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A 32 bit address bus allows access to a total memory of CPU
1. 64 MB
2. 1 GB
3. 4 GB
4. 6 GB
5. 2  GB

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Correct Answer - Option 3 : 4 GB

Concept:

1 GB = 1024 MB

1 MB = 1024 KB

1 KB = 1024 B

1 B = 8 bits

So, 32-bit address bus has access to 232  locations with 8 bits(1 Byte) of information.

232 locations = 22 * 210 * 210 * 210 locations

232 locations = 22 * 210 * 210 * 1024 bits

232 locations = 22 * 210 * 1024 Kbits

232 locations = 22  * 1024 Mbits

232 locations = 4 Gbits of location containing 8 bits(1 Byte) of information.

Therefore, 4 Gbits * 8 bits = 4 GB 

Correct Option is 3

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