UGC NET CSE | June 2019 | Part 2 | Question: 49

Question

A processor can support a maximum memory of $4$ GB where memory is word addressable and a word is $2$ bytes. What will be the size of the address bus of the processor?

• A. At least $28$ bits
• B. At least $2$ bytes
• C. At least $31$ bits
• D. Minimum $4$ bytes

Answer 1

Memory Size = 4GB

Word Size = 2B

So,Unique addresses = $ 2^(35) / 2^4 $

                                  = $2^(31)$

Hence (3) At least 31 bits are required.

Comments

4gb = 2^10 bits. so how do we get 2^35?

Answer 2

total addresses--->mm size/word size                                                                                                                  number of bits needed----> log(mm size/word size )    in base 2                                                                      ------->log(4GB/2B)--->31                                                                                                                                       so option 3 is correct