McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electronics and Communication Engineering .
| 151. |
Following are coordinates of clipping window : Lower Left Corner (10,10) and Upper Right Corner (1,1). A line has end coordinates as (50,5) and (150,50). What will be the outcodes associated with line segment? |
| A. | 01 , 0010 |
| B. | 0001, 0000 |
| C. | 0000, 0010 |
| D. | 0010, 0110 |
| Answer» A. 01 , 0010 | |
| 152. |
Following are coordinates of clipping window : Lower Left Corner (10,10) and Upper Right Corner (1,1). A line has end coordinates as (50,5) and (150,50). The given line segment is ___________. Use Cohen – Sutherland Outcode Algorithm. |
| A. | completely visible |
| B. | partially visible |
| C. | completely invisible |
| D. | none of above |
| Answer» B. partially visible | |
| 153. |
Following are coordinates of clipping window : Lower Left Corner (20,20) and Upper Right Corner (200,200). A line has end coordinates as (5,50) and (75,90). The given line segment is ___________. Use Cohen – Sutherland Outcode Algorithm. |
| A. | completely visible |
| B. | partially visible |
| C. | completely invisible |
| D. | none of above |
| Answer» B. partially visible | |
| 154. |
Following are coordinates of clipping window : Lower Left Corner (20,20) and Upper Right Corner (200,200). A line has end coordinates as (25,50) and (175,190). The given line segment is ___________. Use Cohen – Sutherland Outcode Algorithm. |
| A. | completely visible |
| B. | partially visible |
| C. | completely invisible |
| D. | none of above |
| Answer» A. completely visible | |
| 155. |
A clipping window has coordinates as A(50,10), B(80,10), C(80,40), D(50,40). A line segment has end coordinates (40,15) and (75,45). What will be the end points of clipped line? Use Cohen – Sutherland Outcode Algorithm. |
| A. | (23.67,50) and (69.06,40) |
| B. | (50,23.67) and (69.06, 40) |
| C. | (50,23.67) and (40,69.06) |
| D. | none of above |
| Answer» B. (50,23.67) and (69.06, 40) | |
| 156. |
A clipping window has coordinates as A (50, 10), B (80, 10), C (80, 40), D (50, 40). A line segment has end coordinates (70, 20) and (1, 10). What will be the end points of clipped line? Use Cohen – Sutherland Outcode Algorithm. |
| A. | (80, 16.66) |
| B. | (80, 16.66) and (20, 70) |
| C. | (80, 16.66) and (70, 20) |
| D. | (16.66, 80) and (70, 20) |
| Answer» C. (80, 16.66) and (70, 20) | |
| 157. |
Clip a line starting from (-13, 5) and ending at (17, 11) against the window having lower left corner at (-8, -4) and upper right corner at ( 12,8). What will be the end points of clipped line? Use Cohen – Sutherland Outcode Algorithm. |
| A. | (-8,6) and (2,8) |
| B. | (-8,6) and (8,2) |
| C. | (6,-8) and (2,8) |
| D. | (8,-6) and (8,2) |
| Answer» A. (-8,6) and (2,8) | |
| 158. |
Following are coordinates of clipping window : Lower Left Corner (20,20) and Upper Right Corner (200,200). A line has end coordinates as (5,50) and (15,150). The given line segment is ___________. Use Cohen – Sutherland Outcode Algorithm. |
| A. | completely visible |
| B. | partially visible |
| C. | completely invisible |
| D. | none of above |
| Answer» C. completely invisible | |
| 159. |
Following are coordinates of clipping window : Lower Left Corner (20,20) and Upper Right Corner (200,200). A line has end coordinates as (0,10) and (250,15). The given line segment is ___________. Use Cohen – Sutherland Outcode Algorithm. |
| A. | completely visible |
| B. | partially visible |
| C. | completely invisible |
| D. | none of above |
| Answer» C. completely invisible | |
| 160. |
A rectangular clipping window whose lower left corner is at (-3,1) and upper right corner is at (2,6). If line segment has end coordinates (-4,2) and (-1,7). What will the end coordinates of clipped line (use Cohen – Sutherland Outcode Algorithm) |
| A. | (-3, 11/3) and (-8/5, 6) |
| B. | (3/11, -3) and ( -5/8, 6) |
| C. | (-3, 3/11) and (-8/5,6) |
| D. | (-11/3, 3) and (-6, 8/5) |
| Answer» A. (-3, 11/3) and (-8/5, 6) | |
| 161. |
In normalization transformation for window to viewport, window is lower left corner (1,1) and upper right corner at (3,5) to a view point with lower left corner at (0,0) and upper right corner at(1/2,1/2) .Scaling factor Sx = |
| A. | 0.25 |
| B. | 0.125 |
| C. | 4 |
| D. | 0.5 |
| Answer» A. 0.25 | |
| 162. |
In normalization transformation for window to viewport, window is lower left corner (1,1) and upper right corner at (3,5) to a view point with lower left corner at (0,0) and upper right corner at(1/2,1/2) .Scaling factor Sy = |
| A. | 0.25 |
| B. | 0.125 |
| C. | 4 |
| D. | 0.5 |
| Answer» B. 0.125 | |
| 163. |
In normalization transformation for window to viewport, window is lower left corner (1,1) and upper right corner at (3,5) to a view point with lower left corner at (0,0) and upper right corner at(1/2,1/2) .Scaling factors Sx =___ & Sy =___ |
| A. | 0.25 & 0.125 |
| B. | 0.125 & 0.25 |
| C. | 4 & 8 |
| D. | 0.5 & 1 |
| Answer» A. 0.25 & 0.125 | |
| 164. |
In normalization transformation for window to viewport, window is lower left corner (2,2) and upper right corner at (6,10) to a view point with lower left corner at (0,0) and upper right corner at(1, 1) .Scaling factor Sx =___ |
| A. | 0.25 |
| B. | 0.125 |
| C. | 0.5 |
| D. | 4 |
| Answer» A. 0.25 | |
| 165. |
In normalization transformation for window to viewport, window is lower left corner (2,2) and upper right corner at (6,10) to a view point with lower left corner at (0,0) and upper right corner at(1, 1) .Scaling factor Sy=___ |
| A. | 0.25 |
| B. | 0.125 |
| C. | 0.5 |
| D. | 4 |
| Answer» B. 0.125 | |
| 166. |
In normalization transformation for window to viewport, window is lower left corner (2,2) and upper right corner at (6,10) to a view point with lower left corner at (0,0) and upper right corner at(1, 1) .Scaling factors Sx =___ & Sy= ___ |
| A. | 0.25 & 0.5 |
| B. | 0.25 & 0.125 |
| C. | 0.5 & 0.25 |
| D. | 0.125 & 0.25 |
| Answer» B. 0.25 & 0.125 | |
| 167. |
In viewing from a window in world coordinates with x extent 3 to 12 & y extent 2 to 10 onto a viewport with x extent ¼ to 3/4and y extent 0 to ½ in normalized device space .Scaling factors Sx= |
| A. | (1/18) |
| B. | (1/16) |
| C. | 18 |
| D. | 16 |
| Answer» A. (1/18) | |
| 168. |
In viewing from a window in world coordinates with x extent 3 to 12 & y extent 2 to 10 onto a viewport with x extent ¼ to 3/4and y extent 0 to ½ in normalized device space .Scaling factor Sy= |
| A. | (1/18) |
| B. | (1/16) |
| C. | 18 |
| D. | 16 |
| Answer» B. (1/16) | |
| 169. |
In viewing from a window in world coordinates with x extent 3 to 12 & y extent 2 to 10 onto a viewport with x extent ¼ to 3/4and y extent 0 to ½ in normalized device space .Scaling factor Sx= & Sy = ___ |
| A. | 1/18 & 16 |
| B. | 1/16 & 1/18 |
| C. | 18 & 1/16 |
| D. | 1/18 & 1/16 |
| Answer» D. 1/18 & 1/16 | |
| 170. |
In 8 connected region if one pixel is (x,y) then remaining neighboring pixels are |
| A. | (x+1, y ) (x-1,y) (x,y+1) (x,y-1) (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| B. | (x+1, y ) (x-1,y) (x,y+1) (x,y) (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| C. | (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| D. | (x+1, y ) (x-1,y) (x,y+1) (x,y-1) (x-1,y-1) |
| Answer» A. (x+1, y ) (x-1,y) (x,y+1) (x,y-1) (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) | |
| 171. |
In 4 connected region if one pixel is (x,y) then remaining neighboring pixels are |
| A. | (x+1, y ) (x-1,y) (x,y+1) (x,y-1) (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| B. | (x+1, y ) (x-1,y) (x,y+1) (x,y) (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| C. | (x-1,y-1) (x-1, y+1) (x+1, y-1) (x+1, y+1) |
| D. | (x+1, y ) (x-1,y) (x,y+1) (x,y-1) |
| Answer» D. (x+1, y ) (x-1,y) (x,y+1) (x,y-1) | |
| 172. |
‘Scan-line coherence ‘ means |
| A. | if a pixel on a scan line lies within a polygon, pixels near it will most likely lie within the polygon |
| B. | if edge of the polygon intersects a given polygon , pixels near it will most likely lie within the polygon |
| C. | if a pixel on a edge lies within a polygon, pixels near it will most likely lie within the polygon |
| D. | none of above |
| Answer» A. if a pixel on a scan line lies within a polygon, pixels near it will most likely lie within the polygon | |
| 173. |
The function of scan line polygon fill algorithm are |
| A. | find intersection point of the boundary of polygon and scan line |
| B. | find intersection point of the boundary of polygon and point |
| C. | both a & b |
| D. | none of these |
| Answer» A. find intersection point of the boundary of polygon and scan line | |
| 174. |
In scan line algorithm, using Edge coherence property the next incremental x-intersection(xs+1) can be calculated as: ------------Note: Xs is the previous x-intersection. |
| A. | xs+1= xs + m |
| B. | xs+1= xs + 1 |
| C. | xs+1= xs + 1/m |
| D. | xs+1= xs + m/2 |
| Answer» C. xs+1= xs + 1/m | |
| 175. |
In scan line algorithm, Let an edge is represented by Formula y = mx + b and for scan_line s value of y = s. What will be the Xs ? Note : Scan line is intersecting with edge at (s,Xs) |
| A. | xs = (s-b)+m |
| B. | xs = (s-b) |
| C. | xs = (s-b)/m |
| D. | xs = (s-b)*m |
| Answer» C. xs = (s-b)/m | |
| 176. |
In winding number method if net winding is --------then point is outside otherwise it is inside. |
| A. | odd |
| B. | two |
| C. | one |
| D. | zero |
| Answer» D. zero | |
| 177. |
In winding number method if net winding is zero then point is ---------- otherwise it is inside. |
| A. | outside |
| B. | inside |
| C. | colorful |
| D. | none of above |
| Answer» A. outside | |
| 178. |
In winding number method if net winding is non-zero then point is --------. |
| A. | outside |
| B. | inside |
| C. | colorful |
| D. | none of above |
| Answer» B. inside | |
| 179. |
what will be the intersection points for current scan line as shown in figure using scan line polygon filling algorithm |
| A. | (p0,p1,p2,p3) |
| B. | (p0,p1, p1,p2,p3) |
| C. | (p0,p1,p2,p3, p3) |
| D. | (p0, p3) |
| Answer» A. (p0,p1,p2,p3) | |
| 180. |
What will be the intersection points for current scan line as shown in figure using scan line polygon filling algorithm. |
| A. | (p0 ,p0,p1,p1,p2) |
| B. | (p0,p1,p2) |
| C. | (p0,p2) |
| D. | (p0,p1,p1,p2) |
| Answer» D. (p0,p1,p1,p2) | |
| 181. |
In 4 connected region if one pixel is (2,3) then remaining neighboring pixels are |
| A. | (3, 3 ) (1,3) (2,4) (2,2) (1,2) (1, 4) (3,2) (3, 4) |
| B. | (3, 3 ) (1,1) (2,4) (2,2) (2,2) (1, 4) (3,2) (3, 4) |
| C. | (3, 3 ) (1,2) (2,3) (2,2) |
| D. | (3, 3 ) (1,3) (2,4) (2,2) |
| Answer» D. (3, 3 ) (1,3) (2,4) (2,2) | |
| 182. |
In 8 connected region if one pixel is (3,2) then remaining neighboring pixels are |
| A. | (2, 2 ) (4,2) (3,3) (3,1) (4,3) (4,1) (2,3) (2, 1) |
| B. | (3, 3 ) (1,1) (2,4) (2,2) (2,2) (1, 4) (3,2) (3, 4) |
| C. | (3, 3 ) (1,2) (2,3) (2,2) |
| D. | (2, 2 ) (4,2) (3,3) (3,1) |
| Answer» A. (2, 2 ) (4,2) (3,3) (3,1) (4,3) (4,1) (2,3) (2, 1) | |
| 183. |
In 8 connected region if one pixel is (2,3) then remaining neighboring pixels are |
| A. | (1, 3 ) (3,3) (2,4) (2,2) (3,4) (3,2) (1,4) (1, 2) |
| B. | (3, 3 ) (1,1) (2,4) (2,2) (2,2) (1, 4) (3,2) (3, 4) |
| C. | (3, 3 ) (1,2) (2,3) (2,2) |
| D. | (2, 2 ) (4,2) (3,3) (3,1) |
| Answer» A. (1, 3 ) (3,3) (2,4) (2,2) (3,4) (3,2) (1,4) (1, 2) | |
| 184. |
If one edge of the polygon has end coordinates (10,20) and (15,40) and current scan line is scanning at y=25. What will be the intersection point? |
| A. | (11.2,25) |
| B. | (12,25) |
| C. | (12.2,25) |
| D. | (25,11.2) |
| Answer» A. (11.2,25) | |
| 185. |
In 8 connected region if one pixel is (5,3) then remaining neighboring pixels are |
| A. | (6, 3 ) (4,3) (5,4) (5,2) (6,4) (6,2) (4,4) (4, 2) |
| B. | (6, 3 ) (1,1) (2,4) (2,2) (6,2) (1, 4) (3,3) (3, 4) |
| C. | (3, 3 ) (6,2) (2,3) (2,2) |
| D. | (2, 2 ) (6,2) (6,3) (3,1) |
| Answer» A. (6, 3 ) (4,3) (5,4) (5,2) (6,4) (6,2) (4,4) (4, 2) | |
| 186. |
In 8 connected region if one pixel is (5,5) then remaining neighboring pixels are |
| A. | (2, 2 ) (6,2) (6,3) (3,1) |
| B. | (6, 3 ) (1,1) (2,4) (2,2) (6,2) (1, 4) (3,3) (3, 4) |
| C. | (3, 3 ) (6,2) (2,3) (2,2) |
| D. | (6, 5 ) (4,3) (5,6) (5,4) (6,6) (6,4) (4,6) (4, 4) |
| Answer» D. (6, 5 ) (4,3) (5,6) (5,4) (6,6) (6,4) (4,6) (4, 4) | |
| 187. |
In 4 connected region if one pixel is (3,3) then remaining neighboring pixels are |
| A. | (3, 3 ) (1,3) (2,4) (2,2) |
| B. | (2,2) (1, 4) (3,2) (3, 4) |
| C. | (3, 3 ) (1,2) (2,3) (2,2) |
| D. | (4, 3 ) (2,3) (3,4) (3,2) |
| Answer» D. (4, 3 ) (2,3) (3,4) (3,2) | |
| 188. |
In 4 connected region if one pixel is (5,5) then remaining neighboring pixels are |
| A. | (3, 3 ) (6,3) (2,4) (2,2) |
| B. | (2,2) (5, 4) (3,2) (6, 4) |
| C. | (3, 3 ) (4,2) (2,3) (2,2) |
| D. | (6, 5 ) (4,5) (5,4) (5,6) |
| Answer» D. (6, 5 ) (4,5) (5,4) (5,6) | |
| 189. |
In scan line polygon filling algorithm for current scan line the x-intersections got are 20,10,50,30. How pairing will be formed? |
| A. | (10,20)and (30,50) |
| B. | (10,20)and (20,30)and (30,50) |
| C. | (20,10)and (10,50) |
| D. | (20,10)and (10,50)and (50,30) |
| Answer» A. (10,20)and (30,50) | |
| 190. |
In scan line polygon filling algorithm for current scan line the x-intersections obtained are 20,10,50,30. How these x-intersections are stored in intersection array? |
| A. | (10,20,30,50) |
| B. | (20,10,50,30) |
| C. | (50,30,20,10) |
| D. | (10,50,20,30) |
| Answer» A. (10,20,30,50) | |
| 191. |
One edge of the polygon has coordinates (10,20) and (15,40). In scan line polygon filling at the ith step x-value of the intersection point of the scan line ‘i’ and above mentioned edge is 12. What will be the x-value of the intersection point of the scan line in i+1 th step? |
| A. | 12.2 |
| B. | 12.5 |
| C. | 13 |
| D. | 13.5 |
| Answer» A. 12.2 | |
| 192. |
One edge of the polygon has coordinates (10,20) and (15,40). In scan line polygon filling at the ith step x-value of the intersection point of the scan line ‘i’ and above mentioned edge is 14. What will be the x-value of the intersection point of the scan line in i+1 th step? |
| A. | 16.2 |
| B. | 14.2 |
| C. | 16.2 |
| D. | 14.2 |
| Answer» D. 14.2 | |
| 193. |
In winding number method , constructed horizontal line between point Q and point P intersects two edges of the polygon with winding no. +1 and -1. Tell whether point Q is inside or outside and what is the net winding ? Note : Q is the point to be tested for inside test.P is point outside the polygon. |
| A. | outside , zero |
| B. | inside , zero |
| C. | outside , nonzero |
| D. | inside , nonzero |
| Answer» A. outside , zero | |
| 194. |
In winding number method , constructed horizontal line between point Q and point P intersects edge of the polygon with winding no. +1 . Tell whether point Q is inside or outside and what is the net winding ? Note : Q is the point to be tested for inside test. P is point outside the polygon. |
| A. | outside , zero |
| B. | inside , zero |
| C. | outside , nonzero |
| D. | inside , nonzero |
| Answer» D. inside , nonzero | |
| 195. |
___________ is basic change in shape & size of object. |
| A. | transformation |
| B. | deformation |
| C. | illumination |
| D. | none of above |
| Answer» A. transformation | |
| 196. |
. Transformation is ___________ in nature. |
| A. | flexible |
| B. | rigid |
| C. | composite |
| D. | all of above of above |
| Answer» B. rigid | |
| 197. |
There are ________________ basic transformations. |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 5 |
| Answer» C. 3 | |
| 198. |
There are total __________ types of transformations |
| A. | 1 |
| B. | 3 |
| C. | 5 |
| D. | 2 |
| Answer» C. 5 | |
| 199. |
__________ is a process of changing position of an object in a straight line path. |
| A. | translation |
| B. | scaling |
| C. | rotation |
| D. | reflection |
| Answer» A. translation | |
| 200. |
Translation distance pair is also called as ____________________. |
| A. | scaling factor |
| B. | shear value |
| C. | pivot point |
| D. | shift vector |
| Answer» D. shift vector | |
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