Computer-Graphics
Question 1 |
Perform window to viewport transformation for the point (20, 15). Assume that (Xwmin, Ywmin) is (0,0); (Xwmax, Ywmax) is (100,100); (Xvmin, Yvmin) is (5,5); (Xvmax, Yvmax) is (20,20). The value of x and y in the viewport is
A | x = 4 , y = 4 |
B | x = 3 , y = 3 |
C | x = 8 , y = 7.25 |
D | x = 3 , y = 4 |
Question 1 Explanation:


Question 2 |
A frame buffer array is addressed in row major order for a monitor with pixel locations starting from (0,0) and ending with (100,100). What is address of the pixel(6,10)? Assume one bit storage per pixel and starting pixel location is at 0.
A | 1016 |
B | 1006 |
C | 610 |
D | 616 |
Question 2 Explanation:
Given data,
Pixel location starts from =(0,0)
Pixel location ends from =(100,100)
Address of the pixel= ?
Hint: Frame buffer array is addressed in row major order
. Step-1: Row major order is= 0+1((6-0)+101(10-0))
= 0+6+1010
= 1016
Pixel location starts from =(0,0)
Pixel location ends from =(100,100)
Address of the pixel= ?
Hint: Frame buffer array is addressed in row major order
. Step-1: Row major order is= 0+1((6-0)+101(10-0))
= 0+6+1010
= 1016
Question 3 |
Which of the following statements is/are True regarding the solution to the visibility problem in 3D graphics ?
S1 : The Painter’s algorithm sorts polygons by depth and then paints (scan – converts) each Polygon onto the screen starting with the most nearest polygon. S2 : Backface Culling refers to eliminating geometry with back facing normals. Code :A | S1 only |
B | S2 only
|
C | Both S1 and S2 |
D | Neither S1 Nor S2 |
Question 3 Explanation:
Visibility problem in 3D graphics
1. Painter’s algorithm
– A depth sorting method
– Surfaces are sorted in the order of decreasing depth
– Surfaces are drawn in the sorted order, and overwrite the pixels in the frame buffer – Subtle difference from depth buffer approach: entire face drawn
– Two problems:
• It can be nontrivial to sort the surfaces
• There can be no solution for the sorting order
2. Back Face Culling
– Back faces: faces of opaque object which are “pointing away” from viewer
– Back face culling – remove back faces (supported by OpenGL)
How to detect back faces
– If we find backface, do not draw, save rendering resources
– There must be other forward face(s) closer to eye
– F is face of object we want to test if backface
– P is a point on F
– Form view vector, V as (eye – P)
– N is normal to face F
3. View-Frustum Culling
– Remove objects that are outside the viewing frustum
– Done by 3D clipping algorithm (e.g. Liang-Barsky)
4. Ray Tracing
– Ray tracing is another example of image space method
– Ray tracing: Cast a ray from eye through each pixel to the world
5. Z(Depth buffer algorithm)
Reference: https://web.cs.wpi.edu/~emmanuel/courses/cs543/slides/lecture08_p2.pdf
1. Painter’s algorithm
– A depth sorting method
– Surfaces are sorted in the order of decreasing depth
– Surfaces are drawn in the sorted order, and overwrite the pixels in the frame buffer – Subtle difference from depth buffer approach: entire face drawn
– Two problems:
• It can be nontrivial to sort the surfaces
• There can be no solution for the sorting order
2. Back Face Culling
– Back faces: faces of opaque object which are “pointing away” from viewer
– Back face culling – remove back faces (supported by OpenGL)
How to detect back faces
– If we find backface, do not draw, save rendering resources
– There must be other forward face(s) closer to eye
– F is face of object we want to test if backface
– P is a point on F
– Form view vector, V as (eye – P)
– N is normal to face F
3. View-Frustum Culling
– Remove objects that are outside the viewing frustum
– Done by 3D clipping algorithm (e.g. Liang-Barsky)
4. Ray Tracing
– Ray tracing is another example of image space method
– Ray tracing: Cast a ray from eye through each pixel to the world
5. Z(Depth buffer algorithm)
Reference: https://web.cs.wpi.edu/~emmanuel/courses/cs543/slides/lecture08_p2.pdf
Question 4 |
Memory mapped displays
A | are utilized for high resolution graphics such as maps |
B | uses ordinary memory to store the display data in character form |
C | stores the display data as individual bits |
D | are associated with electromechanical teleprinters |
Question 4 Explanation:
● Graphs can be displayed on a screen by writing character values into a special area of RAM within the video controller.
● Prior to cheap RAM that enabled bit-mapped displays, this character cell method was a popular technique for computer video displays.
Question 5 |
Which of the following compression algorithms is used to generate a .png file?
A | LZ78 |
B | Deflate |
C | LZW |
D | Huffman |
Question 5 Explanation:
→ Deflate is a lossless data compression algorithm and associated file format that uses a combination of the LZ77 algorithm and Huffman coding.
→ An algorithm producing Deflate files is widely thought to be implementable in a manner not covered by patents.
→ This has led to its widespread use, for example in gzip compressed files, PNG image files and the ZIP file format for which Katz originally designed it.
→ An algorithm producing Deflate files is widely thought to be implementable in a manner not covered by patents.
→ This has led to its widespread use, for example in gzip compressed files, PNG image files and the ZIP file format for which Katz originally designed it.
Question 6 |
Which of the following is not an image type used in MPEG?
A | A frame |
B | B frame |
C | D frame |
D | P frame |
Question 6 Explanation:
A Group Of Pictures can contain the following picture types:
I picture (or) I frame (intra coded picture) – a picture that is coded independently of all other pictures. Each GOP begins (in decoding order) with this type of picture.
P picture (or) P frame (predictive coded picture) – contains motion-compensated difference information relative to previously decoded pictures. Each P picture can only reference one picture, and that picture must precede the P picture in display order as well as in decoding order and must be an I or P picture.
B picture (or) B frame (bi predictive coded picture) – contains motion-compensated difference information relative to previously decoded pictures. Each B picture can only reference two pictures, the one which precedes the B picture in display order and the one which follows, and all referenced pictures must be I or P pictures.
D picture or D frame (DC direct coded picture) – serves as a fast-access representation of a picture for loss robustness or fast-forward. D pictures are only used in MPEG-1 video.
I picture (or) I frame (intra coded picture) – a picture that is coded independently of all other pictures. Each GOP begins (in decoding order) with this type of picture.
P picture (or) P frame (predictive coded picture) – contains motion-compensated difference information relative to previously decoded pictures. Each P picture can only reference one picture, and that picture must precede the P picture in display order as well as in decoding order and must be an I or P picture.
B picture (or) B frame (bi predictive coded picture) – contains motion-compensated difference information relative to previously decoded pictures. Each B picture can only reference two pictures, the one which precedes the B picture in display order and the one which follows, and all referenced pictures must be I or P pictures.
D picture or D frame (DC direct coded picture) – serves as a fast-access representation of a picture for loss robustness or fast-forward. D pictures are only used in MPEG-1 video.
Question 7 |
While making bubbled lists, which of the following options are available?
A | Square,disc,tringle |
B | Triangle,disc,circle |
C | Triangle,square,circle |
D | Disc,square,circle |
Question 7 Explanation:
→ Disc: A filled circle
→ Circle : An unfilled circle
→ square : A filled square
→ Circle : An unfilled circle
→ square : A filled square
Question 8 |
MPEG involves both spatial compression and temporal compression. The spatial compression is similar to JPEG and temporal compression removes_____frames.
A | Voice |
B | Spatial |
C | Temporal |
D | Redundant |
Question 8 Explanation:
MPEG: The MPEG codec can only be used when the final video file will be in MPEG format (it is not compatible with other file types). It uses a lossy compression scheme (although it may be lossless at high-quality settings) and spatial and temporal compression. MPEG offers the best compression possible, but MPEGs are not yet as widely supported on the Web as other video formats.
Spatial compression:
Spatial (or intraframe) compression takes place on each individual frame of the video, compressing the pixel information as though it were a still image. JPEG, PNG and PICT files are an example of spatial compression.
It is known as the intraframe method.
Temporal compression:
It happens over a series of frames of the video. Temporal (or interframe) compression happens over a series of frames and takes advantage of areas of the image that remain unchanged from frame to frame, throwing out data for repeated pixels.
It is also called the interframe method.
Spatial compression:
Spatial (or intraframe) compression takes place on each individual frame of the video, compressing the pixel information as though it were a still image. JPEG, PNG and PICT files are an example of spatial compression.
It is known as the intraframe method.
Temporal compression:
It happens over a series of frames of the video. Temporal (or interframe) compression happens over a series of frames and takes advantage of areas of the image that remain unchanged from frame to frame, throwing out data for repeated pixels.
It is also called the interframe method.
Question 9 |
In which of the following, the density of the core remains constant from the center to the edges?
A | Single mode fiber |
B | Multimedia step-index fiber |
C | Multimode graded index fiber |
D | Single mode step index fiber |
Question 9 Explanation:
Multimode Step-index: In multimode step-index fiber, the density of the core remains constant from the center to the edges. A beam of light moves through this constant density in a straight line until it reaches the interface of the core and the cladding
Question 10 |
Which of the following is a special effect in motion pictures and animation that changes one image/picture into another through a seamless transition?
A | Modeling |
B | Morphing |
C | Animating |
D | Wrapping |
Question 10 Explanation:
→ Morphing is a special effect in motion pictures and animations that changes (or morphs) one image or shape into another through a seamless transition.
→ Morphing means stretching or as part of a fantasy or surreal sequence.
→ Morphing means stretching or as part of a fantasy or surreal sequence.
Question 11 |
Even when the screen is completely dark while the film is in motion, commercial motion pictures use
A | 32 frames per second or 101 screen illuminations per second |
B | 72 frames per second or 234 screen illuminations per second |
C | 8 frames per second or 32 screen illuminations per second |
D | 24 frames per second or 72 screen illuminations per second |
Question 11 Explanation:
→ Motion picture, also called film or movie, series of still photographs on film, projected in rapid succession onto a screen by means of light.
→ When a motion picture film is projected on a screen at the rate of at least 16 illuminations per second but the commercial motion pictures uses 24 frames per second or 72 screen illuminations per second.
→ Film Technology standard is 24 frames per second, a three bladed shutter and some dreamy motion blur, all projected as shadow and light on the side of a wall.
→ When a motion picture film is projected on a screen at the rate of at least 16 illuminations per second but the commercial motion pictures uses 24 frames per second or 72 screen illuminations per second.
→ Film Technology standard is 24 frames per second, a three bladed shutter and some dreamy motion blur, all projected as shadow and light on the side of a wall.
Question 12 |
The file size of a 640 by 480 pictures of 256 colours in a 8-bit resolution is
A | 300KB |
B | 900KB |
C | 128KB |
D | 1024KB |
Question 12 Explanation:
●Give picture size with 640 by 480 resolution with 256 colors per pixel.
●This implies 640 * 480 = 307,200 pixels. Recall that 256 values requires 8 bits of computer storage.
●Hence (307,200 pixels) * (8 bits per pixel) = 2,457,600 bits of storage.
●A common method for selling computer storage is in 8 bit groupings called bytes.
●Hence the picture would require (2,457,600 bits) / (8 bits per byte) = 307,200 bytes of computer storage.
●This implies 640 * 480 = 307,200 pixels. Recall that 256 values requires 8 bits of computer storage.
●Hence (307,200 pixels) * (8 bits per pixel) = 2,457,600 bits of storage.
●A common method for selling computer storage is in 8 bit groupings called bytes.
●Hence the picture would require (2,457,600 bits) / (8 bits per byte) = 307,200 bytes of computer storage.
Question 13 |
The process of producing bitmapped images from a view of 3-D models in a 3-D scene is called
A | Rendering |
B | Looping |
C | Cross dissolving |
D | Imaging |
Question 13 Explanation:
Rendering or image synthesis is the automatic process of generating a photorealistic or non-photorealistic image from a 2D or 3D model (or models in what collectively could be called a scene file) by means of computer programs.
Question 14 |
Which of the following is interactive?
A | A radio broadcast |
B | A talk show on TV |
C | A newspaper |
D | A computer game |
Question 14 Explanation:
Options A,B and C are not interactive , it will provide only information to the people. Where as Computer game in which computer user should interact with the computer in order to play the game
Question 15 |
Anti aliasing is important to improve readability of text. It deals with the
A | Elimination of “jaggies” |
B | Spacing between two individual characters |
C | Underlining letters |
D | Spacing of a group of characters |
Question 15 Explanation:
● Antialiasing removes jagged edges by adding subtle color changes around the lines, tricking the human eye into thinking that the lines are not jagged.
● The slight changes in color around the edges of an image help the line blend around curves, giving the impression that the line is true.
● These color changes are made on a very small scale that the human eye cannot detect under normal circumstances. In order to be able to see that an image has been antialiased, it would have to be magnified.
● The slight changes in color around the edges of an image help the line blend around curves, giving the impression that the line is true.
● These color changes are made on a very small scale that the human eye cannot detect under normal circumstances. In order to be able to see that an image has been antialiased, it would have to be magnified.
Question 16 |
What is the bitrate for transmitting uncompressed 800×600 pixel color frame with 8 bits/pixel at 40 frames/Second?
A | 1536 Mbps |
B | 2.4 Mbps |
C | 15.36 Mbps |
D | 153.6 Mbps |
Question 16 Explanation:
Step-1: Given data, Uncompressed pixel = 800*600
Step-2: Each of 8 bit = 800*600 *8
Step-3: Bit rate for transmitting = 800*600 *8 *40 =153600000 bits
Step-4: Here, they are given in Mbps. 153600000 bits =153.6Mbps
Step-2: Each of 8 bit = 800*600 *8
Step-3: Bit rate for transmitting = 800*600 *8 *40 =153600000 bits
Step-4: Here, they are given in Mbps. 153600000 bits =153.6Mbps
Question 17 |
The easiest method in flash to draw a heptagon, is to use the___
A | Polystar tool with the “Polygon” style |
B | Polygon Tool |
C | Lasso Tool with the “create star” option |
D | Polystar Tool with the “star” style |
Question 17 Explanation:
There are 3 tools for drawing geometric shapes
1. Rectangle(Rectangle tool)
2. Oval(Oval tool)
3. Polygon(Polystar tool)
Note: We are not using Option A,C and D for drawing heptagon.
1. Rectangle(Rectangle tool)
2. Oval(Oval tool)
3. Polygon(Polystar tool)
Note: We are not using Option A,C and D for drawing heptagon.
Question 18 |
___ is a special effect in motion pictures and animations that changes one image or sharp into another through a seamless transition
A | Tweening |
B | Inverse kinematics |
C | Morphing |
D | Tweaking |
Question 18 Explanation:
Morphing is a special effect in motion pictures and animations that changes (or morphs) one image or shape into another through a seamless transition. morphing means stretching or as part of a fantasy or surreal sequence. Traditionally such a depiction would be achieved through cross-fading techniques on film.
Question 19 |
Given that a 22 inch monitor with an aspect ratio of 16:9 has a monitor of 1920×1080, what is the width of the monitor?
A | 22 inches |
B | 8:53 inches |
C | 10:79 inches |
D | 19:17 inches |
Question 19 Explanation:

Step-1: To check the ratio [width/height= 1920 /1080 ]
Step-2: Ratio: 16/9.
Step-3: Using Pythagoras theorem (9x) 2 +(16x) 2 =(22) 2
x=19.17 inch
Question 20 |
Consider the following statements:
a) 3D Studio Max includes a number of high-level professional tools for character animation, game development, and visual effect production.
b) MAYA is a complete modeling package developed by Microsoft
c) RenderMan is a rendering package developed by Pixar
Which of the above statements are true?
a) 3D Studio Max includes a number of high-level professional tools for character animation, game development, and visual effect production.
b) MAYA is a complete modeling package developed by Microsoft
c) RenderMan is a rendering package developed by Pixar
Which of the above statements are true?
A | Only a and b |
B | Only a and c |
C | Only b and c |
D | a,b and c |
Question 20 Explanation:
MAYA 3D animation software offers a comprehensive creative feature set for 3D computer animation, modeling, simulation, rendering, and compositing on a highly extensible production platform. Maya has next-generation display technology, accelerated modeling workflows, and tools for handling complex data.
Question 21 |
The ____ file format allows storing an animation sequence
A | PNG |
B | GIF |
C | JPG |
D | PDF |
Question 21 Explanation:
→PDF means portable document format.
→Portable Network Graphics is a raster-graphics file-format that supports lossless data compression.
→JPG/JPEG: Joint Photographic Experts Group. JPEG is usually known as JPG. It stands for Joint Photographic Expert Group, a joint working group of International Standardization Organization (ISO) and International Electrotechnical Commission (IEC). It is a standard method of compressing graphic images.
→A GIF is a computer file that is used on the internet for sending images, especially moving images. GIF is an abbreviation for ‘Graphic Interchange Format’.
→Portable Network Graphics is a raster-graphics file-format that supports lossless data compression.
→JPG/JPEG: Joint Photographic Experts Group. JPEG is usually known as JPG. It stands for Joint Photographic Expert Group, a joint working group of International Standardization Organization (ISO) and International Electrotechnical Commission (IEC). It is a standard method of compressing graphic images.
→A GIF is a computer file that is used on the internet for sending images, especially moving images. GIF is an abbreviation for ‘Graphic Interchange Format’.
Question 22 |
If the frame buffer has 10-bits per pixel and 8-bits are allocated for each of the R,G and B components then what would be the size of the color lookup table(LUT)
A | (2 10 +2 11 ) bytes |
B | (2 10 +2 8 ) bytes |
C | (2 10 + 2 24 ) bytes |
D | (2 8 + 2 9 ) bytes |
Question 22 Explanation:
10-bits per pixel it means we will have 2 10 entries in color lookup table.
8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).
24-bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2 10 * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 10 + 2 11 ) Bytes
8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).
24-bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2 10 * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 10 + 2 11 ) Bytes
Question 23 |
Which homogeneous 2D matrix transforms the figure (a) on the left side to the figure (b) on the right ?


A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
Question 23 Explanation:
The homogeneous coordinates of the cartesian point (x,y) is (x,y,1) and the transformation Matrix is a 3×3 Matrix.
The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.
The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.
Question 24 |
In 3D Graphics, which of the following statements about perspective and parallel projection is/are true?
P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.
Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away
R: Perspective projections do not preserve straight lines.
P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.
Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away
R: Perspective projections do not preserve straight lines.
A | P and R only |
B | P,Q and R |
C | Q and R only |
D | P and Q only |
Question 24 Explanation:
Perspective Projection :
Perspective projection is representing or drawing objects which resemble the real thing.
Perspective projection preserves the straight line.
In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.
Parallel lines do not remain parallel
Distance and angles are not preserved.
Parallel Projection :
In this projection drawing objects looks less realistic.
In this projection parallel lines remains parallel.
Angles are not preserved in this projection.
It is good for exact measurements.
Perspective projection is representing or drawing objects which resemble the real thing.
Perspective projection preserves the straight line.
In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.
Parallel lines do not remain parallel
Distance and angles are not preserved.
Parallel Projection :
In this projection drawing objects looks less realistic.
In this projection parallel lines remains parallel.
Angles are not preserved in this projection.
It is good for exact measurements.
Question 25 |
In 3D Graphics, which of the following statement/s is/are true ?
P: Back-face culling is an example of an image-precision visible-surface determination.
Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.
P: Back-face culling is an example of an image-precision visible-surface determination.
Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.
A | P only |
B | Q only |
C | Neither P nor Q |
D | P and Q |
Question 25 Explanation:
Back Face Culling:
→ Back-face culling (an object space algorithm) works on ‘solid’ objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.
→ Back-face culling is not an example of an image-precision visible-surface determination.
→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box – the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.
1. Back faces: faces of opaque object which are “pointing away” from viewer
2. Back face culling – remove back faces (supported by OpenGL)
→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.
→ Back-face culling (an object space algorithm) works on ‘solid’ objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.
→ Back-face culling is not an example of an image-precision visible-surface determination.
→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box – the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.
1. Back faces: faces of opaque object which are “pointing away” from viewer
2. Back face culling – remove back faces (supported by OpenGL)
→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.
Question 26 |
Which API is used to draw a circle ?
A | Circle( ) |
B | Ellipse( ) |
C | Round Rect( ) |
D | Pie( ) |
Question 26 Explanation:
→ When we are giving same height and width to a ellipse it will become circle.
→ API is a protocol intended to be used as an interface by software components to communicate with each other.
→ An API is a library that may include specification for routines, data structures, object classes and variables
→ API is a protocol intended to be used as an interface by software components to communicate with each other.
→ An API is a library that may include specification for routines, data structures, object classes and variables
Question 27 |
The LED’s for their display require
A | A voltage of 1.2v and a current of 20 mA |
B | A voltage of 20v and a current of 1.2 A |
C | A voltage of 1.2v and current of 100 mA |
D | A voltage of 10v and a current of 120 mA |
Question 28 |
Two equal voltages os same frequency applied to the X and Y plates of a CRD, produces a circle on the screen. The phase difference between the two voltages is
A | 30o |
B | 90o |
C | 180o |
D | 0o |
Question 29 |
The spacing between character pairs is called :
A | Kerning |
B | x-height |
C | intercap |
D | serif |
Question 29 Explanation:
→ kerning is the process of adjusting the spacing between characters in a proportional font, usually to achieve a visually pleasing result.
→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.
→ Kerning adjusts the space between individual letter forms, while tracking (letter-spacing) adjusts spacing uniformly over a range of characters.
Question 30 |
Which of the following is not a component of Memory tube display ?
A | Flooding gun |
B | Collector |
C | Ground |
D | Liquid Crystal |
Question 30 Explanation:
Memory tube display components:
1. Connector Pins
2. Electron Gun
3. Base
4. Focusing System
5. Control Grid Voltage
6. X/Y Deflect
7. Phosphor
8. Collector
9. Ground
Note: Liquid crystal is not a component of memory tube display
.
1. Connector Pins
2. Electron Gun
3. Base
4. Focusing System
5. Control Grid Voltage
6. X/Y Deflect
7. Phosphor
8. Collector
9. Ground
Note: Liquid crystal is not a component of memory tube display
.
Question 31 |
Which of the following is not true in case of Oblique Projections?
A | Parallel projection rays are not perpendicular to the viewing plane. |
B | Parallel lines in space appear parallel on the final projected image. |
C | Used exclusively for pictorial purposes rather than formal working drawings. |
D | Projectors are always perpendicular to the plane of projection. |
Question 32 |
With respect to CRT, the horizontal retrace is defined as:
A | The path an electron beam takes when returning to the left side of the CRT. |
B | The path an electron beam takes when returning to the right side of the CRT. |
C | The technique of turning the electron beam off while retracing. |
D | The technique of turning the electron beam on/off while retracing. |
Question 33 |
A 4*4 DFT matrix is given by :
Where values of x and y are _____, _____ respectively.

Where values of x and y are _____, _____ respectively.
A | 1, −1 |
B | −1, 1 |
C | −j, j |
D | j, −j |
Question 34 |
Which of the following statement(s) is/are correct ?
A | Persistence is the term used to describe the duration of phosphorescence. |
B | The control electrode is used to turn the electron beam on and off. |
C | The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT. |
D | All of the above |
Question 34 Explanation:
TRUE: Persistence is the term used to describe the duration of phosphorescence. Phosphorescence is a process in which energy absorbed by a substance is released relatively slowly in the form of light.
TRUE: The control electrode is used to turn the electron beam on and off.
TRUE: The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT.
TRUE: The control electrode is used to turn the electron beam on and off.
TRUE: The electron gun creates a source of electrons which are focused into a narrow beam directed at the face of CRT.
Question 35 |
A segment is any object described by GKS commands and data that start with CREATE SEGMENT and Terminates with CLOSE SEGMENT command. What functions can be performed on these segments ?
A | Translation and Rotation |
B | Panning and Zooming |
C | Scaling and Shearing |
D | Translation, Rotation, Panning and Zooming |
Question 36 |
Match the following:


A | a-i, b-ii, c-iii, d-i |
B | a-i, b-iii, c-ii, d-iv |
C | a-iv, b-iii, c-ii, d-i |
D | a-iv, b-ii, c-i, d-iii |
Question 37 |
Below are the few steps given for scan-converting a circle using Bresenham’s Algorithm. Which of the given steps is not correct ?
A | Compute d = 3 – 2r (where r is radius) |
B | Stop if x > y |
C | If d < 0, then d = 4x + 6 and x = x + 1 |
D | If d ≥, then d = 4 * (x – y) + 10, x = x + 1 and y = y + 1 |
Question 37 Explanation:
Scan converting a circle using Bresenham’s algorithm:
1 Initially X = 0 , Y = R and D = 3 – 2R
2. While (X < Y)
3. Call Draw Circle(Xc, Yc, X, Y)
4. Set X = X + 1
5. If (D < 0) Then
6. D = D + 4X + 6
7. Else
8. Set Y = Y – 1
9. D = D + 4(X – Y) + 10
10. End If
11. Draw Circle(Xc, Yc, X, Y) /* calling function call */
12. End While
1 Initially X = 0 , Y = R and D = 3 – 2R
2. While (X < Y)
3. Call Draw Circle(Xc, Yc, X, Y)
4. Set X = X + 1
5. If (D < 0) Then
6. D = D + 4X + 6
7. Else
8. Set Y = Y – 1
9. D = D + 4(X – Y) + 10
10. End If
11. Draw Circle(Xc, Yc, X, Y) /* calling function call */
12. End While
Question 38 |
Which of the following is/are side effects of scan conversion ?
a. Aliasing
b. Unequal intensity of diagonal lines
c. Overstriking in photographic applications
d. Local or Global aliasing
a. Aliasing
b. Unequal intensity of diagonal lines
c. Overstriking in photographic applications
d. Local or Global aliasing
A | a and b |
B | a, b and c |
C | a, c and d |
D | a, b, c and d |
Question 38 Explanation:
Side effects of scan conversions are
1. Aliasing
2. Unequal intensity of diagonal lines
3. Overstriking in photographic applications
4. Local or Global aliasing
2. Unequal intensity of diagonal lines
3. Overstriking in photographic applications
4. Local or Global aliasing
Question 39 |
Consider a line AB with A=(0,0) and B=(8,4). Apply a simple DDA algorithm and compute the first four plots on this line.
A | [(0, 0), (1, 1), (2, 1), (3, 2)] |
B | [(0, 0), (1, 1.5), (2, 2), (3, 3)] |
C | [(0, 0), (1, 1), (2. 2.5), (3, 3)] |
D | [(0, 0), (1, 2), (2, 2), (3, 2)] |
Question 39 Explanation:

Question 40 |
Consider the Breshenham’s circle generation algorithm for plotting a circle with centre (0, 0) and radius ‘r’ units in first quadrant. If the current point is (xi , yi ) and decision parameter is pi then what will be the next point (xi + 1, yi + 1 + 1) and updated decision parameter pi + 1 for pi ≥ 0?
A | xi + 1 = xi+ 1 yi + 1 = yi pi + 1 = pi + 4xi + 6 |
B | xi + 1 = xi + 1 yi + 1 = yi – 1 pi + 1 = pi + 4(xi – yi) + 10 |
C | xi + 1 = xi yi + 1 = yi – 1 pi + 1 = pi + 4(xi – yi) + 6 |
D | xi + 1 = xi – 1 yi + 1 = yi pi + 1 = pi + 4(xi – yi) + 10 |
Question 40 Explanation:
Write the steps required to scan – convert a circle using Bresenham’s algorithm.
Set the initial values of the variables: (h, k) = coordinates of circle center; x=0; y=circle radius r and d = 3 – 2r.
Test to determine whether the entire circle has been scan-converted. If x>y, stop.
Plot the eight points, found by symmetry with respect to the center (h, k), at the current (x, y) coordinates:
Plot(x+h, y+k) Plot(-x+h, -y+k)
Plot(y+h, x+k) Plot(-y+h, -x+k)
Plot(-y+h, x+k) Plot(y+h, -x+k)
Plot(-x+h, y+k) Plot(x+h, -y+k)
Compute the location of the next pixel. If d<0, then d=d+4x+6 and x=x+1. If d≥0, then d=d+4(x-y)+10, x=x+1 and y=y-1.
Go to step 2.
Set the initial values of the variables: (h, k) = coordinates of circle center; x=0; y=circle radius r and d = 3 – 2r.
Test to determine whether the entire circle has been scan-converted. If x>y, stop.
Plot the eight points, found by symmetry with respect to the center (h, k), at the current (x, y) coordinates:
Plot(x+h, y+k) Plot(-x+h, -y+k)
Plot(y+h, x+k) Plot(-y+h, -x+k)
Plot(-y+h, x+k) Plot(y+h, -x+k)
Plot(-x+h, y+k) Plot(x+h, -y+k)
Compute the location of the next pixel. If d<0, then d=d+4x+6 and x=x+1. If d≥0, then d=d+4(x-y)+10, x=x+1 and y=y-1.
Go to step 2.
Question 41 |
A point P(5, 1) is rotated by 90° about a pivot point (2, 2). What is the coordinate of new transformed point P′ ?
A | (3, 5) |
B | (5, 3) |
C | (2, 4) |
D | (1, 5) |
Question 41 Explanation:
Rotation around a pivot point (2, 2) can be represented as:
(h, k) = (2, 2)

(h, k) = (2, 2)


Question 42 |
Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has a lower left-hand corner at (– 5, 1) and upper right-hand corner at (3, 7). Consider the following three lines for clipping with the given endpoint coordinates:
Line AB : A (– 6, 2) and B (–1, 8)
Line CD : C (– 1, 5) and D (4, 8)
Line EF : E (–2, 3) and F (1, 2)
Which of the following line(s) is/are candidate for clipping?
Line AB : A (– 6, 2) and B (–1, 8)
Line CD : C (– 1, 5) and D (4, 8)
Line EF : E (–2, 3) and F (1, 2)
Which of the following line(s) is/are candidate for clipping?
A | AB |
B | CD |
C | EF |
D | AB and CD |
Question 43 |
In perspective projection, if a line segment joining a point which lies in front of the viewer to a point in back of the viewer is projected to a broken line of infinite extent. This is known as _______.
A | View confusion |
B | Vanishing point |
C | Topological distortion |
D | Perspective foreshortening |
Question 44 |
Let us consider that the original point is (x, y) and new transformed point is (x′, y′). Further, Shx and Shy are shearing factors in the x and y directions. If we perform the y-direction shear relative to x = xref then the transformed point is given by _______.
A | x′ = x + Shx ⋅ (y – yref) y′ = y |
B | x′ = x y′ = y ⋅ Shx |
C | x′ = x y′ = Shy (x – xref) + y |
D | x′ = Shy ⋅ y y′ = y ⋅ (x – xref) |
Question 44 Explanation:

Question 45 |
Which of the following statement(s) is/are correct with reference to curve generation?
I. Hermite curves are generated using the concepts of interpolation.
II. Bezier curves are generated using the concepts of approximation.
III. The Bezier curve lies entirely within the convex hull of its control points.
IV. The degree of Bezier curve does not depend on the number of control points.
I. Hermite curves are generated using the concepts of interpolation.
II. Bezier curves are generated using the concepts of approximation.
III. The Bezier curve lies entirely within the convex hull of its control points.
IV. The degree of Bezier curve does not depend on the number of control points.
A | I, II and IV only |
B | II and III only |
C | I and II only |
D | I, II and III only |
Question 46 |
A triangulation of a polygon is a set of T chords that divide the polygon into disjoint triangles. Every triangulation of n-vertex convex polygon has _____ chords and divides the polygon into _____ triangles.
A | n – 2, n – 1 |
B | n – 3, n – 2 |
C | n – 1, n |
D | n – 2, n – 2 |
Question 47 |
Consider a raster grid having XY-axes in positive X-direction and positive upward Y-direction with Xmax = 10, Xmin = –5, Ymax = 11, and Ymin = 6. What is the address of memory pixel with location (5, 4) in raster grid assuming base address 1 (one) ?
A | 150 |
B | 151 |
C | 160 |
D | 161 |
Question 48 |
Consider a N-bit plane frame buffer with W-bit wide lookup table with W > N. How many intensity levels are available at a time ?
A | 2N |
B | 2 W |
C | 2 N+W |
D | 2 N-1 |
Question 49 |
Consider the Bresenham’s line generation algorithm for a line with gradient greater than one, current point (xi, yi) and decision parameter, di. The next point to be plotted (xi+1, yi+1) and updated decision parameter, di+1, for di < 0 are given as _______.
A | xi+1 = xi +1 yi+1 = yi di+1 = di+ 2 dy |
B | xi+1 = xi yi+1 = yi +1 di+1 = di+ 2 dx |
C | xi+1 = xi yi+1 = yi +1 di+1 = di+ 2(dx -dy) |
D | xi+1 = xi +1 yi+1 = yi +1 di+1 = di+ 2(dy -dx) |
Question 49 Explanation:

Question 50 |
A point P(2, 5) is rotated about a pivot point (1, 2) by 60°. What is the new transformed point P’ ?
A | (1, 4) |
B | (–1, 4) |
C | (1, – 4) |
D | (– 4, 1) |
Question 50 Explanation:


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