MCQs on Digital Electronics

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01․ A read only memory in which the present contents must be erased before the new information can be stored is
ROM
PROM
EAROM
all of the above

The ROM programmed during manufacturing process itself is called MROM. Data is stored by the manufacturer during fabrication. User can not alter. In PROM data is stored by user. Only one time programmable and reprogramming is not possible. EPROM erasing is done using UV light and programming is through electrical.
02․ An EPROM
is of random-access type
is non volatile
is programmable
has all of the above requirements

EPROM is non volatile and random access type. IN EPROM erasing is done using UV light and programming is through electrical.
03․ A PROM
is mask programmed
is erasable by ultraviolet light
can be programmed only once
can be programmed any number of times

The ROM programmed during manufacturing process itself is called MROM. Data is stored by the manufacturer during fabrication. User can not alter. In PROM data is stored by user. Only one time programmable and reprogramming is not possible. EPROM erasing is done using UV light and programming is through electrical.
04․ The process of entering information in EPROM is commonly known as
writing
programming
storing
none of the above

The process of entering information in EPROM is commonly known as programming. In EPROM erasing is done using UV light and programming is through electrical.
05․ An EPROM is
non erasable
volatile
programmable and erasable
erasable but not programmable

The ROM programmed during manufacturing process itself is called MROM. Data is stored by the manufacturer during fabrication. User can not alter. In PROM data is stored by user. Only one time programmable and reprogramming is not possible. EPROM erasing is done using UV light and programming is through electrical. Therefore, programming and erasing is possible with EPROM.
06․ The field, which is never present in an assembly language statement, is
Opcode
Operand
Continue
Comment

There is nothing called “continue” in assembly language statement.
07․ The maximum positive and negative numbers which can be represented in two’s complement form using n bits are respectively,
+(2n-1-1),-(2n-1-1)
+(2n-1-1),-2n-1
+2n-1,-2n-1
+2n-1,-(2n-1+1)

In two’s complement form zero has only one representation i.e +0. So the range of the two’s representation of a n bit system will be +(2n-1-1) to -2n-1.
08․ When two n-bit binary numbers are added then the sum will contain at most
n-bit
(n + 1)-bit
(n + 2)-bit
(n + n)-bit

If two n-bit binary numbers are added then sum will contain (n + 1) bit at most. If summation results a carry then result will contain (n + 1) bit.
09․ The largest positive number that can be stored in a computer that has 16-bit word length and uses two’s complement arithmetic is
32
32,767
32,768
65,536

As we know in two’s complement representation of n-bit number,the largest positive number that can be represented is +(2n-1-1) In the above problem n = 16. So largest positive number is (216-1-1) = (215-1) = 32,767
10․ The maximum positive and negative numbers that can be represented in one’s complement using n-bits are respectively
+(2n-1-1),-(2n-1-1)
+(2n-1-1),-2n-1
+2n-1,-(2n-1-1)
None of these

In one’s complement method zero has two representation i.e. +0 & -0 so maximum positive and negative numbers will be one less. i.e. from +(2n-1-1)to -(2n-1-1).
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