MCQOPTIONS
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MCQOPTIONS
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.
| 451. |
If A is 1011, A" is |
| A. | 1011 |
| B. | 100 |
| C. | 1100 |
| D. | 1010 |
| Answer» B. 100 | |
| 452. |
What will be FSV in 2 bit BCD D/A converter is a weighted resistor type with ER = 1V, R = 1 MΩ and Rf = 10KΩ |
| A. | 0.99 V |
| B. | 0.9 V |
| C. | 0.1 V |
| D. | 0 |
| Answer» B. 0.9 V | |
| 453. |
Time delay of a TTL standard family is about |
| A. | 180 ns |
| B. | 50 ns |
| C. | 18 ns |
| D. | 3 ns |
| Answer» D. 3 ns | |
| 454. |
The ASCII code is |
| A. | 5-bit code |
| B. | 7-bit code |
| C. | 9-bit code |
| D. | 11-bit code |
| Answer» C. 9-bit code | |
| 455. |
Which of the following needs DC forward voltage to emit light? |
| A. | LED |
| B. | LCD |
| C. | Both LED as well as LCD |
| D. | None of the above |
| Answer» B. LCD | |
| 456. |
Which of the following methods is used for solving differential equations numerically? |
| A. | Runge-Kutta method |
| B. | Gauss-elimination method |
| C. | Newton-Raphson method |
| D. | Any one of these |
| Answer» B. Gauss-elimination method | |
| 457. |
Binary 1111 when added to binary 11111, the result in binary is |
| A. | 111111 |
| B. | 1111 |
| C. | 1000 |
| D. | 10000 |
| Answer» E. | |
| 458. |
The density of dynamic RAM is |
| A. | the same as static RAM |
| B. | less than that of static RAM |
| C. | more than that of static RAM |
| D. | either equal or less than that of static RAM |
| Answer» D. either equal or less than that of static RAM | |
| 459. |
For the circuit shown in the figure, what is the frequency of the output Q? |
| A. | Same as input clock frequency |
| B. | Inverse of the propagation delay of the FF |
| C. | A second and fourth |
| D. | Fifth and eight |
| Answer» C. A second and fourth | |
| 460. |
1111 + 11111 = |
| A. | 101111 |
| B. | 101110 |
| C. | 111111 |
| D. | 11111 |
| Answer» C. 111111 | |
| 461. |
For checking the parity of a digital word, it is preferable to use |
| A. | AND gates |
| B. | NAND gates |
| C. | EX-OR gates |
| D. | NOR gates |
| Answer» D. NOR gates | |
| 462. |
A parity check usually can detect |
| A. | one-bit error |
| B. | double-bit error |
| C. | three-bit error |
| D. | any-bit error |
| Answer» B. double-bit error | |
| 463. |
An SR flip flop can be built using NOR gates or NAND gates. |
| A. | 1 |
| B. | |
| C. | 5 V |
| D. | 2.5 V |
| Answer» B. | |
| 464. |
In level clocking the output can change |
| A. | on rising edge of clock cycle |
| B. | on falling edge of clock cycle |
| C. | during entire half cycle of the clock |
| D. | none of the above |
| Answer» D. none of the above | |
| 465. |
In digital circuits Schottky transistors are preferred over normal transistors because of their |
| A. | lower propagation delay |
| B. | lower power dissipation |
| C. | higher propagation delay |
| D. | higher power dissipation |
| Answer» B. lower power dissipation | |
| 466. |
Read the following statements: Dual slope ADC provides higher speed as compared to other ADCvery good accuracygood rejection of power supplybetter resolutions compared to other ADC for the same number of bits Which of the above are correct? |
| A. | 1 and 2 |
| B. | 2 and 3 |
| C. | 1, 2, 3 |
| D. | 1, 2, 3, 4 |
| Answer» C. 1, 2, 3 | |
| 467. |
In case of static storage elements |
| A. | information is permanently stored |
| B. | information is to be periodically refreshed |
| C. | information is to be intermittently refreshed |
| D. | information is lost in case power is removed |
| Answer» E. | |
| 468. |
A 6 bit DAC uses binary weighted resistors. If MSB resistor is 20 k ohm, the value of LSB resistor is |
| A. | 20 k ohm |
| B. | 80 k ohm |
| C. | 320 k ohm |
| D. | 640 k ohm |
| Answer» E. | |
| 469. |
Nibble is |
| A. | a string of 4 bits |
| B. | a string of 8 bits |
| C. | a string of 16 bit |
| D. | a string of 64 bits |
| Answer» B. a string of 8 bits | |
| 470. |
The frequency of the driving network connected between pins 1 and 2 of a 8085 chip must be |
| A. | equal to the desired clock frequency |
| B. | twice the desired clock frequency |
| C. | four times the desired clock frequency |
| D. | eight times the desired clock frequency |
| Answer» C. four times the desired clock frequency | |
| 471. |
A clock signal driving a 6-bit ring counter has a frequency of 1 MHz. How long is each timing bit high? |
| A. | 1 ms |
| B. | 2 ms |
| C. | 3 ms |
| D. | 6 ms |
| Answer» B. 2 ms | |
| 472. |
Binary number 11011.01 when converted to its 2's complement will become |
| A. | 101.11 |
| B. | 1111.1 |
| C. | 1100.1 |
| D. | 100.11 |
| Answer» E. | |
| 473. |
Assertion (A): The advantages of totem pole output are fast switching and low power consumption.Reason (R): IC packages available are DIP, surface mount and J lead surface mount. |
| 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 | |
| 474. |
100101.10112 = __________ . |
| A. | 115.228 |
| B. | 115.548 |
| C. | 215.548 |
| D. | 100.018 |
| Answer» C. 215.548 | |
| 475. |
The input to a parity detector is 1001. The output is |
| A. | 0 |
| B. | 1 |
| C. | 0 or 1 |
| D. | indeterminate |
| Answer» B. 1 | |
| 476. |
Precisely 1 K byte means |
| A. | 1000 bits |
| B. | 1012 bits |
| C. | 1020bits |
| D. | 1024 bits |
| Answer» E. | |
| 477. |
Which of the following flip-flops is used as latch? |
| A. | JK flip-flop |
| B. | D flip-flop |
| C. | RS flip-flop |
| D. | T flip-flop |
| Answer» D. T flip-flop | |
| 478. |
In floating point representation, the accuracy is |
| A. | 8 bit |
| B. | 6 bit |
| C. | (A + B + C) (D E) |
| D. | (A + B + C) (D E) |
| Answer» C. (A + B + C) (D E) | |
| 479. |
Inverter 74 LS04 has following specifications I0H max = - 0.4 mA, I0L max = 8 mA, IIH max = 20 μA, IIL max = 0.1 mAThe fan out of this inverter is |
| A. | 60 |
| B. | 100 |
| C. | 2 |
| D. | 18 |
| Answer» C. 2 | |
| 480. |
Decimal -90 equals __________ in 8 bit 2s complement |
| A. | 1000 1000 |
| B. | 1010 0110 |
| C. | 1100 1100 |
| D. | 0101 0101 |
| Answer» C. 1100 1100 | |
| 481. |
For the binary number 11101000, the equivalent hexadecimal number is |
| A. | F 9 |
| B. | F 8 |
| C. | E 9 |
| D. | E 8 |
| Answer» E. | |
| 482. |
Access time of a storage is the time required |
| A. | to write one word into the memory |
| B. | to read one word from the memory |
| C. | to write one word into and to read one word from the memory |
| D. | either to write one word or to read one word |
| Answer» E. | |
| 483. |
The main drawbacks of EEPROM are |
| A. | low density |
| B. | high cost |
| C. | both low density and high cost |
| D. | none of the above |
| Answer» D. none of the above | |
| 484. |
In a NOT gate the output is always the opposite of the input. |
| A. | 1 |
| B. | |
| C. | for programming the microprocessors |
| D. | for writing small programs efficiently |
| Answer» B. | |
| 485. |
10112 x 1012 = __________ 10 |
| A. | 55 |
| B. | 45 |
| C. | 35 |
| D. | 25 |
| Answer» B. 45 | |
| 486. |
The logic performed by the high-noise immunity logic circuit shown below is |
| A. | OR |
| B. | AND |
| C. | NOR |
| D. | NAND |
| Answer» E. | |
| 487. |
A NOR gate is a combination of |
| A. | OR gate and AND gate |
| B. | AND gate and NOT gate |
| C. | OR gate and NOT gate |
| D. | two NOT gates |
| Answer» D. two NOT gates | |
| 488. |
A half adder adds |
| A. | 2 bits |
| B. | 3 bits |
| C. | 4 bits |
| D. | 2 or 3 bits |
| Answer» B. 3 bits | |
| 489. |
An index register in digital computer is a register to be used for |
| A. | performing arithmetic and logic operations |
| B. | temporary storage of result |
| C. | counting number of times a program is executed |
| D. | address modification purpose |
| Answer» E. | |
| 490. |
The number of inputs and outputs of a full adder are |
| A. | 3 and 2 respectively |
| B. | 2 and 3 respectively |
| C. | 4 and 2 respectively |
| D. | 2 and 4 respectively |
| Answer» B. 2 and 3 respectively | |
| 491. |
The reason for glitches on the outputs of decoding gates on a synchronous counter is |
| A. | FFs changing states together |
| B. | FFs changing states one at a time |
| C. | AND gates not functioning properly |
| D. | none of the above |
| Answer» C. AND gates not functioning properly | |
| 492. |
SSI is used in flip flops. |
| A. | 1 |
| B. | |
| C. | 0011 is subtracted |
| D. | 0011 is added |
| Answer» B. | |
| 493. |
Gray code is used in devices which convert analog quantities to digital signal because it is |
| A. | more error free |
| B. | much simpler than binary code |
| C. | superior to Excess-3 code |
| D. | absolutely error free |
| Answer» B. much simpler than binary code | |
| 494. |
TTL logic is preferred to DRL logic because |
| A. | greater fan-out is possible |
| B. | greater logic levels are possible |
| C. | greater fan-in is possible |
| D. | less power consumption is achieve |
| Answer» B. greater logic levels are possible | |
| 495. |
Assertion (A): Hamming code is commonly used for error correction Reason (R): In Hamming code the number of parity bits increases as the number of information bits increases. |
| 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 | |
| 496. |
Which interrupts are masked? |
| A. | 5.5 |
| B. | 6.5 |
| C. | 7.5 |
| D. | None |
| Answer» C. 7.5 | |
| 497. |
Which of the following is equivalent to AND-OR realization? |
| A. | NAND-NOR |
| B. | NOR-NOR |
| C. | NOR-NAND |
| D. | NAND-NAND |
| Answer» E. | |
| 498. |
The applications of shift registers are Time delayRing counterSerial to parallel data conversionSerial to serial data conversion Which of the above are correct? |
| A. | 1, 2, 3 |
| B. | 2, 3, 4 |
| C. | 1, 2, 3, 4 |
| D. | 1, 2, 4 |
| Answer» D. 1, 2, 4 | |
| 499. |
In a digital circuit, a clock is a |
| A. | crystal type |
| B. | multivibrator |
| C. | flip-flop |
| D. | free running multivibrator |
| Answer» E. | |
| 500. |
The control logic for a binary multiplier is specified by a states diagram. The state diagram has four states and two inputs. To implement it by the sequence register and decoder method. |
| A. | two flip-flop & 3 x 9 decoder are needed |
| B. | four flip-flop & 3 x 9 decoder are needed |
| C. | E 52 F |
| D. | E 52 E |
| Answer» B. four flip-flop & 3 x 9 decoder are needed | |