NTA UGC NET DEC-2022 Paper-2
October 8, 2023UPPCL AE 2022
October 8, 2023UGC NET CS 2013 Sep-paper-2
| Question 1 |
A file is downloaded in a home computer using a 56 kbps MODEM connected to an Internet Service Provider. If the download of file completes in 2 minutes, what is the maximum size of data downloaded ?
| 112 Mbits | |
| 6.72 Mbits | |
| 67.20 Mbits | |
| 672 Mbits |
Question 1 Explanation:
Given data,
— A file downloaded per second=56 kbps
— If the file completes in 2 min then maximum size=?
Step-1: Convert minutes into seconds
Given 2 min= 2*60 sec
= 120 sec
Step-2: Downloaded speed per second= 56 kbps
= 56 kbps * 120 sec
= (56 * 1024 *120) / 1024
= 6720 Kbps
= 6.72 Mbits
— A file downloaded per second=56 kbps
— If the file completes in 2 min then maximum size=?
Step-1: Convert minutes into seconds
Given 2 min= 2*60 sec
= 120 sec
Step-2: Downloaded speed per second= 56 kbps
= 56 kbps * 120 sec
= (56 * 1024 *120) / 1024
= 6720 Kbps
= 6.72 Mbits
Correct Answer: B
Question 1 Explanation:
Given data,
— A file downloaded per second=56 kbps
— If the file completes in 2 min then maximum size=?
Step-1: Convert minutes into seconds
Given 2 min= 2*60 sec
= 120 sec
Step-2: Downloaded speed per second= 56 kbps
= 56 kbps * 120 sec
= (56 * 1024 *120) / 1024
= 6720 Kbps
= 6.72 Mbits
— A file downloaded per second=56 kbps
— If the file completes in 2 min then maximum size=?
Step-1: Convert minutes into seconds
Given 2 min= 2*60 sec
= 120 sec
Step-2: Downloaded speed per second= 56 kbps
= 56 kbps * 120 sec
= (56 * 1024 *120) / 1024
= 6720 Kbps
= 6.72 Mbits