Memory Address and Capacity MCQ Quiz in मल्याळम - Objective Question with Answer for Memory Address and Capacity - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct option is 3

Concept:

To determine the minimum number of bits needed to address 2000 memory locations, we can calculate the binary logarithm (log base 2) of 2000 and round it up to the nearest integer. This will give us the minimum number of bits required to represent 2000 unique memory locations.

Calculating the binary logarithm of 2000:

log₂(2000) ≈ 10.965784

Rounding up to the nearest integer:

⌈10.965784⌉ = 11

Therefore, the correct answer is 11.

• A barcode is a square or rectangular image consisting of a series of parallel black lines and white spaces of varying widths that can be read by a scanner

• Assembly language is mnemonic code

          Example: MOV A, B

• In digital electronics, a binary decoder is a combinational logic circuit that converts binary information from the n coded inputs to a maximum of 2n unique outputs

let \({\rm{A}} = {\rm{\;}}{{\rm{a}}_3}{{\rm{a}}_2}{{\rm{a}}_1}{{\rm{a}}_0}\)

\({\rm{B}} = {{\rm{b}}_3}{{\rm{b}}_2}{{\rm{b}}_1}{{\rm{b}}_0}\)

Then

∴ Number of input bits = 4 + 4 = 8

Number of output = 8 (7 + 1 extra bit for carry)

∴ Size of ROM = 2⁸ × 8

The Correct Answer is "Cache Memory".

Important Points

Cache Memory :

• Cache Memory is a special very high-speed memory.
• It is used to speed up and synchronizing with a high-speed CPU. Cache memory is costlier than main memory or disk memory but economical than CPU registers.
• Cache memory is an extremely fast memory type that acts as a buffer between RAM and the CPU.
• It holds frequently requested data and instructions so that they are immediately available to the CPU when needed.
• Cache memory is used to reduce the average time to access data from the Main memory.

Additional Information

Secondary Memory​ :

• It is non-volatile, i.e. it retains data when power is switched off.
• It is large capacities to the tune of terabytes.
• It is cheaper as compared to the primary memory.
• Depending on whether the Secondary memory device is part of the CPU or not, there are two types of secondary memory – fixed and removable.

 Auxiliary Memory :

• Auxiliary memory is the non-volatile memory lowest-cost, highest-capacity, and slowest-access storage in a computer system.
• It is where programs and data kept for long-term storage or when not in immediate use.
• Such memories tend to occur in two types-sequential access (data must access in a linear sequence) and direct access (data may access in any sequence).
• The most common sequential storage device is the hard disk drives, whereas direct-access devices include rotating drums, disks, CD-ROMs, and DVD-ROMs.
• It used as permanent storage of data in mainframes and supercomputers.

Virtual Memo :

• A computer can address more memory than the amount physically installed on the system.
• This extra memory is actually called virtual memory and it is a section of a hard disk that's set up to emulate the computer's RAM.
• The main visible advantage of this scheme is that programs can be larger than physical memory.
• Virtual memory serves two purposes.
  • First, it allows us to extend the use of physical memory by using the disk.
  • Second, it allows us to have memory protection because each virtual address is translated to a physical address.

A₁₃, A₁₄ and A15 are connected to NAND gate.

The output of NAND gate should be low to select the chip.

⇒ A₁₃ = A₁₄ = A₁₅ = 1

A₁₂ is not specified.

⇒ A₁₂ = Either 0 (or) 1.

If A₁₂ = 0,

\(\begin{array}{*{20}{c}} {\;\begin{array}{*{20}{c}} {}&{{{\rm{A}}_{15}}}&{{A_{14}}}&{{A_{13}}}&{{A_{12}}}&{{A_{11}}}&{{A_{10}}}&{ \ldots \ldots \ldots {A_0}}\\ {Starting\;address:}&1&1&1&0&0&0&{ \ldots \ldots \ldots 0}\\ {Ending\;adress:}&1&1&1&0&1&1&{ \ldots \ldots \ldots 1} \end{array}} \end{array}\) 

Address range = E000 H to EFFF H

The size of a memory register is given by \({\rm{M}} \times {\rm{N}}\) where \({\rm{M}}\) is the number of rows and \({\rm{N}}\) is the number of columns. Now, the given memory array is in the size of a square, i.e, the number of rows is equal to the number of columns. Thus, \({\rm{M}} = {\rm{N}}\). So, we have \({{\rm{M}}^2} = 16384\).

\(\Rightarrow {\rm{M}} = 128\)

Given address range: 1000 H to 2FFF

∴ Total No. of address lines = 13 i.e.  \(\begin{array}{*{20}{c}} {2{\rm{FFF}}}\\ {\underline {1000} }\\ {1{\rm{FFF}}} \end{array}\)

(1FFF)_H = (0001 1111 1111 1111)₂ = 13 address lines

i.e. \(\begin{array}{*{20}{c}} {{{\rm{A}}_{15}}}&{{{\rm{A}}_{14}}}&{{{\rm{A}}_{13}}}&{{{\rm{A}}_{12}}}&{{{\rm{A}}_{11}}}&{{{\rm{A}}_{10}}}& \cdots &{{{\rm{A}}_2}}&{{{\rm{A}}_1}}&{{{\rm{A}}_0}}&{}\\ 0&0&0&1&0&0&{}&0&0&0&{ = {{\left( {1000} \right)}_{\rm{H}}}}\\ {}&{}& \vdots &{}&{}& \vdots &{}&{}&{}&{}&{}\\ 0&0&1&0&1&1&{}&1&1&1&{ = {{\left( {2{\rm{FFF}}} \right)}_{\rm{H}}}} \end{array}\)

i.e. \(\begin{array}{*{20}{c}} {{{\rm{A}}_{15}}}&{{{\rm{A}}_{14}}}&{{{\rm{A}}_{13}}}&{{{\rm{A}}_{12}}}\\ 0&0&0&1\\ 0&0&1&0 \end{array}\)

To provide \(\overline {{\rm{CS}}}\) as low, the condition is

A₁₅ = A₁₄ = 0 and A₁₃ = A₁₂ = (01) or (10)

i.e. A15 = A14 = 0 and A13 , A12 shouldn't be (11) , (00)

Thus it is A15 + A14 + (A 13 . A12 + A̅13 . A̅ 12)

Hence option (1) is correct

So memory counters address 3 = 0100

∴ Decimal equivalent = 4

Explanation:

A memory chip has a capacity of 4 kB. To determine the number of address lines required, we need to understand the relationship between memory capacity and address lines. The capacity of a memory chip is the product of the number of addressable memory locations and the size of each memory location.

Firstly, 4 kB (kilobytes) needs to be converted into bytes. Since 1 kB = 1024 bytes, we have:

4 kB = 4 × 1024 bytes = 4096 bytes

Each memory location in a typical memory chip stores 1 byte of data. Therefore, the memory chip with a capacity of 4096 bytes has 4096 addressable memory locations.

The number of address lines required (n) can be determined using the formula:

Number of addressable memory locations = 2ⁿ

Setting this equal to the number of memory locations we have:

2ⁿ = 4096

To solve for n, we take the logarithm base 2 of both sides:

n = log₂(4096)

Since 4096 is a power of 2 (specifically, 2¹²), we have:

n = 12

Therefore, 12 address lines are required to address 4096 memory locations.

Hence, the correct answer is option 2, which is 12.

The correct answer is Programmable read-only memory

Key Points

• PROM stands for Programmable Read-Only Memory.
• Read-only memory (ROM) is a type of memory that is used in electronics and computers.
• It is a non-volatile memory, that is, information stored in such memory is permanent and cannot be changed.
• The other two types of ROM are
  • Electrically Erasable Programmable ROM (EEPROM)
  • Erasable Programmable ROM (EPROM)

Additional Information

• Primary memory can be divided into RAM (Random Access Memory) and ROM (Read Only Memory)