MCQOPTIONS
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MCQOPTIONS
This section includes 2171 Mcqs, each offering curated multiple-choice questions to sharpen your ENGINEERING SERVICES EXAMINATION (ESE) knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Effective address is calculated by adding or subtracting displacement value to |
| A. | immediate address |
| B. | Relative address |
| C. | absolute address |
| D. | base address |
| Answer» E. | |
| 252. |
A 4 line to 16 line decoder has |
| A. | 16 inputs and 4 outputs |
| B. | 4 inputs and 16 outputs |
| C. | either (a) or (b) |
| D. | neither (a) nor (b) |
| Answer» C. either (a) or (b) | |
| 253. |
The two outputs of RS flip-flop are |
| A. | always low |
| B. | always high |
| C. | either low or high |
| D. | always complementary |
| Answer» E. | |
| 254. |
Which of the following is 'synchronous'? |
| A. | Half adder |
| B. | Full adder |
| C. | R-S flip-flop |
| D. | Clocked R-S flip-flop |
| Answer» E. | |
| 255. |
Which of the following binary product is incorrect? |
| A. | 1100 x 1010 = 1111000 |
| B. | 1.01 x 10.1 = 11.001 |
| C. | 1100110 x 1000 = 1100110000 |
| D. | None of the above |
| Answer» E. | |
| 256. |
1111 + 11111 = |
| A. | 101111 |
| B. | 101110 |
| C. | 111111 |
| D. | 011111 |
| Answer» C. 111111 | |
| 257. |
The value of 2⁵ in octal system is |
| A. | 40 |
| B. | 20 |
| C. | 100 |
| D. | 200 |
| Answer» B. 20 | |
| 258. |
(7BF)₁₆ = __________ ₂ |
| A. | 0111 1011 1110 |
| B. | 0111 1011 1111 |
| C. | 0111 1011 0111 |
| D. | 0111 1011 0011 |
| Answer» C. 0111 1011 0111 | |
| 259. |
The stack is a specialized temporary __________ access memory during __________ and __________ instructions.The 8156 of a figure has RAM locations from 2000 H to 20 FFH. |
| A. | random, store, load |
| B. | random, push, load |
| C. | sequential, store, pop |
| D. | sequential, push, pop |
| Answer» E. | |
| 260. |
Assertion (A): Power drain of CMOS increases with operating frequency Reason (R): All unused CMOS inputs should be tied either to a fixed voltage level (0 or VDD) or to another input. |
| 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 | |
| 261. |
In ASCII, letter B is coded as |
| A. | 1000001 |
| B. | 1000010 |
| C. | 100100 |
| D. | 1001000 |
| Answer» C. 100100 | |
| 262. |
A buffer is |
| A. | always non-inverting |
| B. | always inverting |
| C. | inverting or non-inverting |
| D. | none of the above |
| Answer» D. none of the above | |
| 263. |
A D-flip-flop is |
| A. | dial type flip-flop |
| B. | delay flip-flop |
| C. | differential flip-flop |
| D. | digital flip-flop |
| Answer» C. differential flip-flop | |
| 264. |
Status register in the 8156 contains information about |
| A. | the timer |
| B. | the ports |
| C. | both (a) and (b) |
| D. | none of the above |
| Answer» D. none of the above | |
| 265. |
How many lines are there in address bus of 8085 μp? |
| A. | 6 |
| B. | 8 |
| C. | 12 |
| D. | 16 |
| Answer» E. | |
| 266. |
EEPROM is also known as |
| A. | UVPROM |
| B. | EAPROM |
| C. | both UVPROM and EAPROM |
| D. | none of the above |
| Answer» C. both UVPROM and EAPROM | |
| 267. |
In CCD |
| A. | a small charge is deposited for logical 1 |
| B. | a small charge is deposited for both logical 1 and 0 |
| C. | a small charge is deposited for logical 0 and large charge for logical 1 |
| D. | none of the above |
| Answer» D. none of the above | |
| 268. |
What is the purpose of using ALE signal high? |
| A. | To latch low order address from bus to separate A₀ - A₇ lines |
| B. | To latch data D₀ - D₇ from bus to separate data bus |
| C. | To disable data bus latch |
| D. | All of the above |
| Answer» B. To latch data D₀ - D₇ from bus to separate data bus | |
| 269. |
For the switch circuit, taking open as 0 and closed as 1, the expression for the circuit is Y. Y is given by |
| A. | A + (B + C) D |
| B. | A + BC + D |
| C. | A(BC + D) |
| D. | none of these |
| Answer» D. none of these | |
| 270. |
For the binary number 11101000, the equivalent hexadecimal number is |
| A. | F 9 |
| B. | F 8 |
| C. | E 9 |
| D. | E 8 |
| Answer» E. | |
| 271. |
In Von-Neumann-or Princeton-type computers, the program |
| A. | can appear any where within the memory |
| B. | memory and data memory are clearly distinguished |
| C. | data and instructions are distinguished at the first stage |
| D. | none of the above |
| Answer» B. memory and data memory are clearly distinguished | |
| 272. |
An 8085 microprocessor based system uses a 4 k x 8 bit RAM whose address in AAOOH. The address of the last byte in this RAM is |
| A. | OFFFH |
| B. | 1000H |
| C. | B9FFH |
| D. | BA00H |
| Answer» E. | |
| 273. |
The hexadecimal number BBC7 is equivalent to decimal number |
| A. | 49761 |
| B. | 46791 |
| C. | 47691 |
| D. | 49761 |
| Answer» C. 47691 | |
| 274. |
When two 16-input multiplexers drive a 2-input MUX, what is the result? |
| A. | 2-input MUX |
| B. | 4-input MUX |
| C. | 16-input MUX |
| D. | 32 input MUX |
| Answer» E. | |
| 275. |
In 8085 microprocessor, what is the memory word addressing capability? |
| A. | 32 K |
| B. | 64 K |
| C. | 256 K |
| D. | 512 K |
| Answer» C. 256 K | |
| 276. |
A 4 bit parallel type A/D converter uses a 6 volt reference. How many comparators are required and what is the resolution in volts? |
| A. | 0.375 V |
| B. | 15 V |
| C. | 4.5 V |
| D. | 10 V |
| Answer» B. 15 V | |
| 277. |
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 not specific reason |
| Answer» B. it is slower than software | |
| 278. |
In the expression A + BC, the total number of minterms will be |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» E. | |
| 279. |
The number of bits in a nibble is |
| A. | 8 |
| B. | 4 |
| C. | 2 |
| D. | 16 |
| Answer» C. 2 | |
| 280. |
Some of MOS families are PMOS, CMOS, the family dominating the LSI field, where low power consumption is necessary is |
| A. | NMOS |
| B. | CMOS |
| C. | PMOS |
| D. | both NMOS, CMOS |
| Answer» C. PMOS | |
| 281. |
When all the seven segments of a display are energized, the number shown is |
| A. | 0 |
| B. | 1 |
| C. | 7 |
| D. | 8 |
| Answer» E. | |
| 282. |
Recommended fanout for ECL is |
| A. | 5 |
| B. | 10 |
| C. | 15 |
| D. | 25 |
| Answer» E. | |
| 283. |
The number of logic devices required in a Mod-16 synchronous counter are |
| A. | 4 |
| B. | 5 |
| C. | 6 |
| D. | 7 |
| Answer» D. 7 | |
| 284. |
For the following decimal multiplication the result are given in binary numbers, Spot the incorrect relation, if any |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» E. | |
| 285. |
A five bit binary adder is used to add 01001 and 00111. The adders from LSB position are numbered as FA₀, FA₁, FA₂ and FA₃. The outputs of FA₂ are |
| A. | SUM = 0 |
| B. | SUM = 0, CARRY = 1 |
| C. | SUM = 1, CARRY = 0 |
| D. | SUM = 1, CARRY = 1 |
| Answer» B. SUM = 0, CARRY = 1 | |
| 286. |
The advantages of flash memory over EEPROM are |
| A. | higher density |
| B. | lower cost |
| C. | both higher density and lower cost |
| D. | none of the above |
| Answer» D. none of the above | |
| 287. |
If all the LEDs in a seven segment display are turned on, the number displayed is |
| A. | 1 |
| B. | 3 |
| C. | 0 |
| D. | 8 |
| Answer» E. | |
| 288. |
Consider the following statements 1. ECL has least propagation delay2. TTL has largest fan out3. CMOS has highest noise margin4. TTL has lowest power dissipation Which of these are correct? |
| A. | 1 and 3 |
| B. | 2 and 4 |
| C. | 3 and 4 |
| D. | 1 and 2 |
| Answer» B. 2 and 4 | |
| 289. |
Assertion (A): In a parallel in-serial out shift register data is loaded one bit-at a time.Reason (R): A serial in-serial out shift register can be used to introduce a time delay. |
| 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» E. | |
| 290. |
If the mantissa of a digital computer is 37 bit long then the accuracy of the digital computer will be of |
| A. | 37 decimal places |
| B. | 23 decimal places |
| C. | 11 decimal places |
| D. | accuracy is independent of the length to mantissa |
| Answer» D. accuracy is independent of the length to mantissa | |
| 291. |
Which of them radiates emission? |
| A. | LED only |
| B. | LCD only |
| C. | Both LED and LCD |
| D. | Neither LED nor LCD |
| Answer» B. LCD only | |
| 292. |
If a RAM has 34 bits in its MAR and 16 bits its MAR, then its capacity will be |
| A. | 32 GB |
| B. | 16 GB |
| C. | 32 MB |
| D. | 16 MB |
| Answer» D. 16 MB | |
| 293. |
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 | |
| 294. |
Binary number 1101101101 is equal to decimal number |
| A. | 3289 |
| B. | 2289 |
| C. | 1289 |
| D. | 289 |
| Answer» B. 2289 | |
| 295. |
Decimal number 10 is equal to binary number |
| A. | 1110 |
| B. | 1010 |
| C. | 1001 |
| D. | 1000 |
| Answer» C. 1001 | |
| 296. |
Data from a satellite is received in serial form (1 bit after another). If this data is coming at a 5 MHz rate and if the clock frequency is 5 MHz how long will it take to serially load a word in a 32-bit shift register? |
| A. | 1.6ms |
| B. | 3.2ms |
| C. | 6.4ms |
| D. | 12.8ms |
| Answer» D. 12.8ms | |
| 297. |
Zero suppression is not used in actual practice. |
| A. | True |
| B. | False |
| C. | May be True or False |
| D. | Can't say |
| Answer» C. May be True or False | |
| 298. |
A mod 4 counter will count |
| A. | from 0 to 4 |
| B. | from 0 to 3 |
| C. | from any number n to n + 4 |
| D. | none of the above |
| Answer» C. from any number n to n + 4 | |
| 299. |
In a NOT gate the output is always the opposite of the input. |
| A. | True |
| B. | False |
| C. | May be True or False |
| D. | Can't say |
| Answer» B. False | |
| 300. |
The method used to transfer data from I/O units to memory by suspending the memory-CPU data transfer for one memory cycle is called |
| A. | I/O spooling |
| B. | Cycle stealing |
| C. | Line conditioning |
| D. | Demand paging |
| Answer» C. Line conditioning | |