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This section includes 179 Mcqs, each offering curated multiple-choice questions to sharpen your Microprocessors knowledge and support exam preparation. Choose a topic below to get started.
151. |
Log function is also called |
A. | Gaussian |
B. | gray scale image |
C. | gradient image |
D. | Mexican hat |
Answer» E. | |
152. |
Ideal edges occur over the distance of |
A. | 1 pixel |
B. | 2 pixels |
C. | 3 pixels |
D. | 4 pixels |
Answer» B. 2 pixels | |
153. |
Segmentation algorithms depends on intensity values' |
A. | discontinuity |
B. | similarity |
C. | continuity |
D. | Both a and b |
Answer» E. | |
154. |
If the standard deviation of the pixels is positive, then sub image is labeled as |
A. | black |
B. | green |
C. | white |
D. | red |
Answer» D. red | |
155. |
Sudden changes in intensity produces peak in |
A. | first derivative |
B. | second derivative |
C. | third derivative |
D. | Both a and b |
Answer» B. second derivative | |
156. |
Thresholding gives the |
A. | large image |
B. | gray scale image |
C. | color image |
D. | binary image |
Answer» E. | |
157. |
Textured inner region of the object produces |
A. | good boundary extraction |
B. | excellent boundary extraction |
C. | good boundary deletion |
D. | excellent boundary deletion |
Answer» B. excellent boundary extraction | |
158. |
For point detection we use |
A. | first derivative |
B. | second derivative |
C. | third derivative |
D. | Both a and b |
Answer» C. third derivative | |
159. |
Marr hildreth method was introduced in |
A. | 1980 |
B. | 1981 |
C. | 1982 |
D. | 1983 |
Answer» B. 1981 | |
160. |
If the inner region of the object is textured then approach we use is |
A. | discontinuity |
B. | similarity |
C. | extraction |
D. | recognition |
Answer» C. extraction | |
161. |
Gradient computation equation is |
A. | |Gx|+|Gy| |
B. | |Gx|-|Gy| |
C. | |Gx|/|Gy| |
D. | |Gx|x|Gy| |
Answer» B. |Gx|-|Gy| | |
162. |
For finding horizontal lines we use mask of values |
A. | [-1 -1 -1; 2 2 2; -1 -1 -1] |
B. | [2 -1 -1; -1 2 -1; -1 -1 2] |
C. | [-1 2 -1; -1 2 -1; -1 2 -1] |
D. | [-1 -1 2; -1 2 -1;2 -1 -1] |
Answer» B. [2 -1 -1; -1 2 -1; -1 -1 2] | |
163. |
For finding lines at angle 45 we use mask of values |
A. | [-1 -1 -1; 2 2 2; -1 -1 -1] |
B. | [2 -1 -1; -1 2 -1; -1 -1 2] |
C. | [-1 2 -1; -1 2 -1; -1 2 -1] |
D. | [-1 -1 2; -1 2 -1;2 -1 -1] |
Answer» C. [-1 2 -1; -1 2 -1; -1 2 -1] | |
164. |
First derivative approximation says that values of constant intensities must be |
A. | 1 |
B. | 0 |
C. | positive |
D. | negative |
Answer» C. positive | |
165. |
To avoid the negative values taking absolute values in lapacian image doubles |
A. | thickness of lines |
B. | thinness of lines |
C. | thickness of edges |
D. | thinness of edges |
Answer» B. thinness of lines | |
166. |
The horizontal gradient pixels are denoted by |
A. | Gx |
B. | Gy |
C. | Gt |
D. | Gs |
Answer» B. Gy | |
167. |
Digital functions' derivatives are defined as |
A. | differences |
B. | multiplication |
C. | addition |
D. | division |
Answer» B. multiplication | |
168. |
Second derivative approximation says that values along the ramp must be |
A. | nonzero |
B. | zero |
C. | positive |
D. | negative |
Answer» C. positive | |
169. |
For line detection we use mask that is |
A. | Gaussian |
B. | laplacian |
C. | ideal |
D. | butterworth |
Answer» C. ideal | |
170. |
THE_TOTAL_DESCRIPTORS_THAT_THE_80386_CAN_HANDLE_IS?$ |
A. | 2K |
B. | 8K |
C. | 4K |
D. | 16K |
Answer» E. | |
171. |
THE_STARTING_ADDRESS_OF_THE_SEGMENT_IN_PHYSICAL_MEMORY_IS_DECIDED_BY?$ |
A. | physical memory |
B. | segment descriptors |
C. | operating system |
D. | base address |
Answer» D. base address | |
172. |
The limit field of the descriptor is o? |
A. | 10 bits |
B. | 8 bits |
C. | 16 bits |
D. | 20 bits |
Answer» E. | |
173. |
Which of the following is not a type of segment descriptor? |
A. | system descriptors |
B. | local descriptors |
C. | gate descriptors |
D. | none |
Answer» E. | |
174. |
The segment descriptor contains |
A. | access rights |
B. | limit |
C. | base address |
D. | all of the mentioned |
Answer» E. | |
175. |
If the Default operation size bit, D=1, the code segment operation size selected is |
A. | 8-bit |
B. | 16-bit |
C. | 32-bit |
D. | 64-bit |
Answer» D. 64-bit | |
176. |
The bit that indicates whether the segment is page addressable is |
A. | base address |
B. | attribute bit |
C. | present bit |
D. | granularity bit |
Answer» E. | |
177. |
If the segment descriptor bit, S=0, then the descriptor is |
A. | data segment descriptor |
B. | code segment descriptor |
C. | system descriptor |
D. | all of the mentioned |
Answer» D. all of the mentioned | |
178. |
The TYPE field of a descriptor is used to find the |
A. | descriptor type |
B. | segment type |
C. | descriptor and segment type |
D. | none |
Answer» D. none | |
179. |
The bit that indicates whether the segment has been accessed by the CPU or not is |
A. | base address |
B. | attribute bit |
C. | present bit |
D. | granular bit |
Answer» C. present bit | |