Question

The number of address lines required to address 8 GB memory is a) 8 b) 1024 c) 32 d) 33 . Please help

Answer 1

Answer: Option d) 33

$8GB = 8*2^{30}B = 2^{33}B$, hence $33$ address lines are required to address $8GB$ memory.

Comments

Why did you multiply by 2^30?

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@Jeetmoni saikia Oh this is due to:-

1K (kilo) = 2^10

1M (mega) = 2^20

1G (giga) = 2^30

1T (Tera) = 2^40

1P (Peta) = 2^50

1E (Exa) = 2^60

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Did you wrote 8gb as 2^3?

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$8 = 2^3$ and $GB = 2^{30} B$

Hence $8GB = 2^{30+3} = 2^{33}B$