CUET PG 2024 — Computer PYQ

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Question

CUET PG 2024 — Computer PYQ

CUET PG | Computer | 2024

Consider a system with 1K pages and 512 frames and each page is of size 2KB. How many bits are required to represent the virtual address space memory.

Choose the correct answer:

Explanation

Data Given:

Number of Pages: $1\text{K} = 2^{10}$
Page Size: $2\text{KB} = 2 \times 2^{10}\text{ bytes} = 2^{11}\text{ bytes}$
Number of Frames: $512$ (Ye Physical Address ke liye use hota hai, Virtual Address ke liye nahi).

Equation Solution:

Virtual Address do hisson se banta hai: Page Number aur Page Offset.

1. Bits for Page Number ( $p$ ):

\text{Number of Pages} = 2^{p}

2^{10} = 2^p \implies p = 10\text{ bits}

2. Bits for Page Offset ( $d$ ):

\text{Page Size} = 2^d

2^{11} = 2^d \implies d = 11\text{ bits}

3. Total Virtual Address Bits:

\text{Virtual Address Space} = p + d

\text{Virtual Address Space} = 10 + 11

\mathbf{\text{Virtual Address Space} = 21\text{ bits}}

Answer: Virtual address space ko represent karne ke liye 21 bits ki zaroorat hogi.