McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 251. |
Which of the following is programmed electrically by the user? |
| A. | rom |
| B. | eprom |
| C. | prom |
| D. | eeprom |
| Answer» C. prom | |
| 252. |
PROMs are available in |
| A. | bipolar and mosfet technologies |
| B. | mosfet and fet technologies |
| C. | fet and bipolar technologies |
| D. | mos and bipolar technologies |
| Answer» D. mos and bipolar technologies | |
| 253. |
Which of the following best describes the fusible-link PROM? |
| A. | manufacturer-programmable, reprogrammable |
| B. | manufacturer-programmable, one-time programmable |
| C. | user-programmable, reprogrammable |
| D. | user-programmable, one-time programmable |
| Answer» D. user-programmable, one-time programmable | |
| 254. |
How can ultraviolet erasable PROMs be recognized? |
| A. | there is a small window on the chip |
| B. | they will have a small violet dot next to the #1 pin |
| C. | their part number always starts with a “u”, such as in u12 |
| D. | they are not readily identifiable, since they must always be kept under a small cover |
| Answer» A. there is a small window on the chip | |
| 255. |
Which part of a Flash memory architecture manages all chip functions? |
| A. | program verify code |
| B. | floating-gate mosfet |
| C. | command code |
| D. | input/output pins |
| Answer» B. floating-gate mosfet | |
| 256. |
How much locations an 8-bit address code can select in memory? |
| A. | 8 locations |
| B. | 256 locations |
| C. | 65,536 locations |
| D. | 131,072 locations |
| Answer» B. 256 locations | |
| 257. |
What is a fusing process? |
| A. | it is a process by which data is passed to the memory |
| B. | it is a process by which data is read through the memory |
| C. | it is a process by which programs are burnout to the diode/transistors |
| D. | it is a process by which data is fetched through the memory |
| Answer» C. it is a process by which programs are burnout to the diode/transistors | |
| 258. |
Fusing process is |
| A. | reversible |
| B. | irreversible |
| C. | synchronous |
| D. | asynchronous |
| Answer» B. irreversible | |
| 259. |
The cell type used inside a PROM is |
| A. | link cells |
| B. | metal cells |
| C. | fuse cells |
| D. | electric cells |
| Answer» C. fuse cells | |
| 260. |
How many types of fuse technologies are used in PROMs? |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» B. 3 | |
| 261. |
Metal links are made up of |
| A. | polycrystalline |
| B. | magnesium sulphide |
| C. | nichrome |
| D. | silicon dioxide |
| Answer» C. nichrome | |
| 262. |
Silicon links are made up of |
| A. | polycrystalline silicon |
| B. | polycrystalline magnesium |
| C. | nichrome |
| D. | silicon dioxide |
| Answer» A. polycrystalline silicon | |
| 263. |
During programming p-n junction is |
| A. | avalanche reverse biased |
| B. | avalanche forward biased |
| C. | zener reverse biased |
| D. | zener reverse biased |
| Answer» A. avalanche reverse biased | |
| 264. |
The full form of FAMOS is |
| A. | floating gate avalanche injection mos |
| B. | float gate avalanche injection mos |
| C. | floating gate avalanche induction mos |
| D. | float gate avalanche induction mos |
| Answer» A. floating gate avalanche injection mos | |
| 265. |
PROM is programmed by |
| A. | eprom programmer |
| B. | eeprom programmer |
| C. | prom programmer |
| D. | rom programmer |
| Answer» C. prom programmer | |
| 266. |
The PROM starts out with |
| A. | 1s |
| B. | 0s |
| C. | null |
| D. | both 1s and 0s |
| Answer» B. 0s | |
| 267. |
For implementation of PROM, which IC is used? |
| A. | ic 74187 |
| B. | ic 74186 |
| C. | ic 74185 |
| D. | ic 74184 |
| Answer» B. ic 74186 | |
| 268. |
IC 74186 is of |
| A. | 1024 bits |
| B. | 32 bits |
| C. | 512 bits |
| D. | 64 bits |
| Answer» C. 512 bits | |
| 269. |
How many memory locations are addressed using 18 address bits? |
| A. | 165,667 |
| B. | 245,784 |
| C. | 262,144 |
| D. | 212,342 |
| Answer» C. 262,144 | |
| 270. |
How many address bits are needed to operate a 2K * 8-bit memory? |
| A. | 10 |
| B. | 11 |
| C. | 12 |
| D. | 13 |
| Answer» B. 11 | |
| 271. |
What is the bit storage capacity of a ROM with a 1024 × 8 organization? |
| A. | 1024 |
| B. | 4096 |
| C. | 2048 |
| D. | 8192 |
| Answer» D. 8192 | |
| 272. |
The logical sum of two or more logical product terms is called |
| A. | sop |
| B. | pos |
| C. | or operation |
| D. | nand operation |
| Answer» A. sop | |
| 273. |
The expression Y=AB+BC+AC shows the operation. |
| A. | ex-or |
| B. | sop |
| C. | pos |
| D. | nor |
| Answer» B. sop | |
| 274. |
The expression Y=(A+B)(B+C)(C+A) shows the operation. |
| A. | and |
| B. | pos |
| C. | sop |
| D. | nand |
| Answer» B. pos | |
| 275. |
The canonical sum of product form of the function y(A,B) = A + B is |
| A. | ab + bb + a’a |
| B. | ab + ab’ + a’b |
| C. | ba + ba’ + a’b’ |
| D. | ab’ + a’b + a’b’ |
| Answer» B. ab + ab’ + a’b | |
| 276. |
A variable on its own or in its complemented form is known as a |
| A. | product term |
| B. | literal |
| C. | sum term |
| D. | word |
| Answer» B. literal | |
| 277. |
Canonical form is a unique way of representing |
| A. | sop |
| B. | minterm |
| C. | boolean expressions |
| D. | pos |
| Answer» C. boolean expressions | |
| 278. |
There are Minterms for 3 variables (a, b, c). |
| A. | 0 |
| B. | 2 |
| C. | 8 |
| D. | 1 |
| Answer» C. 8 | |
| 279. |
Why is a demultiplexer called a data distributor? |
| A. | the input will be distributed to one of the outputs |
| B. | one of the inputs will be selected for the output |
| C. | the output will be distributed to one of the inputs |
| D. | single input gives single output |
| Answer» A. the input will be distributed to one of the outputs | |
| 280. |
Most demultiplexers facilitate which type of conversion? |
| A. | decimal-to-hexadecimal |
| B. | single input, multiple outputs |
| C. | ac to dc |
| D. | odd parity to even parity |
| Answer» B. single input, multiple outputs | |
| 281. |
In 1-to-4 demultiplexer, how many select lines are required? |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» A. 2 | |
| 282. |
In a multiplexer the output depends on its |
| A. | data inputs |
| B. | select inputs |
| C. | select outputs |
| D. | enable pin |
| Answer» B. select inputs | |
| 283. |
In 1-to-4 multiplexer, if C1 = 1 & C2 = 1, then the output will be |
| A. | y0 |
| B. | y1 |
| C. | y2 |
| D. | y3 |
| Answer» D. y3 | |
| 284. |
How many select lines are required for a 1-to-8 demultiplexer? |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» B. 3 | |
| 285. |
How many AND gates are required for a 1-to-8 multiplexer? |
| A. | 2 |
| B. | 6 |
| C. | 8 |
| D. | 5 |
| Answer» C. 8 | |
| 286. |
Which IC is used for the implementation of 1-to-16 DEMUX? |
| A. | ic 74154 |
| B. | ic 74155 |
| C. | ic 74139 |
| D. | ic 74138 |
| Answer» A. ic 74154 | |
| 287. |
The word demultiplex means |
| A. | one into many |
| B. | many into one |
| C. | distributor |
| D. | one into many as well as distributor |
| Answer» D. one into many as well as distributor | |
| 288. |
Why is a demultiplexer called a data distributor? |
| A. | the input will be distributed to one of the outputs |
| B. | one of the inputs will be selected for the output |
| C. | the output will be distributed to one of the inputs |
| D. | single input to single output |
| Answer» A. the input will be distributed to one of the outputs | |
| 289. |
In a multiplexer the output depends on its |
| A. | data inputs |
| B. | select inputs |
| C. | select outputs |
| D. | enable pin |
| Answer» B. select inputs | |
| 290. |
In 1-to-4 multiplexer, if C1 = 0 & C2 = 1, then the output will be |
| A. | y0 |
| B. | y1 |
| C. | y2 |
| D. | y3 |
| Answer» B. y1 | |
| 291. |
In 1-to-4 multiplexer, if C1 = 1 & C2 = 1, then the output will be |
| A. | y0 |
| B. | y1 |
| C. | y2 |
| D. | y3 |
| Answer» D. y3 | |
| 292. |
What is the addition of the binary numbers 11011011010 and 010100101? |
| A. | 0111001000 |
| B. | 1100110110 |
| C. | 11101111111 |
| D. | 10011010011 |
| Answer» C. 11101111111 | |
| 293. |
Perform binary addition: 101101 + 011011 = ? |
| A. | 011010 |
| B. | 1010100 |
| C. | 101110 |
| D. | 1001000 |
| Answer» D. 1001000 | |
| 294. |
Binary subtraction of 100101 – 011110 is |
| A. | 000111 |
| B. | 111000 |
| C. | 010101 |
| D. | 101010 |
| Answer» A. 000111 | |
| 295. |
Perform multiplication of the binary numbers: 01001 × 01011 = ? |
| A. | 001100011 |
| B. | 110011100 |
| C. | 010100110 |
| D. | 101010111 |
| Answer» A. 001100011 | |
| 296. |
100101 × 0110 = ? |
| A. | 1011001111 |
| B. | 0100110011 |
| C. | 101111110 |
| D. | 0110100101 |
| Answer» C. 101111110 | |
| 297. |
On multiplication of (10.10) and (01.01), we get |
| A. | 101.0010 |
| B. | 0010.101 |
| C. | 011.0010 |
| D. | 110.0011 |
| Answer» C. 011.0010 | |
| 298. |
Divide the binary numbers: 111101 ÷ 1001 and find the remainder |
| A. | 0010 |
| B. | 1010 |
| C. | 1100 |
| D. | 0011 |
| Answer» D. 0011 | |
| 299. |
Divide the binary number (011010000) by (0101) and find the quotient |
| A. | 100011 |
| B. | 101001 |
| C. | 110010 |
| D. | 010001 |
| Answer» A. 100011 | |
| 300. |
Binary subtraction of 101101 – 001011 = ? |
| A. | 100010 |
| B. | 010110 |
| C. | 110101 |
| D. | 101100 |
| Answer» A. 100010 | |
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