1
GATE ECE 2021
MCQ (Single Correct Answer)
+2
-0.67

The content of the registers are $R_1=25 \mathrm{H}, R_2=30 \mathrm{H}$ and $R_3 =40 \mathrm{H}$. The following machine instructions are executed.

$$ \begin{aligned} & \operatorname{PUSH}\left\{R_1\right\} \\ & \operatorname{PUSH}\left\{R_2\right\} \\ & \operatorname{PUSH}\left\{R_3\right\} \\ & \operatorname{POP}\left\{R_1\right\} \\ & \operatorname{POP}\left\{R_2\right\} \\ & \operatorname{POP}\left\{R_3\right\} \end{aligned} $$

After execution, the content of registers $R_1, R_2, R_3$ are

A

$R_1=40 \mathrm{H}, R_2=30 \mathrm{H}, R_3=25 \mathrm{H}$

B

$R_1=40 \mathrm{H}, R_2=25 \mathrm{H}, R_3=30 \mathrm{H}$

C

$R_1=25 \mathrm{H}, R_2=30 \mathrm{H}, R_3=40 \mathrm{H}$

D

$R_1=30 \mathrm{H}, R_2=40 \mathrm{H}, R_3=25 \mathrm{H}$

2
GATE ECE 2017 Set 1
MCQ (Single Correct Answer)
+2
-0.6
The following five instructions were executed on an 8085 microprocessor.
MVI A, 33H
MVI B, 78H
ADD B
CMA
ANI 32H

The Accumulator value immediately after the execution of the fifth instruction is

A
00H
B
10H
C
11H
D
32H
3
GATE ECE 2016 Set 1
MCQ (Single Correct Answer)
+2
-0.6
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?
A
For POP, the data transceivers remain in the same direction as for instruction fetch (memory to processor), whereas for PUSH their direction has to be reversed.
B
Memory write operations are slower than memory read operations in an 8085 based system.
C
The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer.
D
Order of registers has to be interchanged for a PUSH operation, whereas POP uses their natural order.
4
GATE ECE 2014 Set 4
MCQ (Single Correct Answer)
+2
-0.6
An 8085 microprocessor executes “STA 1234H” with starting address location 1FFEH (STA copies the contents of the Accumulator to the 16-bit address location). While the instruction is fetched and executed, the sequence of values written at the address pins $${A_{15}}$$ − $${A_{8}}$$ is
A
1FH, 1FH, 20H, 12H
B
1FH, FEH, 1FH, FFH, 12H
C
1FH, 1FH, 12H, 12H
D
1FH, 1FH, 12H, 20H, 12H

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