NTA UGC NET JUNE 2023 Paper-1
February 2, 2024Question 15328 – DSSSB PGT 2021
February 3, 2024Bit-Rate
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Question 2
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Suppose we want to download text documents at the rate of 100 pages per second. Assume that a page consists of an average of 24 lines with 80 characters in each line. What is the required bit rate of the channel?
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192 kbps
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512 kbps
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1.248 Mbps
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1.536 Mbps
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Question 2 Explanation:
Given data,
— Number of Pages(P)=100 per second
— Number of Lines(L)= 24
— Number of characters(C) =80
— One character(O) =8 bits
— Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps
— Number of Pages(P)=100 per second
— Number of Lines(L)= 24
— Number of characters(C) =80
— One character(O) =8 bits
— Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps
Correct Answer: D
Question 2 Explanation:
Given data,
— Number of Pages(P)=100 per second
— Number of Lines(L)= 24
— Number of characters(C) =80
— One character(O) =8 bits
— Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps
— Number of Pages(P)=100 per second
— Number of Lines(L)= 24
— Number of characters(C) =80
— One character(O) =8 bits
— Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps