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The semiconductor random access memory of a computer has 65,536 words, each of 8-bits. It can perform two basic operations Read and Write. How many bits are there in the Address Register of this memory?
1. 8
2. 12
3. 16
4. 24

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Correct Answer - Option 3 : 16

Concept:

For an n-bit address, the number of memory locations that could be addressed is:

N = 2n

∴ For the given number of locations (N), the number of address bits required will be:

n = log2 (N)

Calculation:

The total number of memory locations (N) = 65,536

N = 216

The number of bits in the address register will be:

n = log2 (216)

n = 16 × log2 (2)

n = 16 bits

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