Question
Download Solution PDFIn 8085 microprocessors, what is the output of following program?
LDA 8000H
MVI B, 30H
ADD B
STA 8001HAnswer (Detailed Solution Below)
Detailed Solution
Download Solution PDFLDA 8000H – transfers the data from memory location 8000H to Accumulator
MVI B, 30H – moves the value 30H to Register B
ADD B – Adds the value of Register B (30H) to the data in Accumulator which stores data from location 8000H. This instruction basically increments the value as 8000H by 30H.
STA 8001H – stores the content of Accumulator to memory location 8001H
Therefore, the above code reads a number from memory at location 8000H and stores in memory location 8001H after incrementing it by 30H.Last updated on Jul 7, 2025
-> The UGC NET Answer Key 2025 June was released on the official website ugcnet.nta.ac.in on 06th July 2025.
-> The UGC NET June 2025 exam will be conducted from 25th to 29th June 2025.
-> The UGC-NET exam takes place for 85 subjects, to determine the eligibility for 'Junior Research Fellowship’ and ‘Assistant Professor’ posts, as well as for PhD. admissions.
-> The exam is conducted bi-annually - in June and December cycles.
-> The exam comprises two papers - Paper I and Paper II. Paper I consists of 50 questions and Paper II consists of 100 questions.
-> The candidates who are preparing for the exam can check the UGC NET Previous Year Papers and UGC NET Test Series to boost their preparations.