InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
In a D latch |
| Answer» | |
| 252. |
The number of bits in ASCII is |
| Answer» ASCII is a 7 bit code. | |
| 253. |
The resolution of an bit DAC with a maximum input of 5 V is 5 mV. The value of is |
| Answer» 1000 = 5 or N = 10. | |
| 254. |
2's complement of binary number 0101 is |
| Answer» 1's complement of 0101 is 1010 and 2's complement is 1010+1 = 1011. | |
| 255. |
An OR gate has 4 inputs. One input is high and the other three are low. The output |
| Answer» In OR any input high means high output. | |
| 256. |
Decimal number 10 is equal to binary number |
| Answer» 1010 = 8 + 2 = 10 in decimal. | |
| 257. |
Both OR and AND gates can have only two inputs. |
| Answer» Any number of inputs are possible. | |
| 258. |
AB + A = |
| Answer» AB + AB = A(B + B) = A . 1 = A. | |
| 259. |
The number of distinct Boolean expression of 4 variables is |
| Answer» 22n = 224 = 216 . | |
| 260. |
For the gate in the given figure the output will be |
| Answer» If A = 0, Y = 1 and A = 1, Y = 0 Therefore Y = A. | |
| 261. |
In the expression A + BC, the total number of minterms will be |
| Answer» The min terms are ABC + ABC + AB C + ABC + ABC. | |
| 262. |
The circuit in the given figure is |
| Answer» Since V(I) is lower state than V(0) it is a negative logic circuit. Since diodes are in reversed parallel, it is an AND gate. Note: In Boolean algebra it is recognized that a positive logic OR is a negative logic AND. Similarly a positive logic AND is a negative logic OR. | |
| 263. |
The number of digits in octal system is |
| Answer» The octal system has 8 digits 0 to 7. | |
| 264. |
The access time of a word in 4 MB main memory is 100 ms. The access time of a word in a 32 kb data cache memory is 10 ns. The average data cache bit ratio is 0.95. The efficiency of memory access time is |
| Answer» Access time = 0.95 x 10 + 0.05 x 100. | |
| 265. |
The expression Y = M (0, 1, 3, 4) is |
| Answer» This is product of sums expression. | |
| 266. |
An 8 bit DAC has a full scale output of 2 mA and full scale error of ± 0.5%. If input is 10000000 the range of outputs is |
| Answer» 10000000 = 128, 11111111 = 255 If there is no error, output = = 1004μA. Maximum error = Hence range of output 994 to 1014 μA. | |
| 267. |
Decimal 43 in hexadecimal and BCD number system is respectively. |
| Answer» (43)10 = (2B)16 (43)10 = (01000011)2 . | |
| 268. |
The circuit of the given figure realizes the function |
| Answer» or Y = (A + B)C + DE. | |
| 269. |
An AND gate has two inputs A and B and one inhibit input 3, Output is 1 if |
| Answer» All AND inputs must be 1 and inhibit 0 for output to be 1. | |
| 270. |
The greatest negative number which can be stored is 8 bit computer using 2's complement arithmetic is |
| Answer» The largest negative number is 1000 0000 = -128. | |
| 271. |
A JK flip flop has = 12 ns. The largest modulus of a ripple counter using these flip flops and operating at 10 MHz is |
| Answer» Number of flip-flops = = 8.333 say 8 Modulus = 28 = 256. | |
| 272. |
The basic storage element in a digital system is |
| Answer» Storing can be done only in memory and flip-flop is a memory element. | |
| 273. |
In a ripple counter, |
| Answer» In a ripple counter the effect ripples through the counter. | |
| 274. |
A 12 bit ADC is used to convert analog voltage of 0 to 10 V into digital. The resolution is |
| Answer» . | |
| 275. |
For the truth table of the given figure Y = |
| Answer» Y = A B C + A B C + AB C + A B C = A B (C + C) + A B (C + C) = A B + AB = B(A + A) = B. | |
| 276. |
A full adder can be made out of |
| Answer» | |
| 277. |
If the functions are as follows. = R + Q + S , = PQ + PQ + P Then |
| Answer» Use k-map, then it will be easy. | |
| 278. |
The output of a half adder is |
| Answer» | |
| 279. |
Minimum number of 2-input NAND gates required to implement the function F = ( + ) (Z + W) is |
| Answer» F = (x + y) (z + w) = xy.(z + w) = xyz + xyw = minimum no. of 2 input NAND gate. | |
| 280. |
Which device has one input and many outputs? |
| Answer» Demultiplexer takes data from one line and directs it to any of its N output depending on the status of its select lines. | |
| 281. |
A carry look ahead adder is frequently used for addition because |
| Answer» In look ahead carry adder the carry is directly derived from the gates when original inputs are being added. Hence the addition is fast. This process requires more gates and is costly. | |
| 282. |
The counter in the given figure is |
| Answer» When counter is 110 the counter resets. Hence mod 6. | |
| 283. |
In register index addressing mode the effective address is given by |
| Answer» 4 = 22, in up scaling digit will be shifted by two bit in right direction. | |
| 284. |
7BF = __________ |
| Answer» 7BF16 = 7 x 162 + 11 x 161 + 15 x 160 = 1983 in decimal = 0111 1011 1111 in binary. | |
| 285. |
For the minterm designation Y = ∑ m (1, 3, 5, 7) the complete expression is |
| Answer» Decimal number 1 = binary number 001 = A BC Decimal number 7 = binary number 111= ABC, Decimal number 3 = binary number 011= ABC Decimal number 5 = binary number 101= ABC . Hence result. | |
| 286. |
Zero suppression is not used in actual practice. |
| Answer» Zero suppression is commonly used. | |
| 287. |
A counter type A/D converter contains a 4 bit binary ladder and a counter driven by a 2 MHz clock. Then conversion time |
| Answer» | |
| 288. |
The hexadecimal number (3E8) is equal to decimal number |
| Answer» 3 x 162 + 14 x 161 + 8 = 1000. | |
| 289. |
In a 7 segment display, LEDs and lit up. The decimal number displayed is |
| Answer» | |
| 290. |
The fixed count that should be used so that the output register will represent the input for a 6 bit dual slope A/D converter uses a reference of -6v and a 1 MHz clock. It uses a fixed count of 40 (101000). |
| Answer» If is made to be unity, the O/P count N2 = ei, N1 = 6 = 000110. | |
| 291. |
For the K map in the given figure the simplified Boolean expression is |
| Answer» | |
| 292. |
A memory system of size 16 k bytes is to be designed using memory chips which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is |
| Answer» . | |
| 293. |
In a BCD to 7 segment decoder the minimum and maximum number of outputs active at any time is |
| Answer» Minimum number of outputs when input is decimal 1 and maximum number of outputs when input is decimal 8. | |
| 294. |
A three state switch has three outputs. These are |
| Answer» Third state is floating. | |
| 295. |
Maxterm designation for A + B + C is |
| Answer» A + B + C = 000 = M0 . | |
| 296. |
1's complement of 11100110 is |
| Answer» Replace 1 by 0 and 0 by 1. | |