CBSE Class 11 Engineering Graphics — NCERT Solutions

Given: Various features of an engineering drawing and the requirement to identify the correct line type for each.

(a) Axis of a Cone: A Chain thin line (long-dash dot, thin) is used for the axis (centre line) of a cone.

(b) Boundary line: A Continuous thick line (visible outlines/boundary lines) is used for the boundary/outline of an object.

(c) Projection line (Extension line): A Continuous thin line is used for projection/extension lines.

(d) Line for Short Break: A Continuous thin line with zigzags (irregular freehand) is used for short break lines.

(e) Line for Long Break: A Thin straight line with zigzags (ruled line with short zigzag) is used for long break lines.

(f) Dimension line: A Continuous thin line (with arrowheads at both ends) is used for dimension lines.

Given: The concept of 'order of priority of coinciding lines' in Engineering Graphics.

Explanation:
In engineering drawings, it is common for two or more different types of lines to coincide (fall on the same position). In such cases, a definite order of priority is followed to decide which line should be drawn/shown. The order of priority is:

1. Visible outlines and edges (Continuous thick lines) — highest priority
2. Hidden outlines and edges (Dashed thin lines)
3. Cutting plane lines (Chain thin, thick at ends)
4. Centre lines / Axis lines (Chain thin lines)
5. Lines of symmetry (Chain thin lines)
6. Projection/Extension lines (Continuous thin lines) — lowest priority

This means that if a visible outline coincides with a centre line, the visible outline is drawn and the centre line is omitted at that location.

Given: Figure 2.1 showing a circle with labelled parts.

Answer: The fixed point O is the Centre.

The centre is the fixed interior point equidistant from every point on the circumference.

Answer: The distances OA, OB, OC and OP are radii.

All line segments drawn from the centre to any point on the circumference are equal and are called radii (plural of radius).

Given: Major axis = 70 mm, Minor axis = 40 mm
Semi-major axis $a = 35$ mm, Semi-minor axis $b = 20$ mm

Steps of Construction:

1. Draw two concentric circles with centre $O$, one with radius $a = 35$ mm (outer) and one with radius $b = 20$ mm (inner).
2. Draw the major axis $AA'$ (horizontal, 70 mm) and minor axis $BB'$ (vertical, 40 mm) through $O$.
3. Divide both circles into 12 equal parts (every 30°) by drawing radial lines through $O$. Number the division points 1, 2, 3, … 12 on both circles.
4. From each point on the outer circle, draw a line parallel to the minor axis (i.e., vertically downward/upward).
5. From the corresponding point on the inner circle, draw a line parallel to the major axis (i.e., horizontally).
6. The intersection of these two lines gives a point on the ellipse.
7. Repeat for all 12 divisions to get 12 points on the ellipse.
8. Join all the points with a smooth freehand curve (French curve) to complete the ellipse.

Result: The required ellipse with major axis 70 mm and minor axis 40 mm is constructed.

Given: Semi-major axis $a = 40$ mm, Semi-minor axis $b = 25$ mm
Major axis $AA' = 80$ mm, Minor axis $BB' = 50$ mm

Steps of Construction:

1. Draw the major axis $AA' = 80$ mm (horizontal) and bisect it to get centre $O$. Draw the minor axis $BB' = 50$ mm (vertical) through $O$.
2. Locate the foci: The distance of each focus from centre is $$OF = \sqrt{a^2 - b^2} = \sqrt{40^2 - 25^2} = \sqrt{1600 - 625} = \sqrt{975} \approx 31.2 \text{ mm}$$ Mark foci $F_1$ and $F_2$ on the major axis at 31.2 mm on either side of $O$.
3. Divide the major axis $AA'$ into a number of equal parts (say 8–10 parts). Mark points $1, 2, 3, \ldots$ between $A$ and $A'$.
4. With $F_1$ as centre and radius $= A$-to-point-1 (i.e., $A1$), draw arcs above and below the major axis.
5. With $F_2$ as centre and radius $= A'$-to-point-1 (i.e., $A'1$), draw arcs to intersect the previous arcs. The intersections are points on the ellipse. (Note: $A1 + A'1 = AA' = 2a = 80$ mm always.)
6. Repeat for all division points to get sufficient points on the ellipse.
7. Join all points with a smooth curve using a French curve.

Result: The required ellipse with semi-major axis 40 mm and semi-minor axis 25 mm is drawn.

(i) In orthographic projection, the projectors (lines of sight) are perpendicular to the plane of projection.

(ii) In I angle projection, the object comes between the observer and the plane of projection.

(iii) In III angle projection, the plane of projection comes between the observer and the object.

Given/Concept: Sectioning is a drawing technique used in engineering graphics.

Answer: Sectioning is the process of cutting a solid (or an object) with an imaginary cutting plane (called the section plane) in order to reveal its internal details, shape, and construction. The exposed cut surface is shown with hatching lines in the resulting view, which is called a sectional view.

Given: Object placed in II quadrant and IV quadrant separately; H.P. is rotated clockwise to open the box.

Concept: In the standard glass-box method, the H.P. is rotated downward (clockwise when viewed from the right) to bring it into the plane of the V.P.

Observation and Discussion:

1. When the object is in the II quadrant (above H.P., behind V.P.):

• The Front View (projection on V.P.) appears above XY.
• The Top View (projection on H.P.) also appears above XY after rotation.
• Therefore, both views overlap each other on the same side of XY, making it impossible to distinguish them clearly.

2. When the object is in the IV quadrant (below H.P., in front of V.P.):

• The Front View appears below XY.
• The Top View also appears below XY after rotation.
• Again, both views overlap each other on the same side of XY.

Conclusion: In both II and IV quadrants, the Front View and Top View fall on the same side of the reference line XY and overlap each other. This makes the drawing ambiguous and unreadable. This is why the II and IV quadrants are NOT used in practice for orthographic projection. Only I quadrant (First Angle Projection) and III quadrant (Third Angle Projection) are used, because in these cases the two views fall on opposite sides of XY and do not overlap.

Given: Equilateral triangle, base side = 50 mm, plane in V.P. (Vertical Plane).

Concept: When a plane figure lies in the V.P., its isometric projection is drawn on the isometric front face. All lines parallel to the isometric axes are foreshortened by the isometric scale factor $\approx 0.816$ (i.e., multiply true length by 0.816). In isometric drawing (full-size method), true lengths are used directly.

Step-by-step procedure:

Step 1 – Draw the reference axes.
Draw the isometric axes: one vertical axis and two axes at 30° to the horizontal (left and right). Since the triangle is in the V.P., it will appear on the vertical isometric plane (the front face).

Step 2 – Enclose the triangle in a rectangle.
Enclose the equilateral triangle of side 50 mm in a rectangle.

• Height of equilateral triangle $= \dfrac{\sqrt{3}}{2} \times 50 = \dfrac{1.732}{2} \times 50 \approx 43.3$ mm.
• Rectangle dimensions: width = 50 mm, height = 43.3 mm.

Step 3 – Draw the isometric rectangle on the V.P. face.
On the isometric drawing, the V.P. face uses the vertical axis and the 30° right (or left) axis.

• Mark point $A$ at the bottom-left corner.
• Along the 30° axis, mark $B$ at 50 mm from $A$ (base of rectangle).
• From $A$ and $B$, draw vertical lines of height 43.3 mm to get points $D$ and $C$ respectively.
• $ABCD$ is the isometric rectangle enclosing the triangle.

Step 4 – Locate key points of the triangle inside the rectangle.

• The base of the triangle coincides with $AB$.
• The apex $P$ is at the midpoint of $DC$, i.e., midpoint of the top side of the rectangle.
• Mark $M$ = midpoint of $AB$; the apex $P$ is directly above $M$ at height 43.3 mm.

Step 5 – Draw the triangle.
Join:

• $A$ to $B$ (base, along 30° axis).
• $A$ to $P$ (left slant side — draw as a straight line between the two points).
• $B$ to $P$ (right slant side — draw as a straight line between the two points).

Step 6 – Isometric scale (if isometric projection, not drawing).
Multiply all dimensions by 0.816:

• Base $= 50 \times 0.816 = 40.8$ mm.
• Height $= 43.3 \times 0.816 = 35.3$ mm.

Redraw with these scaled dimensions following the same procedure.

Result: The isometric projection of the equilateral triangle (base 50 mm) in V.P. is obtained — it appears as a triangle on the isometric front face with the base along the 30° axis and the apex located vertically above the midpoint of the base.

Answer: True

In the development of surfaces, every line drawn on the development must represent the true length of the corresponding line on the actual surface of the object. If a line is not in its true length (i.e., it is foreshortened in the given views), its true length must first be determined before it can be used in the development. This is a fundamental principle of surface development.