How many address bits are required to uniquely address a 4K ( $4096$ ) memory space?

Multiple Choice Options

A
5
B
9
C
12
✓ Correct Option
D
10

Detailed Technical Solution & Explanation

In computer architecture, the number of address lines $N$ needed to address $M$ memory locations is given by $2^N = M$ . For $4\text{K} = 4 \times 1024 = 4096$ bytes:

2^N = 4096 = 2^{12} \implies N = 12 \text{ bits}

Option Analysis:

•

Option A (5): Addresses

2^5 = 32

bytes.

•

Option B (9): Addresses

2^9 = 512

bytes.

•

Option C (12): Correct.

2^{12} = 4096 = 4\text{K}

•

Option D (10): Addresses

2^{10} = 1024 = 1\text{K}

Reference: Mock Test 03NEC Licensure Syllabus

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ACtE08 · Chapter 8Mock Test 03🏛️ Past Licensure QuestionID: b6ef1a46

How many address bits are required to uniquely address a 4K ( $4096$ ) memory space?

Multiple Choice Options

A
5
B
9
C
12
✓ Correct Option
D
10

Detailed Technical Solution & Explanation

In computer architecture, the number of address lines $N$ needed to address $M$ memory locations is given by $2^N = M$ . For $4\text{K} = 4 \times 1024 = 4096$ bytes:

2^N = 4096 = 2^{12} \implies N = 12 \text{ bits}

Option Analysis:

•

Option A (5): Addresses

2^5 = 32

bytes.

•

Option B (9): Addresses

2^9 = 512

bytes.

•

Option C (12): Correct.

2^{12} = 4096 = 4\text{K}

•

Option D (10): Addresses

2^{10} = 1024 = 1\text{K}

Reference: Mock Test 03NEC Licensure Syllabus

Want to practice this topic interactively?

Test your knowledge with timed mock exams, instant answer evaluation, and score tracking.

Practice Chapter 8 Now →