DIGITAL ELECTRONICS OBJECTIVE QUESTIONS

DIGITAL ELECTRONICS OBJECTIVE QUESTIONS  

1.The number of digits in octal system isa.8b.7c.10d. none2..The number of digits in Hexadecimal system isa.15b.17c.16d. 83.The number of bits in a nibble isa.16b.5c.4d.84.The digit F in Hexadecimal system is equivalent to —— in decimal systema.16b.15c.17d. 85.Which of the following binary numbers is equivalent to decimal 10a.1000b.1100c.1010d.10016.The number FF in Hexadecimal system is equivalent to —— in decimal systema.256b.255c.240d.2397.IC s area. analogb. digitalc. both analog and digitald. mostly analog8.The rate of change of digital signals between High and Low Level isa. very fastb. fastc. slowd. very slow9. Digital circuits mostly usea. Diodesb. Bipolar transistorsc. Diode and Bipolar transistors

d. Bipolar transistors and FETs10.Logic pulsera. generates short duration pulsesb. generate long duration pulsesc. generates long and short durationd. none of above11.What is the output state of an OR gate if the inputs are 0 and 1?a.0b.1c.3d.212.What is the output state of an AND gate if the inputs are 0 and 1?a.0b.1c.3d.213.A NOT gate has…a. Two inputs and one outputb. One input and one outputc. One input and two outputsd. none of above14.An OR gate has…a. Two inputs and one outputb. One input and one outputc. One input and two outputsd. none of above15.The output of a logic gate can be one of two _____?a. Inputsb. Gatesc.Statesd. none16.Logic states can only be ___ or 0.a. 3b. 2c.1d.0

17.The output of a ____ gate is only 1 when all of its inputs are 1a. NORb. XORc. ANDd. NOT18.A NAND gate is equivalent to an AND gate plus a …. gate put together.a. NORb. NOTc. XORd. none19.Half adder circuit is ______?

a. Half of an AND gateb. A circuit to add two bits togetherc. Half of a NAND gated. none of above20. Numbers are stored and transmitted inside a computer ina. binary formb. ASCII code formc. decimal formd. alphanumeric form

21.The decimal number 127 may be represented bya. 1111 1111Bb. 1000 0000Bc. EEHd. 0111 111122.. A byte corresponds toa. 4 bitsb. 8 bitsc. 16 bitsd. 32 bits23.A gigabyte representsa.1 billion bytesb. 1000 kilobytesc. 230 bytesd. 1024 bytes24. A megabyte representsa. 1 million bytesb. 1000 kilobytesc. 220 bytesd. 1024 bytes25.. A Kb corresponds toa. 1024 bitsb. 1000 bytesc.210 bytesd. 210 bits26.A parity bit isa. used to indicate uppercase lettersb. used to detect errorsc. is the first bit in a byted. is the last bit in a byte27. Which of these devices are two state.a. lampb. punched cardc. magnetic taped. all the above28.The output impedance of of a logic pulser isa. lowb. high

c. may be low or highd. none of above28.The number of LED display indicators in logic probe area.1b.2c.1 or 2d.429.In hexadecimal number system,A is equal to decimal numbera.10b.11c.17d.1830.Hexadecimal number F is equal to octal numbera.15b.16c.17d.1831.Hexadecimal number E is equal to binary numbera.1110b.1101c.1001d.111132.Binary number 1101 is equal to octal numbera.15b.16c.17d.1433.Octal number 12 is equal to decimal numbera.8b.11c.9d. none34.Decimal number 10 is equal to binary numbera.1110b.1000c.1001d.101035.Binary number 110011011001 is equal to decimal numbera.3289b.2289c.1289d.28936.1111+11111=a.101111b.101110c.111111d.01111137.Binary multiplication 1*0=

a.1b.0c.10d.1138.110012 -100012=a.10000b.01000c.00100d.0000139.10112*1012=a.55b.45c.35d.2540.1110112*100012=a.111101101b.111101100c.111110d.110011041.4 bits is equal toa. 1 nibbleb.1 bytec. 2 byted. none of above42. which is non-volatile memorya. RAMb. ROMc. bothd. none43. The contents of these chips are lost when the computer is switched off?a. ROM chipsb. RAM chipsc. DRAM chipsd. none of above44.What are responsible for storing permanent data and instructions.?a. RAM chipsb. ROM chipsc. DRAM chipsd. none of above45. Which parts of the computer perform arithmetic calculations?a. ALUb. Registersc. Logic busd. none of above46.How many bits of information can each memory cell in a computer chip hold?a. 0 bitsb. 1 bitc. 8 bits

d. 2 bits47.What type of computer chips are said to be volatile?a. RAM chipsb. ROM chipsc. DRAMd. none of above48.Which generation of computer uses more than one microprocessor?a. Second generationb. Fifth generationc.Third generationd .none of above49.Which generation of computer developed using integrated circuits?a. Second generationb. Fifth generationc. Third generationd. none of above

50.Which generation of computer was developed from microchips?a. Second generationb. Third generationc. Fourth generationd. none of above51.RAM can be expanded to aa. increase word sizeb. increase word numberc. increase word size or increase word numberd. none of above52. Which memory is available in all technologiesa. PROMb. EEPROMc. ROMd. EPROM53. Which memory does not require programming equipmenta. PROMb. EEPROMc. ROMd. EPROM54. In CCDa. small charge is deposited for logical 1b. small charge is deposited for logical 0 or 1c. small charge is deposited for logical 0 and large charge for logical 1d. none of above55. The internal structure of PLA is similar toa. RAMb. ROMc. both RAM or ROMd. neither RAM nor RAM56.An output of combinational ckt depends on

a. present inputsb. previous inputsc. both present and previousd .none of above57.Which are combinational gatesa. NAND & NORb. NOT & ANDc. X-OR & X-NORd. none of above58.. As access time is decreased, the cost of memorya. remains the sameb. increasesc. decreasesd. may increase or decrease59. Which is correct:a. A.A=0b. A+1=Ac. A+A=A’d. A’.A’=060.A counter is aa. Sequential cktb. Combinational cktc. both combinational and sequential cktd. none of above61.The parity bit isa. always 1b. always 0c.1 or 0d.none of above62.In 2 out of 5 code,decimal number 8 isa.11000b.10100c.1100d.1010

63.In number of information bits is 11,the number of parity Bits in hamming code isa.5b.4c.3d.264.For a 4096*8 EPROM ,the number of address lines isa.14b.12c.10d.865. 23.6 10=……….2a.11111.10011b.10111.10011

c.00111.101d.10111.166.BCD number 0110011=…….10a.66b.67c.68d.6967.The total number of input states for 4 input or gate isa.20b.16c.12d.868.In a 4 input OR gate,the total number of High outputs for the 16 input states area.16b.15c.13d. none of above69.In a 4 input AND gate,the total number of High outputs for the 16 input states area.16b.8c.4d.170.a buffer isa. always non-invertingb.always invertingc. inverting or non-invertingd.none of above71.An AND gate has two inputs A and B and ine inhibits input S.Output is 1 ifa.A=1,B=1,S=1b. A=1,B=1,S=0c. A=1,B=0,S=1d. A=1,B=0,S=072. An AND gate has two inputs A and B and ine inhibits input S.Out of total 8 input states,Output is 1 ina. 1 statesb. 2 statesc. 3 statesd. 4 states73.In a 3 input NOR gate,the number of states in which output is 1 equalsa. 1b. 2c. 3d. 474.Which of these are universal gatesa. only NORb. only NANDc. both NOR and NANDd. NOT,AND,OR

75. In a 3 input NAND gate,the number of gates in which output in 1equalsa.8b.7c.6d..576. A XOR gate has inputs A and B and output Y.Then the output equation isa.Y=A+Bb.Y=AB+A’Bc.AB+ AB’d.AB’+A’B’77.A 14 pin NOT gate IC has………..NOT gatesa.8b.6c.5d.478.A 14 pin AND gate IC has………..AND gatesa.8b.6c.4d.279.The first contribution to logic was made bya. George Booleb. Copernicusc. Aristotled. Shannon80.Boolean Alzebra obeysa. commutative lawb. associative lawc. distributive lawd. commutative, associative, distributive law81. A+(B.C)=a. A.B+Cb. A.B+A.Cc. Ad.(A+B).(A+C)82.A.0=a. 1b. Ac. 0d. A or 183.A+A.B=a. Bb. A.Bc. Ad. A or B

84.Demorgan’s first theorem isa. A.A’=0

b. A’’=Ac. (A+B)’=A’.B’d. (AB)’=A’+B’85. Demorgan’s second theorem isa. A.A’=0b. A’’=Ac. (A+B)’=A’.B’d. (AB)’=A’+B’

86. Which of the following is truea. SOP is a two level logicb. POS is a two level logicc. both SOP and POS are two level logicd. Hybrid function is two level logic87.The problem of logic race occurs ina. SOP functionsb. Hybrod functionsc. POS functionsd. SOP and POS functions

88. In which function is each term known as min terma. SOPb. POSc. Hybridd. both SOP and POS

89. In which function is each term known as max terma. SOPb. POSc. Hybridd. both SOP and Hybrid90. In the expression A+BC, the total number of min terms will bea.2b. 3c.4d. 591.The min term designation for ABCD isa.m0b. m10c. m14d. m1592. The function Y=AC+BD+EF isa. POSb. SOPc. Hybridd. none of above

93. The expression Y=∏M(0,1,3,4) is

a. POSb. SOPc. Hybridd. none of above94. AB+AB’=a. Bb. Ac.1d. 095. In a four variable Karnaugh map eight adjacent cells give aa. Two variable termb. single variable termc. Three variable termd. four variable term96.A karnaugh map with 4 variables hasa. 2 cellsb. 4 cellsc. 8 cellsd.16 cells

97.In a karnaugh map for an expression having ‘don’t care terms’ the don’t cares can be treated asa. 0b. 1c. 1 or 0d. none of above98. The term VLSI generally refers to a digital IC havinga. more than 1000 gatesb. more than 100 gatesc. more than 1000 but less than 9999 gatesd. more than 100 but less than 999 gates99.Typical size of an IC is abouta.1”*1”b. 2”*2”c. 0.1”*0.1”d. 0.0001”*0.0001”

100.A digital clock uses…………..chipa. SSIb. LSIc. VLSId. MSI101. Digital technologies being used now-a-days area. DTL and EMOSb. TTL, ECL, CMOS and RTLc. TTL, ECL and CMOSd. TTL, ECL, CMOS and DTL

102. A TTL circuit with totem pole output hasa. high output impedanceb. low output impedancec. very high output impedanced. any of above103. TTL usesa. multi emitter transistorsb. multi collector transistorsc. multi base transistorsd. multi emitter or collector transistors104. Advanced schottky is a part ofa. ECL familyb. CMOS familyc. TTL familyd. none of above105. For wired AND connection we should usea. TTL gates with active pull upb. TTL gates with open collectorc. TTL gates without active pull up and with open collectord. any of above106. Time delay of a TTL family is abouta. 180nsb. 50nsc. 18nsd. 3 ns107. As compared to TTL, ECL hasa. lower power dissipationb. lower propagation delayc. higher propagation delayd. higher noise margin108. As compared to TTL, CMOS logic hasa. higher speed of operationb. higher power dissipationc. smaller physical sized. all of above109. 74HCT00 series isa.NAND ICb. interface between TTL and CMOSc. inverting ICd. NOR IC110.CD 4010 is aa. inverting bufferb. non inverting hex bufferc. NOR ICd. NAND IC111. Current requirement of a piezo buffer is abouta. 100mAb. 20mA

c. 4 mAd. 0.4 mA112. TSL inverter hasa. one inputb. two inputsc. one or two inputsd. three inputs113. Parallel adder isa. sequential circuitsb. combinational circuitsc. either sequential or combinational circuitsd. none of above114. The inputs to a 3 bit binary adder are 1112 and 1102. The output will bea.101b.1101c.1111d.1110115. A half adder can be used only for addinga. 1sb. 2sc. 4sd. 8s116. A 3 bit binary adder should bea. 3 full addersb. 2 full adders and 1 half adderc. 1 full adder and 2 half adderd. 3 half adders117. when two 4 bit parallel adders are cascaded we geta. 4 bit parallel adderb. 8 bit parallel adderc. 16 bit parallel adderd. none of above118. The widely used binary multiplication method isa. repeated additionb. add and shiftc. shift and addd. any of above119.When microprocessor processes both positive and negative numbers, the representation used isa. 1’s complementb. 2’s complementc. signed binaryd. any of above120. Decimal -90 =………….in 8 bit 2s complementa.1000 1000b.1010 0110c.1100 1100d.0101 0101

121. In 2’s complement addition, the carry generated in the last stage isa. added to LSBb. neglectedc. added to bit next to MSBd. added to the bit next to LSB122. The number of inputs and outputs in a full adder area. 2 and 1b. 2 and 2c. 3 and 3d. 3 and 2123.In a 7 segment display the segments a,c,d,f,g are lit. The decimal number displayed will bea. 9b. 5c. 4d. 2124. In a 7 segment display the segments b and c are lit up. The decimal number displayed will bea. 9b. 7c. 3d. 1125 .A device which converts BCD to seven segments is calleda. encoderb. decoderc. multiplexerd. none of these126. Which device use the nematic fluida. LEDb. LCDc. VF displayd. none of these

127. Which of these is the most recent devicea. LEDb. LCDc. VF displayd. a and c128. VF glows with ………. Colour when activateda. redb. orangec. bluish greend. none of these129. Which display device resembles vacuum tubea. LEDb. LCDc. VFd. none of these

130.Which device changes parallel data to serial dataa. decoderb. multiplexerc. demultiplexerd. flip flop131.A 1 of 4 multiplexer requires…… data select linea. 1b. 2c. 3d. 4132. It is desired to route data from many registers to one register. The device needed isa. decoderb. multiplexerc. demultiplexerd. counter133.Which device has one input and many outputsa. flip flopb. multiplexerc. demultiplexerd. counter134.Two 16:1 and one 2:1 multiplexers can be connected to form aa. 16:1 multiplexerb. 32:1 multiplexerc. 64:1 multiplexerd. 8:1 multiplexer135. A flip flop is aa. combinational circuitb. memory elementc. arithmetic elementd. memory or arithmetic136. I n a D latcha. data bit D is fed to S input and D’ to R inputb. data bit D is fed to R input and D’ to S inputc. data bit D is fed to both R and S inputsd. data bit D’ is not fed to any input137. I n a D latcha. a high D sets the latch and low D resets itb. a low D sets the latch and high D resets itc. race can occurd. none of above

138.In a positive edge triggered JK flip flopa. High J and High K produce inactive stateb. Low J and High K produce inactive statec. High J and Low K produce inactive stated. Low J and Low K produce inactive state139.In a positive edge triggered D flip flop

a. D input is called direct setb.Preset is called direct resetc. present and clear are called direct set and reset respectivelyd. D input overrides other inputs140. In a positive edge triggered JK flip flopJ=1,K=0 and clock pulse is rising.Q willa. be 0b. be 1c. show no changed. toggle141. For edge triggering in flip flops manufacturers usea. RC circuitb. direct coupled designc. either RC circuit or direct coupled designd. none of these142. In a JK flip flop toggle meansa. set Q=1 and Q’=0b. set Q=0 and Q’=1c. change the output to the opposite stated. no change in input143. A mod 4 counter will counta. from 0 to 4b. from 0 to 3c. from any number n to n+4d. none of above144.A counter has N flip flops. The total number of states area. Nb. 2Nc. 2Nd. 4N145.A counter has modulus of 10. The number of flip flops area. 10b. 5c. 4d. 3146.In a ripple countera. whenever a flip flop sets to 1,the next higher FF togglesb. whenever a flip flop sets to 0,the next higher FF remains unchangedc. whenever a flip flop sets to 1,the next higher FF faces race conditiond. whenever a flip flop sets to 0,the next higher FF faces race cond147.A counter has 4 flip flops.It divides the input frequency bya.4b. 2c. 8d. 16148. A decade counter skipsa. binary states 1000 to 1111b. binary states 0000 to 0011

c. binary states 1010 to 1111d. binary states 1111 and higher149.The number of flip flops needed for Mod 7 counter area. 7b. 5c. 3d. 1150.A presettable counter with 4 flip flops start counting froma. 0000b. 1000c. any number from 0000 to 1111d. any number from 0000 to 1000151.A 4 bit down counter can count froma. 0000 to 1111b. 1111 to 0000c. 000 to 111d. 111 to 000152. A 3 bit up-down counter can count froma. 000 to 111b. 111 to 000c. 000 to 111 and also from 111 to 000d. none of above153.IC counters area. synchronous onlyb. asynchronous onlyc. both synchronous and asynchronousd. none of above

154. Shifting digits from left to right and vice versa is needed ina. storing numbersb. arithmetic operationsc. countingd. storing and counting155. The basic storage element in a digital system isa. flip flopb. counterc. multiplexerd. encoder156. The simplest register isa. buffer registerb. shift registerc. controlled buffer registerd. bidirectional register157. The basic shift register operations area. serial in serial outb. serial in parallel outc. parallel in serial outd. all of above

158. A universal shift register can shifta. from right to left b. from left to rightc. both from right to left and left to rightd. none of above159. In a shift register, shifting a bit by one bit meansa. division by 2b. multiplication by 2c. subtraction by 2d. any of above160. An 8 bit binary number is to be entered into an 8 bit serial shift register. The number of clock pulses required isa. 1b. 2c. 4d. 8   

1.In which of the following base systems is 123 not a valid number?

(a) Base 10

(b) Base 16

(c)Base8

(d) Base 3

Ans:d

2. Storage of 1 KB means the following number of bytes

(a) 1000

(b)964

(c)1024

(d) 1064.

Ans:d

3. What is the octal equivalent of the binary number:

10111101

(a)675

(b)275

(c) 572

(d) 573.

Ans:b

4. Pick out the CORRECT statement:

(a) In a positional number system, each symbol represents the same value

irrespective of its position

(b) The highest symbol in a position number system as a value equal to the number

of symbols in the system

(c) It is not always possible to find the exact binary

(d) Each hexadecimal digit can be represented as a sequence of three binary

symbols.

Ans:c

5.The binary code of (21.125)10 is

(a) 10101.001

(b) 10100.001

(c) 10101.010

(d) 10100.111.

Ans:a

6.A NAND gate is called a universal logic element because

(a) it is used by everybody

(b) any logic function can be realized by NAND gates alone

(c) all the minization techniques are applicable for optimum NAND gate realization

(d) many digital computers use NAND gates.

Ans:b

7. Digital computers are more widely used as compared to analog computers,

because they are

(a) less expensive

(b) always more accurate and faster

(c) useful over wider ranges of problem types

(d) easier to maintain.

Ans:c

8. Most of the digital computers do not have floating point hardware because

(a) floating point hardware is costly

(b) it is slower than software

(c) it is not possible to perform floating point addition by hardware

(d) of no specific reason.

Ans:a

9. The number 1000 would appear just immediately after

(a) FFFF (hex)

(b) 1111 (binary)

(c) 7777 (octal)

(d) All of the above.

Ans:d

10. (1(10101)2 is

(a) (37)10

(b) ( 69)10

(c) (41 )10

(d) — (5)10

Ans:a

11. The number of Boolean functions that can be generated by n variables is equal

to

(a) 2n

(b) 22 n

(c) 2n-1

(d) — 2n

Ans:b

12. Consider the representation of six-bit numbers by two’s complement, one’s

complement, or by sign and magnitude: In which representation is there overflow

from the addition of the integers 011000 and 011000?

(a) Two’s complement only

(b) Sign and magnitude and one’s complement only

(c) Two’s complement and one’s complement only

(d) All three representations.

Ans:d

13. A hexadecimal odometer displays F 52 F. The next reading will be

(a)F52E

(b)G52F

(c)F53F

(d)F53O.

Ans:d

14. Positive logic in a logic circuit is one in which

(a) logic 0 and 1 are represented by 0 and positive voltage respectively

(b) logic 0 and, -1 are represented by negative and positive voltages respectively

(c) logic 0 voltage level is higher than logic 1 voltage level

(d) logic 0 voltage level is lower than logic 1 voltage level.

Ans:d

15. Which of the following gate is a two-level logic gate

(a) OR gate

(b) NAND gate

(c) EXCLUSIVE OR gate

(d) NOT gate.

Ans:c

16. Among the logic families, the family which can be used at very high frequency

greater than 100 MHz in a 4 bit

synchronous counter is

(a) TTLAS

(b) CMOS

(c)ECL

(d)TTLLS

Ans:c

17. An AND gate will function as OR if

(a) all the inputs to the gates are “1”

(b) all the inputs are ‘0’

(c) either of the inputs is “1”

(d) all the inputs and outputs are complemented.

Ans:d

18. An OR gate has 6 inputs. The number of input words in its truth table are

(a)6

(b)32

(c) 64

(d) 128

Ans:c

19. A debouncing circuit is

(a) an astable MV

(b) a bistable MV

(c) a latch

(d) a monostable MV.

Ans:c

20. NAND. gates are preferred over others because these

(a) have lower fabrication area

(b) can be used to make any gate

(c) consume least electronic power

(d) provide maximum density in a chip.

Ans:b

21. In case of OR gate, no matter what the number of inputs, a

(a) 1 at any input causes the output to be at logic 1

(b) 1 at any input causes the output to be at logic 0

(c) 0 any input causes the output to be at logic 0

(d) 0 at any input causes the output to be at logic 1.

Ans:a

22. The fan put of a 7400 NAND gate is

(a)2TTL

(b)5TTL

(c)8TTL

(d)10TTL

Ans:d

23. Excess-3 code is known as

(a) Weighted code

(b) Cyclic redundancy code

(c) Self-complementing code

(d) Algebraic code.

Ans:c

24. Assuming 8 bits for data, 1 bit for parity, I start bit and 2 stop bits, the number

of characters that 1200 BPS communication line can transmit is

(a)10 CPS

(b)120 CPS

(c) 12CPS

(d) None of the above.

Ans:c

 

25. Indicate which of the following three binary additions are correct?

1.1011 + 1010 = 10101

II. 1010 + 1101 = 10111

III. 1010 + 1101 = 11111

(a) I and II

(b) II and III

(c) III only

(d) II and III

Ans:d