NIMCET 2026 — Computer PYQ

The correct option is 1. 4,096 locations.

Explanation:

To find the maximum addressable memory space, we need to determine how many bits are allocated specifically for holding the memory address inside the instruction layout.

• Total Instruction Length: $16\text{ bits}$

• Bits allocated for Opcode: $4\text{ bits}$

The problem statement mentions that all the remaining bits are used to specify a single memory address field. Therefore, we can calculate the address field size as follows:

$$\text{Bits for Address Field}=\text{Total Instruction Length}-\text{Opcode Bits}$$

$$\text{Bits for Address Field}=16\text{ bits}-4\text{ bits}=12\text{ bits}$$

Calculating Addressable Space:

With $n$ bits allocated for an address field, a CPU can uniquely reference a maximum of $2^n$ distinct memory locations.

Substituting $n=12$ into the formula:

$$\text{Maximum Addressable Memory Space}=2^{12}\text{ locations}$$

We can break down $2^{12}$ algebraically for simpler calculation:

$$2^{12}=2^{10}\times 2^2$$

$$2^{12}=1024\times 4=4096\text{ locations}$$

Thus, the system can reference a maximum of 4,096 distinct memory locations.