Class 12 Engineering Graphics Solved Question Papers: CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68
Review targeted exam resources with the CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68. Built according to official CBSE standards for the 2026-27 academic year, these downloadable Class 12 Engineering Graphics question papers support effective revision and performance tracking.
Download Class 12 Engineering Graphics Question Paper PDF
Access the complete question paper PDF for Class 12 Engineering Graphics below. Regular practice with these targeted exam papers builds familiarity with standard question patterns and helps secure higher marks in final evaluations.
SECTION A
1. Lines composed of closely and evenly spaced short dashes, in a drawing, represents [1 Mark]
(A) Visible edges
(B) Hidden edges
(C) Hatching
(D) Pitch circle
Answer: (B) Hidden edges
Teacher's Note:
a) As per SP:46-2003, short dashes of equal length with equal gaps show hidden edges and outlines.
b) Visible edges are continuous thick lines, and pitch circles are long chain thin lines.
2. Sectioning of objects helps the engineers in clarifying _________. [1 Mark]
(A) The outlook of a complicated object
(B) The exterior details of a simple object
(C) The interior details of a complicated object
(D) The surfaces of a simple object
Answer: (C) The interior details of a complicated object
Teacher's Note:
a) A section removes the front part of the object, so hidden inner details become visible edges.
b) This avoids a confusing drawing full of hidden lines.
3. Orthographic projection of a cube is shown as : [1 Mark]
[Figure: Four drawings. (i) An isometric-type pictorial view of a cube showing three faces. (ii) Two rectangles one above the other, the upper one tall and narrow, the lower one short. (iii) A pictorial view of a tall square block (prism) showing three faces. (iv) Two equal squares one above the other.]
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Answer: (D) (iv)
Teacher's Note:
a) All faces of a cube are equal squares, so its front view and top view are two equal squares.
b) Options (i) and (iii) are pictorial views, not orthographic views.
4. Select the correct statements for the given figure. [1 Mark]
(i) A horizontal pyramid is placed with its axis parallel to both H.P. and V.P.
(ii) The solid has three triangular faces in total.
(iii) A horizontal pyramid is placed with its axis perpendicular to V.P.
(iv) It is an example of polyhedron.
(A) (i) and (iv) only
(B) (ii) and (iv) only
(C) (i) and (iii) only
(D) (ii) and (iii) only
[Figure: Isometric view of a pyramid lying with its axis horizontal and running from left to right; the base is a vertical polygon at the left end and the apex is at the right end; centre lines are drawn through the base, and an arrow at the bottom left shows the direction of viewing.]
Answer: (A) (i) and (iv) only
Teacher's Note:
a) The axis runs from left to right, so it is parallel to both H.P. and V.P.
b) A pyramid has only plane faces, so it is a polyhedron.
5. The above given top view corresponds to [1 Mark]
[Figure: Top view: a regular pentagon with a square drawn inside it, and both diagonals of the square drawn as visible lines. Options: (i) a square pyramid placed inverted (apex down) on the top face of a pentagonal prism, arrow from the right front; (ii) an upright square pyramid (apex up) on a pentagonal prism, arrow from the left front; (iii) an upright square pyramid (apex up) on a pentagonal prism, arrow from the right front; (iv) an inverted square pyramid on a pentagonal prism, arrow from the left front.]
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Answer: (C) (iii)
Teacher's Note:
a) The diagonals of the square are visible, so the slant edges meet at an apex facing upward.
b) Match the position of the pentagon edge nearest to the viewer with the direction of the arrow.
6. Identify the correct figure which represents the isometric projection of a vertical square prism placed on a horizontal square slab. [1 Mark]
[Figure: Four isometric drawings with arrows for direction of viewing. (i) A horizontal rectangular block placed across another horizontal block. (ii) A vertical square prism standing at the centre of a wide, thin square slab. (iii) A horizontal block resting on top of a vertical prism. (iv) A vertical prism standing on a smaller block, set off towards one side.]
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Answer: (B) (ii)
Teacher's Note:
a) A slab is thin and wide; a vertical prism is tall with its axis perpendicular to H.P.
b) Only figure (ii) shows the tall prism standing on the thin, wide square slab.
7. Select the correct option for the given isometric projection of combination of solids. [1 Mark]
(i) The axis of pyramid is parallel to H.P. and one of its base edges is perpendicular to H.P.
(ii) A sphere of diameter 70 mm is placed on the top face of a prism.
(iii) The axis of pyramid is parallel to V.P. and one of its base edges is perpendicular to V.P.
(iv) The axis of prism is perpendicular to V.P. and one of its base edges is parallel to H.P.
(A) (iii) and (iv) only
(B) (iii) only
(C) (ii) and (iii) only
(D) (ii) only
[Figure: Isometric projection of a sphere of \( \phi 50 \) resting centrally on the triangular base of an inverted triangular pyramid (apex pointing down). Dimensions: pyramid height 70, base edge 50. An arrow at the bottom left shows the direction of viewing.]
Answer: (B) (iii) only
Teacher's Note:
a) The lower solid is an inverted triangular pyramid with a vertical axis, not a prism.
b) The sphere is \( \phi 50 \); 70 mm is the height of the pyramid, so statement (ii) is wrong.
8. Match the List-I with List-II for the given isometric projection of combination of solids : [1 Mark]
List-I | List-II
(1) Top solid | (i) Hemisphere
(2) Bottom solid | (ii) Sphere
(3) Radius of sphere | (iii) 40 mm
(4) Radius of hemisphere | (iv) 25 mm
(A) 1-(ii), 2-(i), 3-(iv), 4-(iii)
(B) 1-(i), 2-(ii), 3-(iii), 4-(iv)
(C) 1-(ii), 2-(i), 3-(iii), 4-(iv)
(D) 1-(i), 2-(ii), 3-(iv), 4-(iii)
[Figure: Isometric projection of a sphere of radius R 25 resting centrally on the flat top face of a hemisphere of diameter \( \phi 80 \) whose curved surface is downward.]
Answer: (A) 1-(ii), 2-(i), 3-(iv), 4-(iii)
Teacher's Note:
a) The sphere is on top (R 25), and the hemisphere is at the bottom.
b) Radius of hemisphere = \( 80 \div 2 = 40 \) mm; always halve a diameter before matching.
9. A thread with vertical and parallel flanks is called [1 Mark]
(A) Square thread
(B) Triangular thread
(C) Knuckle thread
(D) V-thread
Answer: (A) Square thread
Teacher's Note:
a) In a square thread the flanks are perpendicular to the axis and parallel to each other.
b) V-threads have inclined flanks at \( 60^{\circ} \); knuckle threads have rounded crests and roots.
10. The length of the metal end of a plain stud of diameter 30 mm is [1 Mark]
(A) 10 mm
(B) 24 mm
(C) 30 mm
(D) 66 mm
Answer: (C) 30 mm
Teacher's Note:
a) The metal end of a stud is taken as equal to the diameter d, so it is 30 mm here.
b) Remember: the nut end length is taken as \( 2d + 6 \).
11. Choose the correct option regarding screw threads. [1 Mark]
(i) A screw thread formed on the surface of a cylinder is known as taper thread.
(ii) In single start thread lead L = 2P.
(iii) In the practical use of the threads, clearance must be provided between the external and internal threads.
(iv) Root is the edge of the thread surface nearest to the axis in case of internal thread.
(A) (i) and (ii) only
(B) (ii) only
(C) (iii) and (iv) only
(D) (iii) only
Answer: (D) (iii) only
Teacher's Note:
a) A thread on a cylinder is a parallel thread, and for a single start thread lead L = P.
b) For an internal thread the root is farthest from the axis, so (iv) is also wrong.
12. In a BUSH BEARING assembly, the inside of the bush is bored as a fit for _________. [1 Mark]
(A) Foundation bolt
(B) Sole
(C) Shaft
(D) Recess
Answer: (C) Shaft
Teacher's Note:
a) The shaft rotates inside the bush, so the bore of the bush is finished to fit the shaft.
b) The bush can be replaced when worn, which saves the main body of the bearing.
13. Due to which of the following reason, usually pipes are made of standard length, not of desired length ? [1 Mark]
(A) Constraints of manufacturing
(B) Policy of government
(C) Existence of friction
(D) Leakage possibilities
Answer: (A) Constraints of manufacturing
Teacher's Note:
a) Pipes are made in standard lengths because of manufacturing and transport limits.
b) Longer lines are made by joining these lengths with pipe joints such as flanged joints.
14. Match the List-I with List-II : [1 Mark]
List-I (Front view of component of assembly) | List-II (Name of the component)
(1) [drawing 1] | (i) Flange
(2) [drawing 2] | (ii) Bush of BUSH BEARING
(3) [drawing 3] | (iii) Gasket
(4) [drawing 4] | (iv) Bush of OPEN BEARING
(A) 1-(iv), 2-(ii), 3-(iii), 4-(i)
(B) 1-(ii), 2-(iii), 3-(i), 4-(iv)
(C) 1-(ii), 2-(i), 3-(iv), 4-(iii)
(D) 1-(i), 2-(iv), 3-(iii), 4-(ii)
[Figure: List-I drawings. (1) Two concentric circles (a ring) with centre lines and a small hole marked at the top. (2) A thin, tall vertical rectangle with a centre line. (3) A flange with a hub on the left and a broken pipe end, with hidden lines for the bore and bolt holes. (4) A U-shaped half ring with hidden lines inside and vertical collar ends.]
Answer: (B) 1-(ii), 2-(iii), 3-(i), 4-(iv)
Teacher's Note:
a) A gasket is a thin packing sheet, so its front view is a thin rectangle.
b) The bush of an open bearing is a half bush, so it looks like a U in front view.
SECTION B
Q.15 to Q.18 : Read the following paragraph and answer the questions given below :
The given image is a fire-extinguisher. It is made up of a bigger cylindrical tank with a hemispherical portion on it. A very small cylindrical portion at the top has a valve assembly and a hose. Assume that the diameter of the hemispherical portion as 140 mm and height of the bottom cylindered portion as 350 mm.
[Figure: Photograph of a fire-extinguisher: a tall cylindrical tank with a rounded top, a valve and handle assembly at the top, and a hose running down the left side held by a clamp band.]
15. The projections which are mostly used in the design process of this product is [1 Mark]
(A) perspective projection and oblique projection
(B) isometric projection and orthographic projection
(C) perspective projection and axonometric projection
(D) oblique projection and multi view projection
Answer: (B) isometric projection and orthographic projection
Teacher's Note:
a) Isometric views show the 3D shape; orthographic views give exact sizes for manufacturing.
b) Perspective views are mainly used for presentation, not for production drawings.
16. Cylinder and hemisphere are examples of [1 Mark]
(A) Solids of revolution
(B) Tetrahedron
(C) Solids of sectioning
(D) Polyhedron
Answer: (A) Solids of revolution
Teacher's Note:
a) A cylinder is made by revolving a rectangle, and a hemisphere by revolving a quarter circle, about an axis.
b) Polyhedra have only flat faces, so curved solids are not polyhedra.
17. The total height in true scale, of the fire-extinguisher excluding the top valve assembly is [1 Mark]
(A) 336 mm
(B) 420 mm
(C) 140 mm
(D) 350 mm
Answer: (B) 420 mm
Teacher's Note:
a) Height of hemisphere = its radius = \( 140 \div 2 = 70 \) mm.
b) Total height = \( 350 + 70 = 420 \) mm; do not add the full diameter 140 mm.
18. If a label with "USE & CARE" instructions is to be pasted on the entire outer circumference of the bottom cylindrical portion, then the surface area of the label to be used will be _________.
(Hint : Lateral surface area of cylinder = \( 2\pi rh \)) [1 Mark]
(A) \( 770 \, cm^2 \)
(B) \( 1400 \, cm^2 \)
(C) \( 3500 \, cm^2 \)
(D) \( 1540 \, cm^2 \)
Answer: (D) \( 1540 \, cm^2 \)
Teacher's Note:
a) Convert first: r = 70 mm = 7 cm and h = 350 mm = 35 cm.
b) Area = \( 2 \times \frac{22}{7} \times 7 \times 35 = 1540 \, cm^2 \).
Q. 19 to Q. 22 : Read the following paragraph and answer the questions given below :
Grub screw, also known as set screw, is a type of fixing screw that is most often used to join one component or part securely to another. Grub screws are used to hold parts like sleeve, collar, gear etc. on a shaft to prevent relative motion. It is represented by thread size such as M8, M10 etc. Where M stands for metric thread and the numeral represents the diameter of screw in millimeter.
19. The front view of a vertical grub screw is represented by which of the following views ? [1 Mark]
[Figure: Four front views. (i) A plain cylindrical rod with a slot at the top and a broken, hatched lower end, with no thread lines. (ii) A screw with a countersunk (inverted cone) slotted head and a threaded shank. (iii) A screw with a trapezoidal slotted head and a plain shank with a broken, hatched lower end. (iv) A headless cylindrical screw with a slot at the top, thin thread lines along its full length and chamfered ends.]
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Answer: (D) (iv)
Teacher's Note:
a) A grub screw has no head; it is threaded along its full length with a slot at one end.
b) Options (ii) and (iii) have heads, and (i) has no thread lines.
20. The complete circle of \( \phi \) d and the incomplete circle of \( \phi \) 0.8 d in the top view of a vertical grub screw with diameter 'd' are drawn to _____ [1 Mark]
(A) represent conventionally the external V-thread.
(B) attain the neatness of the figure.
(C) represent conventionally the internal V-thread.
(D) attain the 3D effect of the circles.
Answer: (A) represent conventionally the external V-thread.
Teacher's Note:
a) For an external thread, the major diameter is a full thick circle and the minor diameter is a thin three-quarter circle.
b) For an internal thread the order is reversed: the minor circle is complete and the major circle is broken.
21. Grub screws are also known as [1 Mark]
(A) Counter sunk head screw
(B) Pan head screw
(C) Headless screw
(D) Threadless screw
Answer: (C) Headless screw
Teacher's Note:
a) A grub screw has no head, so it sits flush or below the surface of the part.
b) It is fully threaded, so "threadless screw" is wrong.
22. Why grub screws are used to hold sleeve on a shaft ? [1 Mark]
(A) To reduce the misalignment of shaft.
(B) To prevent the relative motion.
(C) To reduce the production cost.
(D) To prevent the leakage of joint.
Answer: (B) To prevent the relative motion.
Teacher's Note:
a) The passage states that grub screws hold parts on a shaft to prevent relative motion.
b) Read the case paragraph carefully; many answers are stated in it directly.
23. (a) Fig. 1 shows the details of the parts of a TURN-BUCKLE. Assemble all these parts correctly and then draw to scale 1 : 1 its following views. Keep 54 mm threaded portion of each rod inside the body of turn-buckle.
(i) Front view upper half in section [13 Marks]
(ii) Right side view [8 Marks]
Print the title and scale used. Draw the projection symbol. Give 6 important dimensions. [6 Marks]
[Figure: Fig.-1 TURN-BUCKLE (figure not to scale, use the dimensions given). Front view of Rod A (M.S) 1 off: threaded length 64, thread \( \phi 22 \) LH, core \( \phi 18 \), with a broken (hatched) end on the left. Front view of Rod B (M.S) 1 off: threaded length 64, thread \( \phi 22 \) RH, core \( \phi 18 \), broken end on the right. Body (M.S) 1 off, front view and top view: overall length 160; at each end a conical portion 26 long followed by a collar 14 long of \( \phi 62 \), the cone reducing to \( \phi 32 \) at the extreme end; threaded holes \( \phi 22 \) LH threads (left end) and \( \phi 22 \) RH threads (right end) shown by hidden lines; central portion 32 thick in front view; in the top view a central rectangular opening 44 wide between the two collars.]
Answer:
Title: ASSEMBLY OF TURN-BUCKLE, Scale 1 : 1, first angle projection.
(i) Front view, upper half in section (13 marks):
1. Draw the upper half of the body: overall length 160, conical ends 26 long (\( \phi 32 \) to \( \phi 62 \)), collars 14 long of \( \phi 62 \), and the central bar 32 thick between the collars (3 marks).
2. Hatch the cut (solid) portions of the body in the upper half with thin lines at \( 45^{\circ} \) (1 mark).
3. Draw the lower half of the body in outside view, with the same outline and no hatching (3 marks).
4. Show rod A (LH) from the left and rod B (RH) from the right, each entering 54 mm into the body; the 64 mm threaded length leaves 10 mm outside; draw the arc (curved end) at the inner end of each rod (3 marks).
5. Show the threads conventionally: thick lines for major diameter \( \phi 22 \) and thin lines for core diameter \( \phi 18 \) on both rods (2 marks).
6. Show the outer ends of both rods with the conventional break (hatched) (1 mark). Rods are not hatched as they are cut along their axis.
(ii) Right side view (8 marks):
7. Draw four concentric circles: \( \phi 62 \) (collar), \( \phi 32 \) (end of cone), \( \phi 22 \) full circle and \( \phi 18 \) broken (three-quarter) circle for the external thread of rod B (4 marks).
8. Hatch the conventional end of rod B inside the circles (½ mark) and mark the cutting plane A-A' (½ mark).
9. Draw hidden lines 44 mm apart to show the central opening of the body (3 marks).
Details (6 marks):
10. Print the title and the scale used (1 + 1 mark) and draw the first angle projection symbol (1 mark).
11. Give six important dimensions, for example 160, 64, 54, 26, 14, \( \phi 62 \), \( \phi 32 \), \( \phi 22 \) LH / RH, \( \phi 18 \) and 44 (3 marks).
Teacher's Note:
a) Marks split: front view 13 (body upper half 3, hatching 1, lower half 3, rods 6), side view 8, details 6.
b) A common mistake is hatching the rods; shafts and rods cut along their axis are never sectioned.
c) Take care with 54 mm inside the body; students often push the full 64 mm thread inside.
OR
(b) Fig. 2 shows the assembly of a SLEEVE AND COTTER JOINT. Disassemble the parts correctly and then draw to scale 1 : 1, its following views of the following components. Keeping the same position of both sleeve and cotter-B with respect to both H.P. and V.P.
(i) SLEEVE
(a) Front view lower half in section [8 Marks]
(b) Left side view [6 Marks]
(ii) COTTER-B
(a) Front view [4 Marks]
(b) Left side 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 SLEEVE AND COTTER JOINT (figure not to scale, use dimensions given). Front view: Sleeve (M.S) of overall length 140 with R4 rounded corners, outside diameter \( \phi 52 \), joining Rod A (M.S) on the left and Rod B (M.S) on the right (rod ends \( \phi 20 \) outside the sleeve with R3 fillet; a 2 x \( 45^{\circ} \) chamfer shown at the rod ends inside the sleeve); Cotter A (M.S) and Cotter B (M.S) pass through slots whose outer edges are 25 from each end of the sleeve; cotter width 27 at the top, 25 at the axis, with a 3 mm clearance in the slots; taper 1 in 30 on the inner side ("Taper on this side"); each cotter extends 40 above and 40 below the axis with curved ends. Side view: circle \( \phi 52 \) of the sleeve with the hatched rod section \( \phi 32 \), and the cotter shown vertically, 8 thick, with ends chamfered to 4.]
Answer:
Title: DISASSEMBLY OF SLEEVE AND COTTER JOINT, Scale 1 : 1, first angle projection.
(i) SLEEVE - (a) Front view, lower half in section (8 marks):
1. Draw the outline of the sleeve, length 140 and \( \phi 52 \), with R4 rounds at the corners (3 marks), and a horizontal line in the lower half showing the bore of \( \phi 32 \) (1 mark).
2. Draw the two cotter slots in the lower half, their outer edges 25 from each end, with the taper of 1 in 30 on the inner side and the 3 mm clearance, exactly as in the question (2½ + ½ marks).
3. Hatch the cut solid portions of the lower half (1 mark).
(i) SLEEVE - (b) Left side view (6 marks):
4. Draw the circle of \( \phi 52 \) (2 marks) and the circle of \( \phi 32 \) for the bore (1½ marks).
5. Show both cotter holes as hidden lines 8 apart, vertical through the centre (2 marks).
6. Mark the cutting plane A-A' (½ mark).
(ii) COTTER-B - (a) Front view (4 marks):
7. Draw the vertical straight edge (1 mark) and the tapered edge (1 in 30), with width 27 at the top and 25 at the axis level (1 mark).
8. Draw the upper and lower ends as curves, 40 above and 40 below the axis (2 marks).
(ii) COTTER-B - (b) Left side view (3 marks):
9. Draw two parallel vertical edges 8 apart (1 mark).
10. Draw the upper and lower ends chamfered to a width of 4 (2 marks).
Details (6 marks):
11. Print the titles of both parts and the scale used (1 + 1 mark), and draw the projection symbol (1 mark).
12. Give six important dimensions, for example 140, 25, R4, \( \phi 52 \), \( \phi 32 \), 8, 27, 25, 40 and 4 (3 marks).
Teacher's Note:
a) Marks split: sleeve front view 8, sleeve side view 6, cotter front view 4, cotter side view 3, details 6.
b) Keep the taper of the slot and of the cotter on the same side as shown in Fig. 2; reversing it loses marks.
c) Hatch only the sleeve's lower half; the cotter views are not sectioned.
SECTION C
24. (a) Construct an isometric scale. [4 Marks]
Answer:
1. Draw a horizontal line and, from its left end, draw two lines at \( 30^{\circ} \) and \( 45^{\circ} \) to it (1 mark).
2. On the \( 45^{\circ} \) line mark the true length divisions of 10 mm (0, 10, 20 ... 70), and divide the first part (left of 0) into 1 mm divisions (1 mark).
3. From each division drop vertical projectors (perpendicular to the horizontal line) to meet the \( 30^{\circ} \) line; these points give the isometric lengths, and complete the scale with its divisions (1 mark).
4. Print "TRUE LENGTH / Scale 1:1" on the \( 45^{\circ} \) line and "ISOMETRIC LENGTH / Isometric scale" on the \( 30^{\circ} \) line, and mark the angles \( 30^{\circ} \) and \( 45^{\circ} \) (1 mark).
Teacher's Note:
a) Isometric length is about 0.816 of true length, so 70 mm true gives about 57 mm isometric.
b) A common mistake is forgetting the 1 mm sub-divisions of the first part and the printed labels.
(b) Draw the isometric projection of a triangular prism (base edge 35 mm, length 60 mm). It is placed on H.P. with one of its long edges and its axis is perpendicular to V.P. Indicate the direction of viewing. Give all the dimensions. [9 Marks]
Answer:
1. Draw the helping figure: an equilateral triangle of side 35 with its apex downward (resting on the long edge), enclosed in a rectangle (1 mark).
2. Using the isometric scale, draw the front triangular face as an inverted triangle in isometric (3 marks), with the top edge 35 along the \( 30^{\circ} \) direction and the lowest corner on H.P.
3. Draw the rear triangular face 60 mm behind along the other \( 30^{\circ} \) direction (1½ marks).
4. Join the three corresponding corners with three parallel long edges of length 60 (1½ marks).
5. Draw the axis perpendicular to V.P. (½ mark) and an arrow showing the direction of viewing (½ mark).
6. Give all dimensions: base edge 35 and length 60 (1 mark).
Teacher's Note:
a) Marks split: helping figure 1, triangles 4½, edges 1½, axis and arrow 1, dimensions 1.
b) One mark is deducted for incorrect position, so keep the prism resting on a long edge with the axis perpendicular to V.P.
c) Measure all lengths with the isometric scale, and do not show hidden lines.
25. (a) Draw to scale 1 : 1, the standard profile of EXTERNAL METRIC THREAD taking enlarged pitch as 50 mm. Give standard dimensions. [8 Marks]
Answer:
1. Calculate: P = 50, 0.5P = 25, D = 0.866P = 43.3, d = 0.61P = 30.5, D/8 = 5.4 and D/6 = 7.2 (mm).
2. Draw horizontal lines D = 43.3 apart and mark vertical lines at 0.5P = 25 intervals; mark D/8 below the top line and D/6 above the bottom line (2 marks).
3. Draw the flanks at \( 60^{\circ} \) to each other, flat crests cut at D/8 and rounded roots at D/6, showing at least two threads; the thread depth is d = 30.5 (3 marks).
4. Hatch the thread section and show the conventional break at the bottom (1 mark).
5. Give standard dimensions: P, 0.5P, \( 60^{\circ} \), D, d, D/8 and D/6 (2 marks).
Teacher's Note:
a) Marks split: distances 2, crests, roots and flanks 3, hatching with break 1, dimensions 2.
b) Two marks are deducted in all if the profile is sketched freehand instead of drawn to scale 1 : 1.
OR
(b) Draw to scale 1 : 1, the front view of a vertical hexagonal bolt of diameter 25 mm. The bolt is resting on H.P. with its hexagonal head on it. Give standard dimensions. [8 Marks]
Answer:
1. Proportions for d = 25: head height 0.8d = 20, width across flats 1.5d + 3 = 40.5, width across corners 2d + 6 = 56 (approx.), threaded length 2d + 6 = 56, core diameter 0.8d = 20.
2. Draw the hexagonal head resting on H.P., 20 high, showing either across flats (two faces) or across corners (three faces) (3 marks).
3. Draw the chamfering arcs on the head: arc of R = d on the middle face and smaller arcs on the side faces, with the \( 30^{\circ} \) chamfer (1 mark).
4. Draw the shank of \( \phi 25 \) upward from the head: the threaded portion 56 long with thin lines at \( \phi 0.8d \) (1 mark), the unthreaded portion (½ mark), and the conventional rounded end with R = d (½ mark).
5. Give standard dimensions: \( \phi d \), \( \phi 0.8d \), 0.8d, 1.5d + 3 (or across corners), 2d + 6 and R = d (2 marks).
Teacher's Note:
a) One mark is deducted for incorrect position; the head must be at the bottom on H.P. with the shank vertical.
b) Two marks are deducted in all for a freehand sketch instead of a 1 : 1 drawing.
c) A common mistake is drawing the thread lines up to the head; threads cover only 2d + 6 from the end.
Please click the link below to download pdf file of CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68
Free study material for Engineering Drawing
Download CBSE Question Papers: Class 12 Engineering Graphics
Class 12 Engineering Graphics Past Exam Papers & Resources
Explore downloadable past papers for Class 12 Engineering Graphics. Utilizing the CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68 ensures complete preparedness by offering clear insights into historical question styles and marking expectations.
Importance of Solving CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68
Reviewing official papers clarifies the exact marking scheme and structural layout established by the CBSE, enabling students to structure answers for maximum score potential.
Complete Your Exam Preparation
Wrap up your exam preparation by reviewing detailed answer keys and tackling additional practice sets. All resources on our platform are free to access.
FAQs
The CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68 is available for download on StudiesToday.com. It includes complete set with all sections so that Class 12 students can practice with the exact same paper that came in the CBSE exams.
Yes, the solutions for CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68 are prepared by subject matter experts as per official marking scheme. Class 12 students will understand the structure of answers and 'step-marks' methodology Engineering Graphics.
Solving previous year papers like CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68 is important to understand repeat themes and question difficulty levels of Engineering Graphics. It helps Class 12 students to test their time management skills too.
Yes, where applicable, CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68 is available in both English and Hindi mediums. All students from Class 12 can access Engineering Graphics study material in their preferred language.
No, all previous year question papers on StudiesToday, including CBSE Class 12 Engineering Graphics Question Paper 2025 Solved Code 68, are provided free of charge in mobile-friendly PDF.