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This section includes 1271 Mcqs, each offering curated multiple-choice questions to sharpen your Electronics & Communication Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 1251. |
AB + AB = |
| A. | B |
| B. | A |
| C. | 1 |
| D. | 0 |
| Answer» C. 1 | |
| 1252. |
For an N bit ADC, the percentage resolution is [1/2N - 1)] 100. |
| A. | 1 |
| B. | |
| C. | reduce power consumption |
| D. | increase fan out |
| Answer» B. | |
| 1253. |
In a JK flip flop toggle means |
| A. | set Q = 1 and Q = 0 |
| B. | set Q = 0 and Q = 1 |
| C. | change the output to the opposite state |
| D. | no change in output |
| Answer» D. no change in output | |
| 1254. |
For the gate in the given figure the output will be |
| A. | 0  |
| B. | 1 |
| C. | A |
| D. | A |
| Answer» E. | |
| 1255. |
Assertion (A): A demultiplexer can be used as a decoder.Reason (R): A demultiplexer can be built by using AND gates only. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R are correct but R is not correct explanation of A |
| C. | A is true, R is false |
| D. | A is false, R is true |
| Answer» D. A is false, R is true | |
| 1256. |
Minimum number of 2-input NAND gates required to implement the function F = (x + y) (Z + W) is |
| A. | 3 |
| B. | 4 |
| C. | 5 |
| D. | 6 |
| Answer» C. 5 | |
| 1257. |
The logic operation that will selectively clear bits in register A in those positions where these are 1 's in the bits of register B is given by |
| A. | A ‚Üê A + B |
| B. | A ‚Üê AB |
| C. | A ‚Üê A + B |
| D. | A ‚Üê AB |
| Answer» E. | |
| 1258. |
Assertion (A): The TTL output acts as a current sink in low state Reason (R): The TTL input current is largest in low state. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R are correct but R is not correct explanation of A |
| C. | A is true, R is false |
| D. | A is false, R is true |
| Answer» C. A is true, R is false | |
| 1259. |
The inputs to logic gate are 0. The output is 1. The gate is |
| A. | NAND or XOR |
| B. | OR or XOR |
| C. | AND or XOR |
| D. | NOR or Ex-NOR |
| Answer» E. | |
| 1260. |
A multilevel decoder is more costly as compared to single level decoder. |
| A. | 1 |
| B. | |
| C. | UVPROM |
| D. | EEPROM |
| Answer» B. | |
| 1261. |
An OR gate has 4 inputs. One input is high and the other three are low. The output |
| A. | is low |
| B. | is high |
| C. | is alternately high and low |
| D. | may be high or low depending on relative magnitude of inputs |
| Answer» C. is alternately high and low | |
| 1262. |
A device which converts BCD to seven segment is called |
| A. | encoder |
| B. | decoder |
| C. | multiplexer |
| D. | none of these |
| Answer» C. multiplexer | |
| 1263. |
Monostable multivibrator is called one-shot or single-shot circuit because it |
| A. | can be used once |
| B. | can be used single and not with other circuits. |
| C. | always returns by itself to its single stable state |
| D. | changes to quasistable for a fixed period of time upon receipt of triggering signal. |
| Answer» D. changes to quasistable for a fixed period of time upon receipt of triggering signal. | |
| 1264. |
The number of memory locations in which 14 address bits can access is |
| A. | 1024 |
| B. | 2048 |
| C. | 4096 |
| D. | 16384 |
| Answer» E. | |
| 1265. |
A 12 bit ADC is used to convert analog voltage of 0 to 10 V into digital. The resolution is |
| A. | 2.44 mV |
| B. | 24.4 mV |
| C. | 1.2 V |
| D. | none of the above |
| Answer» B. 24.4 mV | |
| 1266. |
A Read/Write memory chip has a capacity of 64 k bytes. Assuming separate data and address line and availability of chip enable signal, what is the minimum number of pins required in the IC chip? |
| A. | 28 |
| B. | 26 |
| C. | 24 |
| D. | 22 |
| Answer» B. 26 | |
| 1267. |
If we need a low noise device, we should use |
| A. | BJT |
| B. | FET |
| C. | thyristor |
| D. | UJT |
| Answer» C. thyristor | |
| 1268. |
The counter which require maximum number of FF for a given mod counter is |
| A. | Ripple counter |
| B. | BCD counter |
| C. | Ring counter |
| D. | Programmed counter |
| Answer» D. Programmed counter | |
| 1269. |
A voltage DAC is generally slower than current DAC |
| A. | because of more accuracy |
| B. | because of higher resolution |
| C. | because of response time of op-amp current to voltage converter |
| D. | none of the above |
| Answer» D. none of the above | |
| 1270. |
Both OR and AND gates can have only two inputs. |
| A. | 1 |
| B. | |
| Answer» C. | |
| 1271. |
The resolution of an n bit DAC with a maximum input of 5 V is 5 mV. The value of n is |
| A. | 8 |
| B. | 9 |
| C. | 10 |
| D. | 11 |
| Answer» D. 11 | |