Question

If page size of 4MB memory is 1 KB then the number of higher order bits of address bus used to denote page number is
(a) 10
(b) 11
(c) 12
(d) 9

Answer 1

Answer:

(b) 11

Explanation:

Answer 2

Option C : Address Bus of order 12 bits is used to denote the page number according to the given memory sizes.

Given :

Page Size = 1 KB

Memory Used = 4 MB

To Find :

The number of bits of address bus required

Solution :

Both KB i.e. Kilo Bytes ad MB i.e. MegaBytes are units of measurement of memory. The conversion between the two units KB and MB is :

                        $1MB = 2^{10} KB$

The page size of 1KB requires 4MB of memory. So, the memory required for a single bit of page size is :

           $= 4MB / 1KB$

           $= 4 * 2^{10} KB / 1 KB$

           $= (2^{2}) * 2^{10}$

           $=2^{2+10}$

           $=2^{12}$

Thus, the memory size required for a single unit of the page is $2^{12}$.

In general, "n" bits can be used to denote memory sizes up to $2^{n}$ i.e. n bits can represent numbers from 0 to $2^{n}-1$

• The minimum i.e. all n zero's represents 0
• The maximum i.e. all n ones represent $2^{n} -1$

Similarly, the number of bits used to denote a memory size N is given by :

              $n = log_{2}(N)$

                  $= log_{2}(2^{12} )$

                 $= 12 * log_{2}(2)$

                 $= 12 * 1$

                 $=12$

Hence, the Address bus of order 12 bits is used to denote the page number.

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