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in CO and Architecture retagged by
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Answer: Option d) 33

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

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Why did you multiply by 2^30?
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edited by

@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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edited by
Did you wrote 8gb as 2^3?
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$8 = 2^3$ and $GB = 2^{30} B$

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