Computer Graphics
Question 1 |
What does a pixel mask mean?
string containing only 1's | |
string containing only 0's | |
string containing two 0's | |
string containing 1's and 0’s |
Discuss it
Question 1 Explanation:
Pixel counts for the span length and interspan spacing can be specified in a pixel mask, which is a string containing the digits 1 and 0 to indicate which positions to plot along the line path. The mask 1111000, for instance, could be used to display a dashed line with a dash length of four pixels and an interdash spacing of three pixels.
So, option (D) is correct.
Question 2 |
How much memory is required to implement z-buffer algorithm for a 512 x 512 x 24 bit-plane image?
768 KB | |
1 MB | |
1.5 MB | |
2 MB |
Discuss it
Question 2 Explanation:
Z-buffer, which is also known as the Depth-buffer method is one of the commonly used method for hidden surface detection. Z-buffer requires 2 type of buffers to be filled: Depth buffer and Frame buffer
Space required by depth buffer = 512 x 512 x 24 = 6291456 bits.
Space required by frame buffer = 512 x 512 x 24 = 6291456 bits.
Total space required = 6291456 + 6291456 bits = 12582912 bits ≈ 1.5 MB
Question 3 |
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.
1016 | |
1006 | |
610 | |
616 |
Discuss it
Question 4 |
In graphics, the number of vanishing points depends on
the number of axes cut by the projection plane | |
the centre of projection | |
the number of axes which are parallel to the projection plane | |
the perspective projections of any set of parallel lines that are not parallel to the projection plane |
Discuss it
Question 4 Explanation:
A vanishing point is a point on the image plane of a perspective drawing where the two-dimensional perspective projections (or drawings) of mutually parallel lines in three-dimensional space appear to converge.
So, (D) is the correct option.
Question 5 |
Which of the following is true about z-buffer algorithm?
It is a depth sort algorithm | |
No limitation on total number of objects | |
. Comparisons of objects is done | |
z-buffer is initialized to background colour at start of algorithm
|
Discuss it
Question 5 Explanation:
Option 1: False, as it is a depth buffer and not depth sort algorithm.
Option 2: True, as the size of objects can be large.
Option 3: False, as only the depth of object is compared and not the entire object.
Option 4: False, no background colour is initiated even at the start of algorithm.
Option (B) is correct.
Question 6 |
A system is having 8 M bytes of video memory for bit-mapped graphics with 64-bit colour. What is the maximum resolution it can support?
800 x 600 | |
1024 x 768 | |
1280 x 1024 | |
1920 x 1440 |
Discuss it
Question 7 |
The end points of a given line are (0, 0) and (6, 18). Compute each value of y as x steps from 0 to 3, by using equation of straight line:
For x = 0, y = 0; x = 1, y = 3; x = 2, y = 6; x = 3, y = 9 | |
For x = 0, y = 1; x = 1, y = 3; x = 2, y = 4; x = 3, y = 9 | |
For x = 0, y = 2; x = 1, y = 3; x = 2, y = 6; x = 3, y = 9 | |
For x = 0, y = 0; x = 1, y = 3; x = 2, y = 4; x = 3, y = 6 |
Discuss it
Question 8 |
Which of the following graphic primitives are considered as the basic building blocks of computer graphics ?
(a)Points
(b)Lines
(c)Polylines
(d)Polygons
Codes :
(a) only | |
(a) and (b) | |
(a), (b) and (c) | |
(a), (b), (c) and (d) |
Discuss it
Question 8 Explanation:
Points(Co-ordinate) and lines are the basic building blocks of computer graphics.
So, Option (B) is correct.
Question 9 |
What is the bit rate of a video terminal unit with 80 characters/line, 8 bits/character and horizontal sweep time of 100 μs (including 20 μs of retrace time)?
8 Mbps | |
6.4 Mbps | |
0.8 Mbps | |
0.64 Mbps |
Discuss it
Question 9 Explanation:
Refer: GATE-IT-2004 | Question 11
Option (B) is correct.
Question 10 |
Consider the two class classification task that consists of the following points:
Class C1 : [21, 21], [21, 1], [1, 21]
Class C2 : [1, 1]
The decision boundary between the two classes C1 and C2 using single perceptron is given by:
x1 - x2 - 0.5 = 0 | |
- x1 + x2 - 0.5 = 0 | |
0.5(x1 + x2) - 1.5 = 0 | |
x1 + x2 - 0.5 = 0 |
Discuss it
There are 23 questions to complete.
