RRB JE Electrical solved question paper 2015 | RRB JE Electrical

Ques.61. UTF-32 is a/an ___

  1. 16-bit fixed-width encoding
  2. 16-bit variable-width encoding
  3. 32-bit variable-width encoding
  4. 32-bit fixed-width encoding

Answer.4. 32-bit fixed-width encoding

Explanation:-

There is a Unicode encoding that uses 4 bytes per character. It’s called UTF-32 because 32 bits = 4 bytes. UTF-32 is a straightforward and fixed-width encoding; it takes each Unicode character (a 4-byte number) and represents the character with that same number. This has some advantages, the most important being that you can find the Nth character of a string in constant time because the Nth character starts at the 4xNth byte.

 

 

Ques.62. The octal representation of (A0E)16 is _________

  1. (5016)8
  2. (5061)8
  3. (1650)8
  4. (5610)8

Answer.1. (5016)8

Explanation:-

The Hexa number system has a base of 16

Hexa to Octal conversion: Following procedure is adapted to convert a given octal integer into its equivalent decimal number:

Conversion of hexadecimal to octal is an indirect method like octal to hexadecimal number. That is, there is no direct way to convert from hexadecimal to octal. Thus, the hexadecimal number goes underway of either decimal or binary then the octal number.

Write the Hexa number, i.e., all its digits in a row. If n is the position of the digit from the right end then multiply each digit with 16n-1

(A0E)16 = 10 × 162 + 0 × 161 + 14 × 160

(A0E)16 = (2574)10

Divide the decimal number with 8

hexa to octal 1

Write the remainders in reverse order and the obtained number is the required result

(2574)10 = (5016)8

 

Ques.63. Considering X and Y as binary variables, the Boolean expression X(X + Y’) is equivalent to

  1. X
  2. 1
  3. 0
  4. Y

Answer.1. X

Explanation:-

According to the fundamental law of Boolean Algebra

XX = X  & XY’ = 0

Now in the given question

X(X + Y’) = XX + XY’

since XX = X & XY’ = 0

∴ X(X + Y’) = X 

The above expression i.e X(X + Y’) = X is also called Law of absorbtion in Boolean Algebra

 

Ques.64. The 2’s complement of the binary numbers (00001111)2

  1. (00001111)2
  2. (11110000)2
  3. (10101010)2
  4. (11110001)2

Answer.4. (11110001)2

Explanation:-

Two complements of Binary Number = One complement + 1

To find the One’s complements just inverse the digit i.e 1 ⇒ 0 & 0 ⇒ 1

00001111  ⇒ (11110000)1

Now adding 1

(11110000 + 1) = (11110001)2

 

Ques.65. Which of the following categories of viruses normally infect executable code, such as .com and .exe files?

  1. File infector viruses
  2. Boot sector viruses
  3. Master boot record viruses
  4. Macro viruses

Answer.1. File infector viruses

Explanation:-

File infector viruses: File infector viruses infect program files. These viruses normally infect executable code, such as .com and .exe files. They can infect other files when an infected program is run from a floppy, hard drive, or from the network Many of these viruses are memory residents. After memory becomes infected, any noninfected executable that runs becomes infected. Examples of known file infector viruses include Jerusalem and Cascade.

 

Ques.66. In class ‘C’ IP addresses, the number of network ID bits used to identify the class is _______

  1. 0
  2. 1
  3. 2
  4. 3

Answer.4. 3

Explanation:-

IP Classes

The number of bits assigned to the network ID and the host ID depends on the number of hosts required on a given network and the number of networks required in an environment. The first 3 bit of the address indicates a class A, B. or C IP address. After a while, you will only need to look at the first octet of all IP addresses in decimal format to determine their class. The class of an IP address governs the number of bits that can be used for network IDs and the number of bits that can be used for host IDs. 

Class A

Class A addresses are typically assigned to very large organizations, universities, and the military. These addresses are identified in binary by the first bit having a value of 0 or in decimal by having a value between 1 and 126. Class A addresses use the first 8 bits to specify the network ID and the last 24 bits to designate the host ID.

Class B

Class B addresses are typically assigned to medium and large organizations. These addresses are identified in binary by the first 2 bin having a value of 10, or in decimal by having a value between 128 and 191. Class B addresses use the first 16 bits to specify the network ID and the last 16 bits to designate the host IDs. Because the first 2 bits of all class B addresses are always 10, however, only 14 bits are available to be used for network IDs.

Class C

Class C addresses are typically assigned for small and medium organizations. These addresses are identified as m binary by the First 3 bits having a value of 110, or in decimal by having a value between 192 and 223. Class C addresses use die first 24 bits to specify the network ID and the last 8 bits to designate the host IDs. Because Me first 3 bits of a Class C address is 110, however, only 21 bits are available to be used for network IDs. 

 

Ques.67. Which of the following is a general-purpose network?

  1. Local Area Network (LAN)
  2. Storage Area Network (SAN)
  3. Enterprise Private Network (EPN)
  4. Virtual Private Network (VPN)

Answer.1. Local Area Network (LAN)

Explanation:-

Local Area Network

A local area network (LAN) is a group of computers and associated devices that share a common communication line and typically share the resources of a single processor or server within a small geographic area (for example, within an office building). Usually, the server has applications and data storage that are shared in common by multiple computer users. A local area network may serve as few as two or three users (for example, in a home network) or may as thousands of users.

Features of Local Area Network

  • Inexpensive transmission media;
  • Inexpensive devices(modems, repeaters, and transceivers) to interface to the media;
  • The easy physical connection of devices to the media;
  • High data transmission rates;
  • Network data transmission is independent of the rates used by the attached devices, making it easier for devices of one-speed information to devices of another speed.
  • A high degree of interconnection between devices;
  • Every attached device has the potential to communicate with every other device on the network;
  • There is seldom a central controlling processor which polls the attached devices on the network;
  • In the majority of cases, each attached device hears (but does not process) messages intended for other devices as well as for itself.

 

Ques.68. The term “plan” in orthographic projection refers to:

  1. Top view
  2. Front view
  3. Side view
  4. Sectioned view

Answer.1. Top view

Explanation:-

Architectural drawings are typically not labeled the same as engineering drawings. However, in reality, they are still created based on the same principles. In an architectural drawing, the top view is not referred to as a top view but instead is called a plan. In addition, architectural drawings do not refer to left and right side views as left and right side views; instead, they are referred to as elevations.

 

Ques.69. A thread profile having included 55° angle is

  1. V thread
  2. Acme thread
  3. Square thread
  4. Whitworth thread

Answer.4. Whitworth thread

Explanation:-

Whitworth thread: The most widely used thread form in British practice is the British Standard Whitworth thread (BSW).  In this category, the thread angle is 55° and the angular depth H of the fundamental triangle is equal to 0.96 times the pitch.

 

Ques.70. The common map of “road network” is drawn to scale RF:-

  1. Less than one
  2. Greater than one
  3. Equal to one
  4. RF is not relevant for such maps

Answer.1. Less than once

A road map, route map, or street map is a map that primarily displays roads and transport links rather than natural geographical information. The common map of “road network” is drawn to scale RF is Less than one.

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