Question 1

In microprocessor execution, do all instructions take the same number of machine cycles, or can different instructions require different numbers of machine cycles?

Accepted Answer

Introduction:Microprocessors execute instructions through a sequence of well-defined phases (fetch, decode, execute, memory/IO access, write-back). Different instructions stress these phases differently. This question checks whether you know that machine-cycle counts vary by instruction type.Given Data / Assumptions:Context: general-purpose microprocessors (e.g., 8085/8086-era and beyond).Options claim fixed one-cycle execution or even sub-cycle execution versus variable cycles.Machine cycle is a basic timing unit comprising several clock periods (T-states).Concept / Approach:Simple register-to-register operations may complete in fewer cycles than memory or I/O operations, which require additional bus transactions. Therefore, machine-cycle counts are instruction-dependent. No real ISA guarantees that every instruction finishes in exactly one machine cycle, and the notion of 'less than one machine cycle' is not meaningful because cycles are the smallest architected timing unit for bus/control sequencing.Step-by-Step Solution:Step 1: Identify instruction classes: register ALU ops, memory read/write, control transfer, I/O.Step 2: Recognize memory/I-O instructions add bus cycles for address and data transfers.Step 3: Conclude cycle counts differ across instructions; some use one cycle, others require multiple.Verification / Alternative check:Consult any instruction timing table (e.g., 8085 MOV vs. LDA/JMP): memory-referencing and branching typically take more cycles than register-only operations.Why Other Options Are Wrong:'all instructions need only one machine cycle' – contradicted by memory/I-O ops and branches.'all instructions need only a fraction of one machine cycle' – cycles are atomic timing blocks; cannot have 'fraction' in this context.'some instructions need less than one machine cycle' – not meaningful; cycles are indivisible architectural units.Common Pitfalls:Confusing 'clock period' with 'machine cycle' and assuming superscalar/pipelined overlap implies sub-cycle completion (it does not change per-instruction cycle accounting).Final Answer:some instructions need only one machine cycle while some other instructions need more than one machine cycle

Question 2

In Java naming conventions and rules, which of the following statements are correct: (1) the first letter of a variable is lower case, (2) each subsequent word in the name begins with a capital letter, (3) all other letters are lower case, (4) a Jav…

Accepted Answer

Introduction:This question distinguishes between Java’s style conventions (camelCase for variables) and the syntactic rules enforced by the compiler (identifier formation). Understanding both prevents compile-time errors and improves code readability.Given Data / Assumptions:(1) First letter lower case.(2) Each subsequent word begins with a capital letter (lower camelCase).(3) All other letters are lower case.(4) A variable can start with a digit.Concept / Approach:Java identifiers may contain letters, digits, underscores, and dollar signs, but must not start with a digit. Conventions recommend variables in lower camelCase: first word starts lower case; subsequent words are capitalized (e.g., totalAmount). While (3) is often true in practice, it is a stylistic simplification; camelCase typically keeps internal letters of non-leading words capitalized but can include digits or underscores when needed. Crucially, (4) contradicts Java’s rule that identifiers cannot begin with a digit.Step-by-Step Solution:Step 1: Evaluate (1): matches standard convention.Step 2: Evaluate (2): matches standard convention (camelCase).Step 3: Evaluate (3): stylistically true for simple names; the key point is that the bulk remains lower case aside from word-leading capitals.Step 4: Evaluate (4): false because starting with a digit is illegal.Verification / Alternative check:Compilation test: int 2count = 0; // errorint count2 = 0; // OKWhy Other Options Are Wrong:'All' – includes (4), which is invalid. '1, 2, 4 only' and '2, 3, 4 only' both include the invalid (4).Common Pitfalls:Confusing conventions with rules; assuming digits can appear at the start; over-restricting names to letters only (digits and _/ are permitted after the first character).Final Answer:1, 2, 3 only

Question 3

For static RAM (SRAM) devices used in digital systems, what supply voltage do most memory chips typically require?

Accepted Answer

Introduction:Choosing SRAMs requires awareness of typical power rails. Historically and in many legacy/embedded designs, SRAM ICs were specified for standard logic-level supplies. This item checks recognition of the common nominal supply voltage for mainstream SRAM chips (especially TTL/CMOS-compatible parts).Given Data / Assumptions:Technology: conventional asynchronous SRAMs as found in microprocessor-based systems.Answer choices: no supply, 2 V, 5 V, 12 V.Concept / Approach:For decades, the de facto supply for TTL-compatible logic and memory has been 5 V. While modern low-power SRAMs exist at 3.3 V or lower, the broad, classic answer remains 5 V for “most” standard SRAM chips from the traditional catalogs.Step-by-Step Solution:Step 1: Eliminate impossible choices: 'no supply' is impossible; digital ICs require power.Step 2: Compare common rails: 2 V is atypical; 12 V is used for older EPROM programming, not SRAM operation.Step 3: Select 5 V as the standard nominal rail for classic SRAM families.Verification / Alternative check:Review of representative SRAM datasheets (e.g., 6264, 62256 families) shows Vcc = 5 V ± tolerance.Why Other Options Are Wrong:2 V – below typical operating ranges; 12 V – used for programming certain memories, not SRAM supply; 'no supply' – impossible.Common Pitfalls:Generalizing modern low-voltage SRAMs to all SRAMs; conflating EPROM/Flash programming voltages with SRAM operation.Final Answer:need 5 V supply

Question 4

In DRAM devices, the approximate power requirements in active mode and in standby mode are respectively:

Accepted Answer

Introduction:Dynamic RAM (DRAM) consumes significantly more power during active access (row/column activation, refresh) than when placed in standby/idle modes. This question checks recognition of the order-of-magnitude difference between active and standby power.Given Data / Assumptions:Legacy/general DRAM power figures.Relative comparison across modes rather than a specific vendor part.Concept / Approach:Active DRAM cycles involve wordline and bitline charging/discharging plus sense amplifiers and refresh logic, leading to higher mW consumption. In standby, only retention/refresh and minimal biasing are maintained, drastically reducing power.Step-by-Step Solution:Step 1: Recall representative figures: hundreds of mW active vs. a few mW standby for classic parts.Step 2: Compare options; only one shows a large gap from hundreds of mW to single-digit mW.Step 3: Select '350 mW and 5 mW respectively' as the closest standard textbook pair.Verification / Alternative check:Vendor datasheets historically list active power in the several-hundred mW range and standby in the few-mW to tens-of-mW range depending on density and process.Why Other Options Are Wrong:'350 mW each' and '5 mW each' ignore mode-dependent differences; '350 mW and 100 mW' understates the standby reduction.Common Pitfalls:Assuming standby equals zero; overlooking self-refresh vs. active-refresh nuances, which still remain far below active power.Final Answer:350 mW and 5 mW respectively

Question 5

Regarding the Intel 8085 CPU, does it have six sign flags in its flag register?

Accepted Answer

Introduction:The 8085 microprocessor maintains status information in its flag register. This question checks familiarity with the specific flags present and dispels the misconception that multiple 'sign' flags exist.Given Data / Assumptions:Architecture: Intel 8085.Claim: '8085 has 6 sign flags.'Concept / Approach:The 8085 flag register contains five user-visible flags: Sign (S), Zero (Z), Auxiliary Carry (AC), Parity (P), and Carry (CY). The other bits are either fixed or unused. There is exactly one sign flag (S). Therefore, claiming 'six sign flags' is incorrect.Step-by-Step Solution:Step 1: List the flags: S, Z, AC, P, CY.Step 2: Recognize only one of these is a sign flag (S).Step 3: Conclude the statement is false.Verification / Alternative check:8085 diagrams show the 8-bit flag register with positions for S, Z, AC, P, CY and non-flag bits (some reserved or constant).Why Other Options Are Wrong:'True' implies multiple sign flags, which do not exist in 8085.Common Pitfalls:Confusing 'status flags' with 'sign flags'; misreading documentation tables and counting internal/reserved bits as distinct sign indicators.Final Answer:False

Question 6

For interfacing an Intel 8086 microprocessor to a 16-bit Analog-to-Digital Converter (ADC), which peripheral interface chip is most appropriate?

Accepted Answer

Introduction:Interfacing an 8086 with an external 16-bit ADC requires general-purpose parallel I/O lines to read conversion results and control the ADC (start/End-of-Conversion). This question tests knowledge of standard Intel 82xx peripherals.Given Data / Assumptions:CPU: 8086 (16-bit).Peripheral: 16-bit ADC (parallel data output assumed).Candidate devices: 8259 (PIC), 8255 (PPI), 8253 (timer), 8251 (USART).Concept / Approach:The 8255 Programmable Peripheral Interface provides three 8-bit ports configurable for input/output strobes. It is commonly used to interface digital sensors/ADCs/DACs by mapping data lines and control signals to its ports. By contrast, the 8259 handles interrupts, the 8253 provides timers/counters, and the 8251 is for serial communication – none of which directly provide the parallel I/O needed to read a 16-bit ADC bus (two ports can be combined for 16 bits).Step-by-Step Solution:Step 1: Identify the required resource: parallel data lines plus control strobes.Step 2: Map ADC D0–D15 to two 8-bit ports of 8255; use remaining lines for SOC/EOC.Step 3: Memory-map or I/O-map the 8255 to the 8086 and implement read sequence.Verification / Alternative check:Textbook interfacing examples pair 8255 with ADC/DAC devices for parallel data transfer.Why Other Options Are Wrong:8259 – interrupt controller; 8253 – timing only; 8251 – serial I/O; none directly supply the needed parallel ports.Common Pitfalls:Confusing the role of timers/interrupts with data-path I/O; overlooking the need to read two 8-bit halves for a 16-bit word.Final Answer:8255

Question 7

Which of the following is not a valid Java variable name (apply Java identifier rules, no spaces; may include letters, digits, underscore, or dollar sign; must not start with a digit)?

Accepted Answer

Introduction:Java variable names must follow strict lexical rules. This question tests whether you can distinguish valid identifiers from invalid ones, especially with digits and the dollar sign.Given Data / Assumptions:Identifiers cannot contain spaces.Allowed characters after the first: letters, digits, underscore, dollar sign.First character cannot be a digit.Concept / Approach:Check each name against the rules: start character validity, presence of illegal characters (space), and general composition. The names 'lot', 'c1t', and 's2' all start with a letter and contain only permitted characters afterward. The name '2s' starts with a digit, which violates the starting-character rule.Step-by-Step Solution:Step 1: Inspect starting character of each candidate.Step 2: Verify remaining characters belong to the allowed set.Step 3: Mark any starting with a digit as invalid.Verification / Alternative check:Compile-time test: int 2s = 0; // illegalint s2 = 0; // legalWhy Other Options Are Wrong:'lot' – letters only; valid.'c1t' – digit allowed after first character; valid.'s2' – dollar sign is allowed in identifiers (though discouraged by style guides); valid.Common Pitfalls:Thinking the dollar sign is forbidden (it is allowed, but not recommended); assuming digits are allowed at the start; including spaces inside identifiers.Final Answer:2s

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