BT-105 · Engineering Graphics/Unsolved PYQ Paper
BT-105 Engineering Graphics - Dec 2023 Question Paper
Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.
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7 MarksaFill in the blanks.
- i)The number of mutually perpendicular planes that may surround an object in space is _______.
- ii)In the orthographic projection, the projectors are _______ to the plane of projection.
- iii)In the first angle projection, the object is imagined to be placed _______.
- iv)In _______ projection, any view is so placed that it represents the side of the object nearer to it.
- v)In half sectional view, _______ of the object is imagined to be removed.
- vi)Sectional views reveal _______.
- vii)Graphics can be converted into hard copy with a _______.
How to answer
- Fill blank (i): 6 (or three pairs of planes)
- Fill blank (ii): perpendicular / normal
- Fill blank (iii): in the first quadrant / between observer and plane of projection
- Fill blank (iv): third angle
- Fill blank (v): one-fourth (one quarter)
- Fill blank (vi): internal/hidden details
- Fill blank (vii): plotter / printer
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7 MarksbDraw an angle of $45^\circ$ and $135^\circ$ with the help of Scale of Chords.
How to answer
- Construct a quadrant with a convenient radius R and divide the arc into chord lengths representing degrees (scale of chords)
- Draw a baseline AB and mark radius R
- Set divider to the chord length of 45° from the scale and cut an arc to draw 45°
- Construct 135° either by adding chord lengths (e.g. 90° + 45°) or using 180° - 45°
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7 MarksaThe distance of $352\text{ km}$ between two cities, on a road map, is represented by a line of length $70.4\text{ mm}$. Draw a diagonal scale to read up to single kilometer and long enough to measure up to $900\text{ km}$. Show on the scale a distance of $649\text{ km}$.
How to answer
- Calculate RF: $\text{RF} = \frac{70.4\text{ mm}}{352 \times 10^6\text{ mm}} = \frac{1}{5,000,000}$
- Calculate Length of Scale (LOS): $\text{LOS} = \text{RF} \times 900\text{ km} = \frac{900 \times 10^6}{5,000,000} = 180\text{ mm} = 18\text{ cm}$
- Divide LOS (18 cm) into 9 equal parts (each 100 km)
- Subdivide the first part into 10 divisions (each 10 km)
- Erect vertical division with 10 diagonal subdivisions (each 1 km)
- Mark the distance of 649 km (600 km + 40 km + 9 km)
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7 MarksbConstruct a Conic section having eccentricity of $3/4$ and focus $25\text{ mm}$ from directrix. Measure its major and minor axes and the distance between two foci.
How to answer
- Identify conic as an ellipse since eccentricity $e = 3/4 < 1$
- Draw directrix $DD'$ and axis perpendicular to it; mark focus $F$ at $25\text{ mm}$ from directrix point $C$
- Divide $CF = 25\text{ mm}$ into $3+4 = 7$ equal parts to find vertex $V_1$ ($CV_1 = 4$ parts, $V_1F = 3$ parts)
- Determine the other vertex $V_2$ using $V_2F/V_2C = 3/4$ to fix the major axis length
- Construct the ellipse using arcs of circles / eccentric method points
- Measure and state the lengths: major axis, minor axis, and distance between two foci ($F_1F_2$)
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7 MarksaThe plan of a line $PQ$, $75\text{ mm}$ long, measures $54\text{ mm}$. The midpoint of the line is $50\text{ mm}$ from $VP$ and $15\text{ mm}$ from the $HP$. The point $Q$ is $24\text{ mm}$ from the $VP$. Draw its projections and find the inclinations with $HP$ and $VP$. Also locate its traces.
How to answer
- Calculate inclination with HP: $\cos\theta = \frac{\text{Plan}}{\text{TL}} = \frac{54}{75} \Rightarrow \theta = \arccos(0.72) \approx 43.95^\circ$
- Locate midpoint $M$: $m'$ at $15\text{ mm}$ above $xy$, $m$ at $50\text{ mm}$ below $xy$
- Locate locus of $q$ at $24\text{ mm}$ below $xy$, find $p$ using midpoint relation ($m$ is midpoint of $pq$ of length $54\text{ mm}$)
- Determine the elevation (front view) $p'q'$ using true length $75\text{ mm}$ and inclination $\theta$
- Find true inclination $\phi$ with VP by rotating plan or elevation to parallel to $xy$
- Locate HT and VT by extending front and top views to intersect reference line $xy$
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7 MarksbA regular hexagonal lamina $30\text{ mm}$ side has a corner on HP. Its surface is inclined at $45^\circ$ to the HP and the plan of the diagonal through the corner, which is in the HP makes an angle of $45^\circ$ with the VP. Draw its projections.
How to answer
- Stage 1: Draw true shape of hexagon of side 30 mm in top view resting on a corner on HP; project front view on $xy$ line
- Stage 2: Tilt the front view at $45^\circ$ to $xy$ line keeping the corner on $xy$; project new top view (apparent shape)
- Stage 3: Rotate the plan such that the top view of the diagonal passing through the resting corner is inclined at $45^\circ$ to $xy$ line (VP)
- Project the final front view by intersecting projectors from Stage 3 top view and Stage 2 front view
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7 MarksaWhat is meant by Projection? Explain the principle of projection and Differentiate between First Angle and Third Angle Projections.
How to answer
- Define Projection (rays from observer through points on object intersecting a projection plane)
- State Principle of Orthographic Projection
- Tabulate differences between 1st Angle and 3rd Angle projections (Object position, Nature of planes, Positions of top view / front view / side views)
- Draw the standard graphical symbols for both projection systems
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7 MarksbA triangular prism of side of base $30\text{ mm}$ and axis $55\text{ mm}$ long lies on one of its rectangular faces in HP with its axis parallel to VP. Draw its Projection.
How to answer
- Analyze orientation: Axis is parallel to both HP (since a face lies in HP) and VP
- The side view or end view will show the true triangular shape (equilateral triangle of side 30 mm resting on one side on HP)
- Draw side view on profile plane / front view showing a rectangle of length 55 mm and height equal to equilateral triangle altitude ($h = 30 \times \frac{\sqrt{3}}{2} \approx 25.98\text{ mm}$)
- Draw top view showing a rectangle of $55\text{ mm} \times 30\text{ mm}$ with top edge visible along center
- Label all vertices appropriately in front view and top view
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14 MarksA right regular square pyramid, edge of base $35\text{ mm}$ and height $50\text{ mm}$, rest on its base on HP with its base edges equally inclined to VP. A section plane perpendicular to VP and inclined to HP on $32^\circ$, cuts the pyramid bisecting its axis. Draw the projections and true shape of the section of truncated pyramid.
How to answer
- Draw top view of base as a square with edges inclined at $45^\circ$ to $xy$ line
- Project the front view of the pyramid with axis height $50\text{ mm}$
- Locate midpoint of axis at $25\text{ mm}$ from base in front view; draw cutting plane line at $32^\circ$ to $xy$ bisecting axis
- Mark cutting points on slant edges and base edges (if any)
- Project section points down to top view to obtain the sectional top view with hatching
- Project the true shape of the section onto an auxiliary plane parallel to the cutting plane
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14 MarksA bicycle has a $650\text{ mm}$ diameter wheels. Draw the locus of a point $P$ on the circumference of a wheel for its complete revolution when it passes over a segmental arched culvert of radius $1950\text{ mm}$.
How to answer
- Identify curve as an Epicycloid (circle rolling outside a circular arc)
- Choose an appropriate scale (e.g. 1:10 or 1:20)
- Calculate subtended angle $\theta = \frac{r}{R} \times 360^\circ = \frac{325}{1950} \times 360^\circ = \frac{1}{6} \times 360^\circ = 60^\circ$
- Draw directing arc of radius $R = 1950\text{ mm}$ subtending $60^\circ$
- Draw center locus arc of radius $R + r = 2275\text{ mm}$
- Divide generating circle and angle $\theta$ into 12 equal parts
- Locate positions of center $C_1, C_2, \dots, C_{12}$ and draw arcs to find points on the epicycloid, connecting them with a smooth curve
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9 MarksaA cube $30\text{ mm}$ edge is placed centrally on the top of a cylindrical block of $\phi\ 52\text{ mm}$ and $20\text{ mm}$ height. Draw the isometric drawing of the solid.
How to answer
- Draw isometric axes at $30^\circ$, $90^\circ$, $30^\circ$
- Construct bottom ellipse for cylinder base ($\\phi 52\text{ mm}$) using four-center method inside an isometric rhombus of $52\text{ mm} \times 52\text{ mm}$
- Offset center upwards by $20\text{ mm}$ and draw the top ellipse of cylinder
- Draw common tangents to represent the cylindrical outline
- Locate center of top face of cylinder and construct an isometric rhombus of $30\text{ mm} \times 30\text{ mm}$ centrally
- Draw the isometric view of the cube of edge $30\text{ mm}$ standing on top of the cylinder, making visible lines dark and hidden lines removed or dashed
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5 MarksbWhat is CAD? Name two CAD softwares. Give advantages and limitations of using CAD.
How to answer
- Define CAD (Computer Aided Design: use of computer systems to assist in creation, modification, analysis, or optimization of a design)
- List two software packages (e.g., AutoCAD, SolidWorks, CATIA, Creo)
- State advantages: high accuracy, easy editing/revisions, faster drawing speed, 3D visualization, easy storage and transfer
- State limitations: high initial software/hardware cost, training required, risk of data loss/cyber issues
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14 MarksWrite short note on: (any two)
- i)View ports
- ii)Layering concept
- iii)Edit commands used in CAD
How to answer
- View ports: Definition, model space vs layout/paper space viewports, displaying multiple standard views simultaneously
- Layering concept: Organizing drawings, controlling visibility, color, linetype, line weight, freeze/thaw, lock/unlock
- Edit commands used in CAD: Key modification commands like TRIM, EXTEND, OFFSET, FILLET, CHAMFER, MOVE, COPY, ROTATE, SCALE, MIRROR with syntax/functions
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