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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.
| 201. |
The first machine cycle of an instruction is always |
| Answer» Fetch cycle is always first machine cycle. | |
| 202. |
A counter has N flip flops. The total number of states are |
| Answer» One flip-flop means 2 states and N flip-flops means 2N states. | |
| 203. |
Out of S, R, J, K, Preset, Clear inputs to flip flops, the synchronous inputs are |
| Answer» Preset and clear inputs are not applied in any fixed sequence. | |
| 204. |
In a mod-12 counter the input clock frequency is 10 kHz. The output frequency is |
| Answer» Mod-12 counter is divide by 12 counter. Output frequency = = 0.833 kHz. | |
| 205. |
Inputs A and B of the given figure are applied to a NAND gate. The output is LOW |
| Answer» NAND gate gives Low output if all inputs are High. For other combinations of inputs, output is High. | |
| 206. |
AECF1 + 15ACD = __________ . |
| Answer» Convert to decimal, add and change the result to hexadecimal. | |
| 207. |
A XOR gate has inputs A and B and output Y. Then the output equation is |
| Answer» XOR gate recognises odd number of 1's. | |
| 208. |
Wired AND connection can be used in TTL with totem pole output. |
| Answer» F, No it cannot be used. | |
| 209. |
A X 2 __________ . |
| Answer» A16 = 10, 216 = 2, 10 x 2 = 20 in decimal = 14 in hexadecimal. | |
| 210. |
For the binary number 11101000, the equivalent hexadecimal number is |
| Answer» 11 01000 = 128 + 64 + 32 + 8 = 232 in decimal = E 8 in hexadecimal. | |
| 211. |
Boolean expression for the output of XNOR (Equivalent) logic gate with inputs A and B is |
| Answer» (A + B) (A + B) = A ⊕ B. | |
| 212. |
An SR flip flop can be built using NOR gates or NAND gates. |
| Answer» | |
| 213. |
In the given figure R = R = 1 kΩ, then V = |
| Answer» Transistor is off. . | |
| 214. |
A 14 pin NOT gate 1C has __________ NOT gates. |
| Answer» 6 Input pins, 6 output pins, 1 supply pin and 1 ground pin. Hence 6 NOT gates. | |
| 215. |
F's complement of (2BFD) is |
| Answer» | |
| 216. |
A Karnaugh map with 4 variables has |
| Answer» 24= 16. | |
| 217. |
An 8 bit data is to be entered into a parallel in register. The number of clock pulses required is |
| Answer» In a parallel in register only one pulse is needed to enter data. | |
| 218. |
The total number of input words for 4 input OR gate is |
| Answer» 24 = 16. | |
| 219. |
In digital circuits Schottky transistors are preferred over normal transistors because of their |
| Answer» Schottky transistors have low switching time and hence low propagation delay. | |
| 220. |
A flip flop is a |
| Answer» It is a memory element used in digital circuits. | |
| 221. |
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? |
| Answer» No. of comparators = 2n - 1 = 24 - 1 = 15 . | |
| 222. |
Quantization error occurs in |
| Answer» Analog input can have any value but digital value can have only 2N discrete levels (for N bits). Hence quantization error in A/D conversion. | |
| 223. |
A universal shift register can shift |
| Answer» Both left to right and right to left operations are possible in universal shift register. | |
| 224. |
Available multiplexer IC package can have a maximum of 8 inputs. |
| Answer» 16 : 1 1C multiplexer is also available. | |
| 225. |
For the logic circuit of the given figure the simplified Boolean equation |
| Answer» = (A + B) (C + D) (E + F). | |
| 226. |
A combination circuit is one in which the output depends on |
| Answer» Combinational circuit does not have memory. | |
| 227. |
71 = __________ . |
| Answer» 71 in octal = 7 x 8 + 1= 57 in decimal = 111001 in binary. | |
| 228. |
1011 x 101 = __________ |
| Answer» 1011 = 11 in decimal and 101 = 5 in decimal, 11 X 5 = 55. | |
| 229. |
A mod 4 counter will count |
| Answer» Mod 4 counter has 4 states, 0 to 3. | |
| 230. |
In the given figure shows a 4 bit serial in parallel out right shift register. The initial contents as shown are 0110. After 3 clock pulses the contents will be |
| Answer» Output of XOR gate is input to register. | |
| 231. |
It is desired to display the digit 7 using a seven segment display. The LEDs to be turned on are |
| Answer» | |
| 232. |
For a MOD-12 counter, the FF has a = 60 ns The NAND gate has a of 25 n sec. The clock frequency is |
| Answer» For a proper working, the clock period should be equal to or greater than tpd = Mod 12 - 4FFs = 4 x 60 = 240 nsec. Total tpd = 240 + 25 = 265 nsec. = fc and fc = 3.774 MHz. | |
| 233. |
100101 is equal to decimal number |
| Answer» 32 + 4 + 1 = 37 in decimal. | |
| 234. |
In the figure, the LED |
| Answer» To emit the light, it is necessary NAND gate output is zero, for NAND O/P zero. Both the I/O must be height. And it is not possible in any case. | |
| 235. |
A 4 bit ripple counter uses flip flops with propagation delay of 50 ns each. The maximum clock frequency which can be used is |
| Answer» Time delay = 50 x 4 x 10-9s, . | |
| 236. |
An AND gate has two inputs A and B and one inhibit input S. Out of total 8 input states, output is 1 in |
| Answer» Only one input, i.e., A = 1, B = 1 and S = 0 gives output 1. | |
| 237. |
268 = __________ . |
| Answer» 10C in hexadecimal = 1 x 162 + 12 = 268 in decimal. | |
| 238. |
In a JK Master slave flip flop |
| Answer» | |
| 239. |
In a 7 segment display the segments are lit. The decimal number displayed will be |
| Answer» | |
| 240. |
The minimum number of comparators required to build an 8 bit flash ADC is |
| Answer» 2n - 1 = 28 - 1 = 255. | |
| 241. |
For a Mod-64 synchronous counter the number of flip flops and AND gates needed is |
| Answer» 26 = 64 Hence 6 flip-flops. The number of AND gates is 6 - 2 = 4 see in the given figure | |
| 242. |
A 4 bit DAC gives an output of 4.5 V for input of 1001. If input is 0110, the output is |
| Answer» | |
| 243. |
In 2's complement form, - 2 is |
| Answer» A = 1110, A = 0001, A + 1 = 0001 + 1 = 0010 = 2 Therefore A = -2. | |
| 244. |
The Boolean function/implemented in the figure using two I/P multiplexers is |
| Answer» when B = 1, f1 = ABC B = 0, f2 = ABC f = f1 + f2 = ABC + ABC. | |
| 245. |
Am equivalent 2's complement representation of the 2's complement number 1101 is |
| Answer» 2's complement of (1101)2 = 0011 | |
| 246. |
The number of digits in hexadecimal system |
| Answer» It has 16 digits 0 to 15. | |
| 247. |
Which of these is the most recent display device? |
| Answer» VF display can operate at very low voltages, has low power consumption and very long life. | |
| 248. |
A NOR gate has 3 inputs A, B, C. For which combination of inputs is output HIGH |
| Answer» NOR gate gives High output when all inputs are Low. | |
| 249. |
9's complement of 56 is |
| Answer» 99 - 56 = 43. | |
| 250. |
4 bit 2's complement representation of a decimal number is 1000. The number is |
| Answer» (a) and (d) both are option, But there is meaning to represent a positive number in 2's complement form, we take complement representation for negative number only. Therefore most appropriate number is "-8". | |