# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2014-16 - UGC NET Computer Science Paper 3 July 2016

>>>>>>>>UGC NET Computer Science Paper 3 July 2016

• Option : D
• Explanation :
• Frequency Division Multiplexing is a technique that can be applied when the bandwidth of a link is greater than combined bandwidth of signals to be transmitted.Correct
• Wavelength Division Multiplexing (WDM) is an analog multiplexing Technique to combine optical signals.Correct
• WDM is a Digital Multiplexing Technique IncorrectIt is wavelength-division multiplexing
• TDM is a Digital Multiplexing Technique.Correct
• So, option (D) is correct.

• Option : B
• Explanation : Throughput(S) in pure aloha is G * e-2G: It is maximum when G = 1 / 2. i.e. S = 1 / 2 * (2.71)-2 * 1 / 2 =1 / 2 * (2.71)-1 = 0.184. So, option (B) is correct.

 (a) Line coding (i) A technique to change analog signal to digital data. (b) Block coding (ii) Provides synchronization without increasing number of bits. (c) Scrambling (iii) Process of converting digital data to digital signal. (d) Pulse code modulation (iv) Provides redundancy to ensure synchronization and inherits error detection.

Codes:

 (a) (b) (c) (d) (1) (iv) (iii) (ii) (i) (2) (iii) (iv) (ii) (i) (3) (i) (iii) (ii) (iv) (4) (ii) (i) (iv) (iii)

• Option : B
• Explanation :
• Line coding is a Process of converting digital data to digital signal.
• Block coding provides redundancy to ensure synchronization and inherits error detection.
• Scrambling provides synchronization without increasing number of bits.
• Pulse code modulation is a technique to change analog signal to digital data.
So, option (B) is correct.

• Option : B
• Explanation :
100 pages , each page contain 24 line, each line have 80 character and each character weighted 8 bit, we have to download 100 pages per minute. i.e. 100 * 24 * 80 * 8 / 60 = 25.6 kbps. So, option (D) is correct.

 3 5 2 1 4 (Cipher text) 1 2 3 4 5 (Plain text)

• Option : A
• Explanation :
According to given question and key , Z is a bogus character Encrypted message will be:Plain text E X T R A N E T Z Z 1 2 3 4 5 1 2 3 4 5

Cipher text
 T A X E R T Z E N Z 1 2 3 4 5 1 2 3 4 5

So, option (A) is correct.