Free UGC NET Paper 2 CS Mock Test 1 – Data Communication & Physical Layer (Part 1) (Unit 9)

0 / 16 answered 00:00

Q1. What is the maximum bit rate of a noiseless channel with a bandwidth of 2 KHz, transmitting a signal with 8 levels?

A12000 bps B16000 bps C4000 bps D32000 bps

Q2. The one way propagation delay for a single bit of information (travelling at the speed of light i.e. 3×10⁸ meters per second) over a fiber optic cable from point A to point B, 12000 kilometers apart would be :

A20 ms B5 ms C50 ms D40 ms

Q3. The problem(s) in piggybacking acknowledgement frames is :
A. There is no need to send separate acknowledgement frames
B. It requires more bandwidth
C. To ascertain, how long the data link layer must wait for a packet onto which the acknowledgement is attached
D. Numbers of acknowledgement frames decrease
Choose the correct answer from the options given below :

AA and D only BB only CB and C only DC only

Q4. In a Stop-and-wait system, the bandwidth of the line is 1 Mbps, and 1 bit takes 30 milliseconds to make a round trip. The bandwidth-delay product is

A15,000 bits B20,000 bits C30,000 bits D60,000 bits

Q5. Which of the following is NOT a characteristic of packet switching?

AData is broken into packets BEach packet may take a different route CRequires a dedicated path DSupports dynamic routing

Q6. Which of the following are examples of data encoding schemes?
A. Non-Return to Zero (NRZ)
B. Manchester Encoding
C. Amplitude Modulation
D. Hamming code
E. Bipolar AMI
Choose the correct answer from the options given below:

AA, C and D Only BA, B and D Only CC, D and E Only DA, B and E Only

Q7. Consider a noiseless channel with a bandwidth of 5000 Hz transmitting a signal with two signal levels. The maximum bit rate is

A2500 bps B10000 bps C5000 bps D20000 bps

Q8. A network using CSMA/CD has a bandwidth of 20 Mbps. If the maximum propagation time is 25 μs, what is the minimum size of the frame?

A500 bits B1000 bits C1500 bits D2000 bits

Q9. Consider a CSMA/CD network that transmits data at 100 Mbps over a 1 km cable with no repeater. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?

A8000 B16000 C12000 D20000

Q10. A multiplexer combines four 100 Kbps channels using a time slot of 2 bits. What is the bit rate?

A100 Kbps B200 Kbps C400 Kbps D1000 Kbps

Q11. Which of the following statements are CORRECT?
(A) NRZ is a bipolar scheme in which a positive voltage defines bit 0.
(B) NRZ-L and NRZ-I both have an average signal rate of N/2.
(C) The ideas of RZ and NRZ-L are combined in the Manchester scheme.
(D) NRZ-L and NRZ-I both have DC component problems.
(E) The minimum bandwidth of Manchester and differential Manchester is 3 times that of NRZ.
Choose the correct answer from the options given below:

A(A), (B) and (C) Only B(A), (C), (D) and (E) Only C(B), (C) and (D) Only D(A), (B), (C) and (E) Only

Q12. Match List I (line-coding scheme) with List II (average bandwidth)

List I	List II
A. NRZ-L	I. B = N/4
B. 2B1Q	II. B = N/8
C. 4D-PAM5	III. B = N/3
D. MLT-3	IV. B = N/2

AA-II, B-I, C-IV, D-III BA-IV, B-II, C-I, D-III CA-IV, B-I, C-II, D-III DA-III, B-IV, C-II, D-I

Q13. What is the total delay (latency) for a frame of 10 million bits sent on a link with 20 routers, each having a queuing time of 2 μs and a processing time of 1 μs? The link is 5000 km long, the speed of light in the link is 2 × 10⁸ m/s and the bandwidth is 6 Mbps.

A0.624050 s B1.691726 s C2.425080 s D1.714030 s

Q14. According to the Nyquist theorem, to reproduce the original analog signal, a necessary condition is that:

AThe sampling rate must be at least 2 times the highest frequency contained in the signal BThe sampling rate must be at least 2 times the lowest frequency contained in the signal CThe sampling rate must be at least ½ times the highest frequency contained in the signal DThe sampling rate must be at least ½ times the lowest frequency contained in the signal

Q15. In standard Ethernet with a transmission rate of 10 Mbps, assume the length of the medium is 2500 m and the size of a frame is 512 bytes. The propagation speed of a signal in the cable is 2 × 10⁸ m/s. Calculate the transmission delay and propagation delay.

A25.25 μs and 51.2 μs B51.2 μs and 12.5 μs C10.24 μs and 50.12 μs D12.5 μs and 51.2 μs

Q16.

Which of the following is not the characteristics of packet switching?

AData is broken into packets BEach packet may take a different rout CRequires a dedicated path DSupports dynamic routing

Next: Mock 2 – Data Communication & Physical Layer (Part 2) →

Mock 1 – Data Communication & Physical Layer (Part 1)