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.
| 1751. |
Binary 101010 is equivalent to decimal number |
| A. | 24 |
| B. | 42 |
| C. | 44 |
| D. | 64 |
| Answer» C. 44 | |
| 1752. |
Consider the Boolean expression X = ABCD + AB̅CD + A̅BCD + A̅CB̅D The simplified form of X is |
| A. | C̅ + D̅ |
| B. | BC |
| C. | CD |
| D. | B̅ |
| Answer» D. B̅ | |
| 1753. |
In a RS flip-flop no change occurs during |
| A. | prohibited mode |
| B. | set mode |
| C. | reset mode |
| D. | disable mode |
| Answer» E. | |
| 1754. |
Which of the following subtraction operations do not result in F₁₆?(BA)₁₆ - (AB)₁₆ (BC)₁₆ - (CB)₁₆(CB)₁₆ - (BC)₁₆(CB)₁₆ - (BC)₁₆ Select the correct answer |
| A. | 1 and 2 |
| B. | 1 and 3 |
| C. | 2 and 3 |
| D. | 1, 2 and 3 |
| Answer» B. 1 and 3 | |
| 1755. |
ICs are |
| A. | analog |
| B. | digital |
| C. | both analog and digital |
| D. | mostly analog |
| Answer» D. mostly analog | |
| 1756. |
Logic pulser |
| A. | generates short duration pulses |
| B. | generates long duration pulses |
| C. | generates sinusoidal voltage |
| D. | generates ideal pulses |
| Answer» B. generates long duration pulses | |
| 1757. |
The decimal equivalent of the binary number 10110.0101011101 is |
| A. | 22.2408216500 |
| B. | 22.3408216750 |
| C. | 22.3408213125 |
| D. | 22.3408203125 |
| Answer» E. | |
| 1758. |
In the 2's complement Adder-Subtractor |
| A. | sign magnitude numbers represent negative numbers |
| B. | the sign magnitude numbers have "1" as leading bit |
| C. | the sign magnitude numbers have '0' as leading bit |
| D. | 2's complement represents positive numbers |
| Answer» D. 2's complement represents positive numbers | |
| 1759. |
In two level logic, logic race does not occur. |
| A. | True |
| B. | False |
| C. | May be True or False |
| D. | Can't say |
| Answer» B. False | |
| 1760. |
In the TTL circuit in the figure, S₂ to S₀ are select lines and X₇ to X₀ are input lines. S₀ and X₀ are LSBs. The output Y is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» C. C | |
| 1761. |
An XOR gate with 6 variables is as follows A ⊕ B ⊕ C ⊕ D ⊕ E ⊕ F. The number of minterms in the Boolean expression is |
| A. | 6 |
| B. | 12 |
| C. | 64 |
| D. | 32 |
| Answer» E. | |
| 1762. |
In 8085 microprocessor, what is the length of status Word? |
| A. | 6 bits |
| B. | 8 bits |
| C. | 12 bits |
| D. | 16 bits |
| Answer» E. | |
| 1763. |
(1111.11)₂ is |
| A. | (1.01)₁₀ |
| B. | (0.75)₁₀ |
| C. | (15.3)₁₀ |
| D. | (15.75)₁₀ |
| Answer» E. | |
| 1764. |
For the K map of the given figure the simplified Boolean expression is |
| A. | A̅ + BC |
| B. | A + BC |
| C. | A̅ + B C̅ |
| D. | A̅ + B C̅ |
| Answer» B. A + BC | |
| 1765. |
Assuming accumulator contain A 64 and the carry is set (1). What will register A and (CY) contain after CMA? |
| A. | 6 AH, 1 |
| B. | 6 AH, 0 |
| C. | 59 H, 0 |
| D. | 9 H, 0 |
| Answer» D. 9 H, 0 | |
| 1766. |
A crystal is frequently used in digital circuits for timing purpose because of its |
| A. | low cost |
| B. | high frequency stability |
| C. | simple circuitry |
| D. | ability to set frequency at desired value |
| Answer» C. simple circuitry | |
| 1767. |
An interrupt in which the external device supplies its address as well as the interrupt is known as |
| A. | vectored interrupt |
| B. | maskable interrupt |
| C. | polled interrupt |
| D. | non-maskable interrupt |
| Answer» B. maskable interrupt | |
| 1768. |
The inputs to logic gate are 0. The output is 1. The gate is |
| A. | NAND or XOR |
| B. | OR or XOR |
| C. | AND or XOR |
| D. | NOR or Ex-NOR |
| Answer» E. | |
| 1769. |
A function table is required in very large numbers the memory most switched for this purpose would be |
| A. | ROM |
| B. | PROM |
| C. | EPROM |
| D. | EAROM |
| Answer» B. PROM | |
| 1770. |
Race condition occurs in |
| A. | synchronous circuit |
| B. | asynchronous circuit |
| C. | combinational circuit |
| D. | all the Digital circuit |
| Answer» C. combinational circuit | |
| 1771. |
Precisely 1 K byte means |
| A. | 1000 bits |
| B. | 1012 bits |
| C. | 1020bits |
| D. | 1024 bits |
| Answer» E. | |
| 1772. |
Out of multiplexer and demultiplexer, which has select inputs? |
| A. | Multiplexer only |
| B. | Demultiplexer only |
| C. | Both multiplexer and demultiplexer |
| D. | None |
| Answer» D. None | |
| 1773. |
A counter has N flip flops. The total number of states are |
| A. | N |
| B. | 2 N |
| C. | 2ᴺ |
| D. | 4 N |
| Answer» D. 4 N | |
| 1774. |
The number of flip flops needed for a Mod 7 counter are |
| A. | 7 |
| B. | 5 |
| C. | 3 |
| D. | 1 |
| Answer» D. 1 | |
| 1775. |
The Boolean expression for the sub-class (Q) of all electronic instruments which are measuring instruments or are non-digital instruments with battery supply is |
| A. | Q = X(Y + Z̅) |
| B. | Q = X + Y̅Z |
| C. | Q = XY̅ + Z |
| D. | Q = XY + Z |
| Answer» C. Q = XY̅ + Z | |
| 1776. |
A counter type A/D converter contains a 4 bit binary ladder and a counter driven by a 2 MHz clock. Then conversion time |
| A. | 8 μ sec |
| B. | 10 μ sec |
| C. | 2 μ sec |
| D. | 5 μ sec |
| Answer» B. 10 μ sec | |
| 1777. |
Assertion (A): A demultiplexer cannot be used as a decoder Reason (R): A multiplexer selects one of many outputs whereas a decoder selects on output corresponding to coded 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» E. | |
| 1778. |
Which interrupts are masked? |
| A. | 5.5 |
| B. | 6.5 |
| C. | 7.5 |
| D. | None |
| Answer» C. 7.5 | |
| 1779. |
The counter in the given figure is |
| A. | Mod 3 |
| B. | Mod 6 |
| C. | Mod 8 |
| D. | Mod 7 |
| Answer» C. Mod 8 | |
| 1780. |
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 |
| A. | 0000 |
| B. | 0101 |
| C. | 1010 |
| D. | 1111 |
| Answer» D. 1111 | |
| 1781. |
For the minterm designation Y = ∑ m (1, 3, 5, 7) the complete expression is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» C. C | |
| 1782. |
For the K map in the given figure the simplified Boolean expression is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» B. B | |
| 1783. |
The Boolean expression for the circuit of the given figure |
| A. | A {F + (B + C) (D + E)} |
| B. | A [F + (B + C) (DE)] |
| C. | A + F + (B + C) (D + E)] |
| D. | A [F + (BC) (DE)] |
| Answer» B. A [F + (B + C) (DE)] | |
| 1784. |
For the K map of the given figure, the simplified Boolean expression is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» B. B | |
| 1785. |
The K-map for a Boolean function is shown in figure. The number of essential prime implicants for this function is |
| A. | 4 |
| B. | 5 |
| C. | 6 |
| D. | 8 |
| Answer» B. 5 | |
| 1786. |
The circuit of the given figure gives the output Y = |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» E. | |
| 1787. |
The circuit of the given figure is |
| A. | full adder |
| B. | full subtractor |
| C. | shift register |
| D. | decade counter |
| Answer» C. shift register | |
| 1788. |
In the given figure, A = B = 1 and C = D = 0. Then Y = |
| A. | 1 |
| B. | 0 |
| C. | either 1 or 0 |
| D. | indeterminate |
| Answer» C. either 1 or 0 | |
| 1789. |
The gates G1 and G2 in the figure have propagation delays of 10 n sec. and 20 n sec. respectively. If the input Vi makes an output change from logic 0 to 1 at time t = t₀ then the output wavefrom V₀ is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» E. | |
| 1790. |
In the circuit of the given figure, Y = |
| A. | 0 |
| B. | 1 |
| C. | X |
| D. | X̅ |
| Answer» B. 1 | |
| 1791. |
For the logic circuit of the given figure the simplified Boolean expression is |
| A. | A + BC |
| B. | A̅ + BC |
| C. | A̅ + B̅ C |
| D. | A + BC̅ |
| Answer» C. A̅ + B̅ C | |
| 1792. |
The number of product terms in the minimized sum of product expression obtained through the following K-map, where d, don't care. |
| A. | 2 |
| B. | 3 |
| C. | 4 |
| D. | 5 |
| Answer» B. 3 | |
| 1793. |
If A = B = 1, the outputs P and Q in the given figure are |
| A. | P = Q = 0 |
| B. | P = 0, Q = 1 |
| C. | P = 1, Q = 0 |
| D. | P = Q = 1 |
| Answer» C. P = 1, Q = 0 | |
| 1794. |
The minimized version of logic circuit in the given figure is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» B. B | |
| 1795. |
In the given figure, Y = |
| A. | (A + B)C + DE |
| B. | AB + C(D + E) |
| C. | (A + B)C + D + E |
| D. | none of the above |
| Answer» B. AB + C(D + E) | |
| 1796. |
In the given figure shows a negative logic AND gate. If positive logic is used this gate is equivalent to |
| A. | AND gate |
| B. | OR gate |
| C. | NOR gate |
| D. | NAND gate |
| Answer» D. NAND gate | |
| 1797. |
For the logic circuit of the given figure the simplified Boolean equation |
| A. | Y = (A + B) (C + D) (E + F) |
| B. | Y = A = B + C + D + E + F |
| C. | ABCDEF |
| D. | ABC + DEF |
| Answer» B. Y = A = B + C + D + E + F | |
| 1798. |
The Boolean function/implemented in the figure using two I/P multiplexers is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» B. B | |
| 1799. |
The modulus of counter in the given figure is |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| 1800. |
In the given figure, the flip flop is |
| A. | negative edge triggered |
| B. | positive edge triggered |
| C. | level triggered |
| D. | either (a) or (c) |
| Answer» B. positive edge triggered | |