2.
b) 1 mark is deducted for incorrect position (axis must be vertical), and 2 marks in all if sketched freehand.
c) Draw the top view first, then project the front view from it.

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