0 votes 0 votes CO and Architecture co-and-architecture digital-logic memory-interfacing + – Jeetmoni saikia asked Oct 11, 2022 • retagged Oct 11, 2022 by makhdoom ghaya Jeetmoni saikia 896 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Answer: Option d) 33 $8GB = 8*2^{30}B = 2^{33}B$, hence $33$ address lines are required to address $8GB$ memory. Abhrajyoti00 answered Oct 11, 2022 Abhrajyoti00 comment Share Follow See all 4 Comments See all 4 4 Comments reply Jeetmoni saikia commented Oct 11, 2022 reply Follow Share Why did you multiply by 2^30? 0 votes 0 votes Abhrajyoti00 commented Oct 11, 2022 i edited by Abhrajyoti00 Oct 11, 2022 reply Follow Share @Jeetmoni saikia Oh this is due to:-1K (kilo) = 2^101M (mega) = 2^201G (giga) = 2^301T (Tera) = 2^401P (Peta) = 2^501E (Exa) = 2^60 1 votes 1 votes Jeetmoni saikia commented Oct 11, 2022 i edited by Jeetmoni saikia Oct 11, 2022 reply Follow Share Did you wrote 8gb as 2^3? 0 votes 0 votes Abhrajyoti00 commented Oct 11, 2022 reply Follow Share $8 = 2^3$ and $GB = 2^{30} B$ Hence $8GB = 2^{30+3} = 2^{33}B$ 1 votes 1 votes Please log in or register to add a comment.
