MCQOPTIONS
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MCQOPTIONS
This section includes 358 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Moore machine has states than a mealy machine. |
| A. | fewer |
| B. | more |
| C. | equal |
| D. | negligible |
| Answer» C. equal | |
| 52. |
Output values of mealy type FSM are determined by its |
| A. | input values |
| B. | output values |
| C. | both input values and current state |
| D. | current state |
| Answer» D. current state | |
| 53. |
. In the FSM diagram, what does the information below the line in the circle represent? |
| A. | change of state |
| B. | state |
| C. | output value |
| D. | initial state |
| Answer» D. initial state | |
| 54. |
In the FSM diagram, what does arrow between the circles represent? |
| A. | change of state |
| B. | state |
| C. | output value |
| D. | initial state |
| Answer» B. state | |
| 55. |
. In FSM diagram what does circle represent? |
| A. | change of state |
| B. | state |
| C. | output value |
| D. | initial state |
| Answer» C. output value | |
| 56. |
What happens if the input is low in FSM? |
| A. | change of state |
| B. | no transition in state |
| C. | remains in a single state |
| D. | invalid state |
| Answer» C. remains in a single state | |
| 57. |
What happens if the input is high in FSM? |
| A. | change of state |
| B. | no transition in state |
| C. | remains in a single state |
| D. | invalid state |
| Answer» B. no transition in state | |
| 58. |
Output values of Moore type FSM are determined by its |
| A. | input values |
| B. | output values |
| C. | clock input |
| D. | current state |
| Answer» E. | |
| 59. |
How many different states does a decade counter count? |
| A. | eight |
| B. | nine |
| C. | ten |
| D. | eleven |
| Answer» D. eleven | |
| 60. |
Synchronous counter use global clock, unlike asynchronous counter. |
| A. | one |
| B. | two |
| C. | three |
| D. | zero |
| Answer» B. two | |
| 61. |
In counter universal clock is not used. |
| A. | synchronous counter |
| B. | asynchronous counter |
| C. | decade counter |
| D. | ring counter |
| Answer» C. decade counter | |
| 62. |
How many types of the data type are there in the ring counter? |
| A. | one |
| B. | two |
| C. | three |
| D. | more than three |
| Answer» E. | |
| 63. |
How many types of shift operators are there in VHDL? |
| A. | three |
| B. | four |
| C. | five |
| D. | six |
| Answer» E. | |
| 64. |
‘shift_reg’ is used to initialize the |
| A. | lsb |
| B. | msb |
| C. | register type |
| D. | register bits |
| Answer» C. register type | |
| 65. |
Which of the following flip-flop is used by the ring counter? |
| A. | d flip-flops |
| B. | sr flip-flops |
| C. | jk flip-flops |
| D. | t flip-flops |
| Answer» B. sr flip-flops | |
| 66. |
Divide the binary number (011010000) by (0101) and find the quotient |
| A. | 100011 |
| B. | 101001 |
| C. | 110010 |
| D. | 010001 |
| Answer» B. 101001 | |
| 67. |
Binary subtraction of 101101 – 001011 = ? |
| A. | 100010 |
| B. | 010110 |
| C. | 110101 |
| D. | 101100 |
| Answer» B. 010110 | |
| 68. |
Divide the binary numbers: 111101 ÷ 1001 and find the remainder |
| A. | 0010 |
| B. | 1010 |
| C. | 1100 |
| D. | 0011 |
| Answer» E. | |
| 69. |
On multiplication of (10.10) and (01.01), we get |
| A. | 101.0010 |
| B. | 0010.101 |
| C. | 011.0010 |
| D. | 110.0011 |
| Answer» D. 110.0011 | |
| 70. |
Perform multiplication of the binary numbers: 01001 × 01011 = ? |
| A. | 001100011 |
| B. | 110011100 |
| C. | 010100110 |
| D. | 101010111 |
| Answer» B. 110011100 | |
| 71. |
Binary subtraction of 100101 – 011110 is |
| A. | 000111 |
| B. | 111000 |
| C. | 010101 |
| D. | 101010 |
| Answer» B. 111000 | |
| 72. |
100101 × 0110 = ? |
| A. | 1011001111 |
| B. | 0100110011 |
| C. | 101111110 |
| D. | 0110100101 |
| Answer» D. 0110100101 | |
| 73. |
Perform binary addition: 101101 + 011011 = ? |
| A. | 011010 |
| B. | 1010100 |
| C. | 101110 |
| D. | 1001000 |
| Answer» E. | |
| 74. |
What is the addition of the binary numbers 11011011010 and 010100101? |
| A. | 0111001000 |
| B. | 1100110110 |
| C. | 11101111111 |
| D. | 10011010011 |
| Answer» D. 10011010011 | |
| 75. |
In 1-to-4 multiplexer, if C1 = 0 & C2 = 1, then the output will be |
| A. | y0 |
| B. | y1 |
| C. | y2 |
| D. | y3 |
| Answer» C. y2 | |
| 76. |
The word demultiplex means |
| A. | one into many |
| B. | many into one |
| C. | distributor |
| D. | one into many as well as distributor |
| Answer» E. | |
| 77. |
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» B. ic 74155 | |
| 78. |
How many AND gates are required for a 1-to-8 multiplexer? |
| A. | 2 |
| B. | 6 |
| C. | 8 |
| D. | 5 |
| Answer» D. 5 | |
| 79. |
How many select lines are required for a 1-to-8 demultiplexer? |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» C. 4 | |
| 80. |
In 1-to-4 multiplexer, if C1 = 1 & C2 = 1, then the output will be |
| A. | y0 |
| B. | y1 |
| C. | y2 |
| D. | y3 |
| Answer» E. | |
| 81. |
In a multiplexer the output depends on its |
| A. | data inputs |
| B. | select inputs |
| C. | select outputs |
| D. | enable pin |
| Answer» C. select outputs | |
| 82. |
In 1-to-4 demultiplexer, how many select lines are required? |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» B. 3 | |
| 83. |
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» C. ac to dc | |
| 84. |
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» B. one of the inputs will be selected for the output | |
| 85. |
There are Minterms for 3 variables (a, b, c). |
| A. | 0 |
| B. | 2 |
| C. | 8 |
| D. | 1 |
| Answer» D. 1 | |
| 86. |
Canonical form is a unique way of representing |
| A. | sop |
| B. | minterm |
| C. | boolean expressions |
| D. | pos |
| Answer» D. pos | |
| 87. |
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» C. sum term | |
| 88. |
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» C. ba + ba’ + a’b’ | |
| 89. |
The expression Y=(A+B)(B+C)(C+A) shows the operation. |
| A. | and |
| B. | pos |
| C. | sop |
| D. | nand |
| Answer» C. sop | |
| 90. |
The expression Y=AB+BC+AC shows the operation. |
| A. | ex-or |
| B. | sop |
| C. | pos |
| D. | nor |
| Answer» C. pos | |
| 91. |
The logical sum of two or more logical product terms is called |
| A. | sop |
| B. | pos |
| C. | or operation |
| D. | nand operation |
| Answer» B. pos | |
| 92. |
What is the bit storage capacity of a ROM with a 1024 × 8 organization? |
| A. | 1024 |
| B. | 4096 |
| C. | 2048 |
| D. | 8192 |
| Answer» E. | |
| 93. |
How many address bits are needed to operate a 2K * 8-bit memory? |
| A. | 10 |
| B. | 11 |
| C. | 12 |
| D. | 13 |
| Answer» C. 12 | |
| 94. |
How many memory locations are addressed using 18 address bits? |
| A. | 165,667 |
| B. | 245,784 |
| C. | 262,144 |
| D. | 212,342 |
| Answer» D. 212,342 | |
| 95. |
IC 74186 is of |
| A. | 1024 bits |
| B. | 32 bits |
| C. | 512 bits |
| D. | 64 bits |
| Answer» D. 64 bits | |
| 96. |
For implementation of PROM, which IC is used? |
| A. | ic 74187 |
| B. | ic 74186 |
| C. | ic 74185 |
| D. | ic 74184 |
| Answer» C. ic 74185 | |
| 97. |
The PROM starts out with |
| A. | 1s |
| B. | 0s |
| C. | null |
| D. | both 1s and 0s |
| Answer» C. null | |
| 98. |
PROM is programmed by |
| A. | eprom programmer |
| B. | eeprom programmer |
| C. | prom programmer |
| D. | rom programmer |
| Answer» D. rom programmer | |
| 99. |
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» B. float gate avalanche injection mos | |
| 100. |
Silicon links are made up of |
| A. | polycrystalline silicon |
| B. | polycrystalline magnesium |
| C. | nichrome |
| D. | silicon dioxide |
| Answer» B. polycrystalline magnesium | |