Q2GATE 2021MCQ

The content of the registers are $R_{1} = 25H, R_{2} = 30H$ and $R_{3} = 40H$ . The following machine instructions are executed. $\text{PUSH } \{R_{1}\}$ $\text{PUSH } \{R_{2}\}$ $\text{PUSH } \{R_{3}\}$ $\text{POP } \{R_{1}\}$ $\text{POP } \{R_{2}\}$ $\text{POP } \{R_{3}\}$ After execution, the content of registers $R_{1},\:R_{2},\:R_{3}$ are

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📖 Explanation

For PUSH SP is decremented and for POP SP is incremented.

\begin{aligned} \therefore \qquad [\mathrm{R} 1]&=40 \\ [\mathrm{R} 2]&=30 \\ [\mathrm{R} 3]&=25 \end{aligned}

Q3GATE 2017MCQ

The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 $\mu s$ , then the number of T-states needed for executing the instruction is

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📖 Explanation

Given than, $f_{\mathrm{CLK}}=5 \mathrm{MHz}$ Execution time $=1.4 \mu \mathrm{s}$ Execution time $=n(T-\text { state })$ $n=$ number of T-states required to execute the instruction T- state (or) $T_{\mathrm{CLK}}=\frac{1}{f_{\mathrm{CLK}}}=0.2 \mu \mathrm{s}$ So, $\quad n=\frac{1.4 \mu \mathrm{s}}{T_{\mathrm{CLK}}}=\frac{1.4}{0.2}=7$

Q4GATE 2016MCQ

In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the following options is the correct reason for this?

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Q5GATE 2016MCQ

In an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0, respectively. If the instruction RLC is executed, then the contents of the accumulator (in hex) and the carry flag, respectively, will be

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📖 Explanation

RLC: Rotate accumulator left by 1 bit without carry Before RLC operation:

\begin{aligned} A &=A 7 H=(10100111)_{2} \\ C Y &=0 \end{aligned}

After RLC operation:

\begin{aligned} A &=(01001111)_{2}=4 \mathrm{FH} \\ \mathrm{CY} &=1 \end{aligned}

Q6GATE 2015MCQ

In an 8085 microprocessor, which one of the following instructions changes the content of the accumulator?

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Q7GATE 2011MCQ

An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is

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📖 Explanation

$MVI \;A,07\Rightarrow A=07H=(0000111)_{2}$

\begin{aligned} \text { After } \mathrm{RLC}, \mathrm{A}&=(00001110)_{2}=\mathrm{OEH} \\ \mathrm{MOV} \mathrm{B}, \mathrm{A} &\Rightarrow \mathrm{B} \leftarrow \mathrm{A} \Rightarrow \mathrm{B}=\mathrm{OEH} \\ \mathrm{RLC} \quad &\Rightarrow \mathrm{A}=(00011100)_{2}=1 \mathrm{CH} \\ \mathrm{RLC} \quad &\Rightarrow \mathrm{A}=(00111000)_{2}=38 \mathrm{H} \\ \mathrm{ADDB} &\Rightarrow \mathrm{A} \leftarrow \mathrm{A}+\mathrm{B} \\ &\Rightarrow \mathrm{A}=38 \mathrm{H}+\mathrm{OEH}=46 \mathrm{H} \\ &\qquad=(01000110)_{2} \\ \mathrm{RRC} \quad &\Rightarrow \mathrm{A}=(00100011)_{2}=23 \mathrm{H} \end{aligned}

Q8GATE 2010MCQ

For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

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📖 Explanation

$\mathrm{MVI} \quad \mathrm{A}, 45 \mathrm{H} \Rightarrow \mathrm{A}=45 \mathrm{H}=01000101$ $\mathrm{MOV} \mathrm{B}, \mathrm{A} \Rightarrow \mathrm{B}=45 \mathrm{H}$ $\mathrm{STC} \Rightarrow \mathrm{CY}=1$ $\mathrm{CMC} \Rightarrow \mathrm{CY}=0$ $RAR$

\begin{aligned} A &=00100010 \\ XR A B & \Rightarrow A \leftarrow(00100010) \oplus(01000101)\\ A&=01100111=67 \mathrm{H} \end{aligned}

Therefore, the content of the accumulator after the execution of the program is 67H.

Q9GATE 2008MCQ

An 8085 executes the following instructions 2710 LXI H, 30A0 H 2713 DAD H 2414 PCHL All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct ?

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📖 Explanation

\begin{aligned} \mathrm{LXIH}, 30 \mathrm{AOH} & \Rightarrow \mathrm{HL}=30 \mathrm{AOH} \\ \mathrm{DADH} & \Rightarrow \mathrm{HL}=6140 \mathrm{H}\\ &(i.e. 30 \mathrm{AOH}+3 \mathrm{OAOH})\\ \mathrm{PCH} & \Rightarrow \mathrm{PC}=6140 \mathrm{H}\\ \end{aligned}

Therefore, contents are $P C=6140 H$ $H L=6140 H$

Q10GATE 2005MCQ

Consider an 8085-microprocessor system. The following program starts at location 0100H. LXI SP, OOFF LXI H, 0701 MVI A, 20H SUB M If in addition following code exists from 019H onwards. ORI 40 H ADD M What will be the result in the accumulator after the last instruction is executed ?

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📖 Explanation

\begin{array}{l} 0109 H \text { ORI } 40 \mathrm{H} \\ 010 \mathrm{BH} \mathrm{ADD} \mathrm{M} \\ \text { Intial: } \mathrm{A}=00 \mathrm{H} \\ 0109 \mathrm{H}: \text { ORI } 40 \mathrm{H} \Rightarrow \mathrm{A} \leftarrow \mathrm{A}(\mathrm{OR}) 40 \mathrm{H}=40 \mathrm{H} \\ 010 \mathrm{BH}: \mathrm{ADDM} \Rightarrow \mathrm{A} \leftarrow \mathrm{A}+\mathrm{M}=40 \mathrm{H}+20 \mathrm{H}=60 \mathrm{H} \\ \therefore \quad \mathrm{A}=60 \mathrm{H} \end{array}

Q11GATE 2004MCQ

It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below : MVI A, 00H LOOP: ------ ------ ----- HLT END The sequence of instructions to complete the program would be

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Q12GATE 2004MCQ

Consider the sequence of 8085 instructions given below LXI H, 9258 MOV A, M CMA MOV M, A Which one of the following is performed by this sequence ?

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📖 Explanation

$\mathrm{LXIH}, 9258 \mathrm{H} \Rightarrow \mathrm{HL} \leftarrow 9258 \mathrm{H}$ MOV A, M $\Rightarrow A \leftarrow$ contents of add. 9258 H $\mathrm{CMA} \Rightarrow$ Complement Accumulator MOV MA $\Rightarrow$ Complement of $\mathrm{A}$ is stored in Memory (9258H)

Q13GATE 2004MCQ

The number of memory cycles required to execute the following 8085 instructions (i) LDA 3000 H (ii) LXI D, FOF1H would be

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📖 Explanation

Memory cycles $LDA 3000 H \rightarrow \text{ Fetch, } \frac{\text { Re ad, Read, }}{\text { address }} \frac{\text { Read }}{\text { data }}$ $\mathrm{LXID}, \mathrm{FOF} 1 \mathrm{H} \rightarrow \text{Fetch, }\frac{\text { Read, Read }}{\text { data }}$

Q14GATE 2003MCQ

In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of register B. As a result

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📖 Explanation

CMP B $\rightarrow$ contents of B and A are compared and result is Indicated by flag. Given that, $A \lt B$ $\therefore CY= 1, Z = 0$

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