Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > Data are recorded on a 2, 400 ft reel of magn...
Start Learning for Free
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor is
- a)0.85
- b)0.67
- c)0.19
- d)0.08
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of...
Tape Utilization Factor Calculation
The tape utilization factor is the ratio of the actual data recorded on the tape to the total available space on the tape. To calculate the tape utilization factor in this scenario, we need to consider the record length, record gap, and the density of characters per inch.
Given Values:
- Record Length: 100 characters
- Record Gap: 0.75 inch
- Density: 556 characters per inch
Step 1: Calculate the total length of each record
The total length of each record is the sum of the record length and the record gap.
Total Record Length = Record Length + Record Gap
Total Record Length = 100 + 0.75
Total Record Length = 100.75 inches
Step 2: Calculate the number of records that can fit on the tape
To calculate the number of records that can fit on the tape, we divide the total length of the tape by the total length of each record.
Number of Records = Total Tape Length / Total Record Length
Step 3: Calculate the total length of data recorded on the tape
To calculate the total length of data recorded on the tape, we multiply the number of records by the record length.
Total Data Length = Number of Records * Record Length
Step 4: Calculate the total available space on the tape
To calculate the total available space on the tape, we multiply the total length of the tape by the density of characters per inch.
Total Available Space = Total Tape Length * Density
Step 5: Calculate the tape utilization factor
The tape utilization factor is the ratio of the total length of data recorded on the tape to the total available space on the tape.
Tape Utilization Factor = Total Data Length / Total Available Space
Now, let's calculate the tape utilization factor using the given values.
Total Tape Length = 2,400 ft = 2,400 * 12 = 28,800 inches
Total Record Length = 100.75 inches
Number of Records = 28,800 / 100.75 ≈ 285.63
Total Data Length = 285.63 * 100 = 28,563 inches
Total Available Space = 28,800 * 556 = 15,964,800 characters
Tape Utilization Factor = 28,563 / 15,964,800 ≈ 0.00179
The tape utilization factor is approximately 0.00179, which is equivalent to 0.18%. Therefore, the correct answer is option C) 0.19.
The tape utilization factor is the ratio of the actual data recorded on the tape to the total available space on the tape. To calculate the tape utilization factor in this scenario, we need to consider the record length, record gap, and the density of characters per inch.
Given Values:
- Record Length: 100 characters
- Record Gap: 0.75 inch
- Density: 556 characters per inch
Step 1: Calculate the total length of each record
The total length of each record is the sum of the record length and the record gap.
Total Record Length = Record Length + Record Gap
Total Record Length = 100 + 0.75
Total Record Length = 100.75 inches
Step 2: Calculate the number of records that can fit on the tape
To calculate the number of records that can fit on the tape, we divide the total length of the tape by the total length of each record.
Number of Records = Total Tape Length / Total Record Length
Step 3: Calculate the total length of data recorded on the tape
To calculate the total length of data recorded on the tape, we multiply the number of records by the record length.
Total Data Length = Number of Records * Record Length
Step 4: Calculate the total available space on the tape
To calculate the total available space on the tape, we multiply the total length of the tape by the density of characters per inch.
Total Available Space = Total Tape Length * Density
Step 5: Calculate the tape utilization factor
The tape utilization factor is the ratio of the total length of data recorded on the tape to the total available space on the tape.
Tape Utilization Factor = Total Data Length / Total Available Space
Now, let's calculate the tape utilization factor using the given values.
Total Tape Length = 2,400 ft = 2,400 * 12 = 28,800 inches
Total Record Length = 100.75 inches
Number of Records = 28,800 / 100.75 ≈ 285.63
Total Data Length = 285.63 * 100 = 28,563 inches
Total Available Space = 28,800 * 556 = 15,964,800 characters
Tape Utilization Factor = 28,563 / 15,964,800 ≈ 0.00179
The tape utilization factor is approximately 0.00179, which is equivalent to 0.18%. Therefore, the correct answer is option C) 0.19.
Free Test
FREE
| Start Free Test |
Community Answer
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of...
Correct is C
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer?
Question Description
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer?.
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer?.
Solutions for Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor isa)0.85b)0.67c)0.19d)0.08Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup