Description : How many address bits are required to represent 4K memory (A) 5 bits. (B) 12 bits. (C) 8 bits. (D) 10 bits.
Last Answer : (B) 12 bits.
Description : Which of the following memories stores the most number of bits (A) a 5M×8 memory. (B) a 1M × 16 memory. (C) a 5M × 4 memory. (D) a 1M ×12 memory.
Last Answer : Ans: A 5Mx8 = 5 x 220 x 8 = 40M (max)
Description : A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively: (1) 14 and 15 (2) 14 and 29 (3) 15 and 14 (4) 16 and 32
Last Answer : (3) 15 and 14
Description : Words having 8-bits are to be stored into computer memory. The number of lines required for writing into memory are (A) 1. (B) 2. (C) 4. (D) 8.
Last Answer : Ans: D Because 8-bit words required 8 bit data lines.
Description : What is the decimal value of 25? 38) A) 10 B) 20 C) 16 D) 32
Last Answer : D) 32
Description : A memory unit stores 2^16 bytes in 32-bit words. How many address bits are necessary in order to retrieve a single word from this memory unit?
Last Answer : Each word is 4 bytes, which occupies 4 consecutive addresses at 16 bits each. The first 14 bits of these addresses will be the same, differing only in the two least significant bits. You only need to specify the first 14 bits, because there are only 2^14 words occupying those 2^16 bytes.
Description : The number of control lines for 32 to 1 multiplexer is (A) 4. (B) 5. (C) 16. (D) 6.
Last Answer : Ans: B The number of control lines for 32 (2^5) and to select one input among them total 5 select lines are required.
Description : The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is (A) 12 bits (B) 15 bits (C) 16 bits (D) 20 bits
Last Answer : (C) 16 bits Explanation: VirtualAddress = 32 bits PageSize = 4KB = 12 bits therefore : VPNTag = 20 bits, OffsetTag = 12 bits TLBEntryLength = VPNTag = 20 bits TotalTLBEntries = 64, 4-way implies ... therefore : TLBIndex = 4 bits TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits
Description : How many Flip-Flops are required for mod–16 counter? (A) 5 (B) 6 (C) 3 (D) 4
Last Answer : (D) 4
Description : The IEEE-754 double-precision format to represent floating point numbers, has a length of ........... bits. (A) 16 (B) 32 (C) 48 (D) 64
Last Answer : (D) 64
Description : A weighted resistor digital to analog converter using N bits requires a total of (A) N precision resistors. (B) 2N precision resistors. (C) N + 1 precision resistors. (D) N – 1 precision resistors.
Last Answer : (A) N precision resistors.
Description : In successive-approximation A/D converter, offset voltage equal to 1/2 LSB is added to the D/A converter's output. This is done to (A) Improve the speed of operation. (B) Reduce the maximum ... Increase the number of bits at the output. (D) Increase the range of input voltage that can be converted
Last Answer : (B) Reduce the maximum quantization error.
Description : The A/D converter whose conversion time is independent of the number of bits is (A) Dual slope (B) Counter type (C) Parallel conversion (D) Successive approximation.
Last Answer : Ans: C The A/D converter whose conversion time is independent of the Number of bits is Parallel conversion. (This type uses an array of comparators connected in parallel and comparators compare the input voltage at a particular ratio of the reference voltage).
Description : A ring counter consisting of five Flip-Flops will have (A) 5 states (B) 10 states (C) 32 states (D) Infinite states.
Last Answer : Ans: A A ring counter consisting of Five Flip-Flops will have 5 states.
Description : Which numbering format is used to represent both numbers and letters on a keyboard? 79) A) octal B) ASCII C) hexadecimal D) BCD
Last Answer : B) ASCII
Description : How many bits internet address isassigned to eachhost on a TCP/IP internet which is used in all communications withthehost? A. 16 - bits B. 32 - bits C. 48 - bits D. 64 - bits E. None of the above
Last Answer : 32 - bits
Description : The processor 80386/80486 and the Pentium processor uses _____ bits address bus: a. 16 b. 32 c. 36 d. 64
Last Answer : b. 32
Description : Convert octal 32 to decimal. 123) A) 3010 B) 2610 C) 1610 D) 3201
Last Answer : B) 2610
Description : The hexadecimal equivalent of decimal 32 is ________. 94) A) 1216 B) 2516 C) 3116 D) 2016
Last Answer : D) 2016
Description : What is the decimal value of 26? A) 128 B) 36 C) 32 D) 64
Last Answer : D) 64
Description : The decimal equivalent of the BCD value 0011 1001 0010 is ________. 32) A) 49010 B) 39210 C) 98210 D) 4131
Last Answer : B) 39210
Description : How many select lines will a 32:1 multiplexer will have (A) 5. (B) 8. (C) 9. (D) 11.
Last Answer : (A) 5.
Description : Which instruction are 32 bits long , with extra 16 bits. a. Memory reference instruction b. Memory reference format c. Both d. None of these
Last Answer : a. Memory reference instruction
Description : Because we wish to allow each ASCII code to occupy one location in memory, most memories are _____ addressable. a. BYTE b. NIBBLE c. WORD (16 bits) d. DOUBLEWORD (32 bits)
Last Answer : a. BYTE
Description : For a concrete mix 1:3:6 and water cement ratio 0.6 both by weight, the quantity of water required per bag, is (A) 10 kg (B) 12 kg (C) 14 kg (D) 16 kg
Last Answer : Answer: Option C
Description : How many bits are needed to address 64 K memory: - a) 3 b) 10 c) 16 d) 32
Last Answer : How many bits are needed to address 64 K memory: - 16
Description : How many flip flops are required to construct a decade counter (A) 10 (B) 3 (C) 4 (D) 2
Last Answer : Ans: C Decade counter counts 10 states from 0 to 9 ( i.e. from 0000 to 1001 ) Thus four FlipFlop's are required.
Description : The hexadecimal equivalent of a decimal 14 is ________. 98) A) "C" B) "D" C) "E" D) "F"
Last Answer : C) "E"
Description : Convert the decimal number 14.875 to binary. 87) A) 1110.0111 B) 1110.1110 C) 1111.1011 D) 1100.1100
Last Answer : B) 1110.1110
Description : Convert the decimal number 6.75 to binary. 14) A) 0110.1100 B) 0111.1100 C) 0110.0110 D) 0110.1010
Last Answer : A) 0110.1100
Description : For JK flipflop J = 0, K=1, the output after clock pulse will be (A) 1. (B) no change. (C) 0. (D) high impedance.
Last Answer : Ans: C J=0, K=1, these inputs will reset the flip-flop after the clock pulse. So whatever be the previous output, the next state will be 0.
Description : The output of a JK flipflop with asynchronous preset and clear inputs is 1'. The output can be changed to 0' with one of the following conditions. (A) By applying J = 0, K = 0 and using a clock. (B) ... (C) By applying J = 1, K = 1 and using the clock. (D) By applying a synchronous preset input.
Last Answer : Ans: C Preset state of JK Flip-Flop =1 With J=1 K=1 and the clock next state will be complement of the present state.
Description : For JK flip flop with J=1, K=0, the output after clock pulse will be (A) 0. (B) 1. (C) high impedance. (D) no change.
Last Answer : (B) 1.
Description : Various machine level components are: a. Address register > Program counter c Data register d. Accumulator register e. Memory of 2K,16 bits/word RAM f. Multiplexers g. Allof these
Last Answer : g. Allof these
Description : A stack pointer is a. a 16-bit register in the microprocessor that indicate the beginning of the stack memory. b. a register that decodes and executes 16-bit arithmetic expression. c. The first memory location where a subroutine address is stored. d. a register in which flag bits are stored
Last Answer : a. a 16-bit register in the microprocessor that indicate the beginning of the stack memory.
Description : For a memory with a 16-bit address space, the addressability is a. 16 bits b. 8 bits c. 2^16 bits d. Cannot be determined
Last Answer : d. Cannot be determined
Description : The binary equivalent of hexadecimal 16 is ________. 126) A) 01110000 B) 1100001 C) 0010011 D) 0001
Last Answer : D) 0001
Description : Convert the binary number 10010.0100 to decimal. 103) A) 16.25 B) 18.25 C) 18.40 D) 24.50
Last Answer : B) 18.25
Description : The decimal equivalent of hexadecimal 16 is ________. 85) A) 2710 B) 2510 C) 3210 D) 2210
Last Answer : D) 2210
Description : Convert octal 16 to decimal. 63) A) 2010 B) 1410 C) 1610 D) 1010
Last Answer : B) 1410
Description : The binary equivalent of hexadecimal DB is ________. 16) A) 10110011 B) 11011100 C) 10111011 D) 11011011
Last Answer : D) 11011011
Description : The number of control lines for 16 to 1 multiplexer is (A) 2. (B) 4. (C) 3. (D) 5.
Last Answer : (B) 4.
Description : How many select lines will a 16 to 1 multiplexer will have (A) 4 (B) 3 (C) 5 (D) 1
Last Answer : (A) 4
Description : The decimal equivalent of Binary number 11010 is (A) 26. (B) 36. (C) 16. (D) 23.
Last Answer : (A) 26.
Description : The Width of a processor’s data path is measured in bits. Which of the following are common data paths? A) 8 bits B) 12 bits C) 16 bits D) 32 bits
Last Answer : Answer : A
Description : The Width of a processor’s data path is measured in bits. Which of the following are common data paths? a. 8 bits b. 12 bits c. 16 bits d. 32 bits
Last Answer : a. 8 bits
Description : Which of the memory is volatile memory (A) ROM (B) RAM (C) PROM (D) EEPROM
Last Answer : Ans: B RAM is a volatile memory (Volatile memory means the contents of the RAM get erased as soon as the power goes off.)
Description : What is the minimum number of bits required in a PCM code for a range of 10,000? A. 12 B. 9 C. 14 D. 10
Last Answer : C. 14
Description : How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB (A) 1, 1 (B) 4, 2 (C) 3, 2 (D) 2, 3
Last Answer : Ans: A There are three product terms, so three AND gates of two inputs are required. As only two input OR gates are available, so two OR gates are required to get the logical sum of three product terms.