Jawaharlal Nehru Technological University Hyderabad
Jawaharlal Nehru Technological University Hyderabad B.Tech I Year ENGINEERING DRAWING Previous Question Paper (September - 2023 - R13)
Answer any five questions
All questions carry equal marks
Main
Construct a rectangular hyperbola when the distance of the focus from the directrix is 65 mm and eccentricity is 3/2.
Answer
To construct a hyperbola with eccentricity e = 3/2 and focus-directrix distance (VF) = 65 mm, follow these steps:
- Draw a vertical line as the directrix (DD') and a horizontal line perpendicular to it as the axis (AB).
- Mark the focus (F) on the axis at a distance of 65 mm from the directrix.
- Divide the distance between the directrix and focus into 5 equal parts (since e = 3/2, 3+2=5). The vertex (V) is located 3 units from the directrix and 2 units from the focus.
- Mark the vertex (V) at 39 mm from the directrix (3/5 of 65 mm).
- Draw a vertical line through the vertex and mark a point P on it such that VF = VP. This establishes the eccentricity ratio.
- Draw a line from the focus through P and extend it.
- Mark several points 1, 2, 3, etc., on the axis beyond the vertex. Draw vertical lines through these points.
- With F as the center and radius equal to the distance from the axis point to the corresponding vertical line (e.g., 1-1'), draw arcs intersecting the vertical line at points P1 and P1'.
- Repeat this process for other points to obtain a series of points on the curve.
- Join these points with a smooth freehand curve to complete the hyperbola.
An area covered by a triangle of base 12 cm and altitude 24 cm, represent an area of 36 km2. Find the scale factor and construct a diagonal scale to read kilometers, hectometers and decameters. Mark the distance of 2.85 km and 4.23 km on it.
Answer
First, calculate the Representative Fraction (RF). The area of the triangle on the drawing is 1/2 × base × altitude = 1/2 × 12 cm × 24 cm = 144 cm2. The actual area is 36 km2 = 36 × (105 cm)2 = 36 × 1010 cm2. The area scale is 144 / (36 × 1010) = 4 / 1010 = 1 / (2.5 × 109). The length scale factor (RF) is the square root of the area scale: RF = sqrt(1 / 2.5 × 109) = 1 / 50,000.
To construct the diagonal scale, assume a maximum length of 5 km. The length of the scale = RF × Maximum Length = (1 / 50,000) × 5 × 105 cm = 10 cm. Draw a line 10 cm long and divide it into 5 equal parts, each representing 1 km. Divide the first part into 10 subdivisions to represent hectometers (1 km = 10 hm). Use the diagonal scale principle to further divide the height into 10 parts to represent decameters (1 hm = 10 dam).
To mark 2.85 km: Place the divider at the 2 km mark on the main scale, move to the 8 hm division on the primary subdivision, and move up 5 units on the diagonal to reach the 0.05 km (5 dam) mark. To mark 4.23 km: Place the divider at the 4 km mark on the main scale, move to the 2 hm division, and move up 3 units on the diagonal to reach the 0.03 km (3 dam) mark.
The front view of line AB measures 65 mm and makes an angle of 45o with xy. A is on HP and VT of the line is 15 mm above HP. The line is inclined at 30o to VP. Draw the projections of the line and find its true length and inclination with HP. Also locate its HT.
Answer
Sign in to view the complete answer paper.
Draw the projections of the regular pentagon of 50 mm side having its surface inclined at 45o with HP. A side of the pentagon lies on HP, inclined at 30o to VP.
Answer
Sign in to view the complete answer paper.
A hexagonal prism, with side of base 30 mm and axis 70 mm long, is resting on one of its longer edges on HP, with its axis inclined at 40o to VP. A rectangular face containing the longer edge, on which the prism rests is inclined at 30o to HP. Draw the projections of the prism.
Answer
Sign in to view the complete answer paper.
A cylinder of base 80 mm diameter and 100 mm long, lies on H.P with its axis parallel to both H.P and V.P. It is cut by a vertical section plane, which intersects the axis of the cylinder at an angle of 30o at its mid point. Draw the sectional front view and determine the true shape of the section.
Answer
Sign in to view the complete answer paper.
A vertical cone of diameter of base 40 mm and height 50 mm, is cut by a section plane, perpendicular to VP and inclined at 30o to HP, so as to bisect the axis. Draw the development of the truncated portion of the cone.
Answer
Sign in to view the complete answer paper.
A square prism of base 50 mm side and height 125 mm stands on the ground with a side of the base inclined at 30o to the VP. It is penetrated by a cylinder, 50 mm diameter and 125 mm long, whose axis is parallel to both the HP and VP and bisects the axis of the prism. Draw the projections of the curves of the intersection.
Answer
Sign in to view the complete answer paper.
A cylindrical block of base 60 mm diameter and height 90 mm, standing on the H.P with its axis perpendicular to the H.P. Draw its isometric view.
Answer
Sign in to view the complete answer paper.
Draw the orthographic projections of the following isometric view.
All dimensions are in mm
Answer
Sign in to view the complete answer paper.