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Quiz Electronics Engineering 2 July 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 02 /07/2020

Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes

Q1. Find the effective access time for the memory for the given data.
Page fault service time = 10 ms
Average memory access time = 20 ns
One page fault is generated for every memory access = 10^6
(a) 29 ns
(b) 33 ns
(c) 28 ns
(d) 30 ns

Q2. _____ Number of leaf nodes in a rooted tree of n nodes, where each node is having 0 or 3 children.
(a) n/2
(b) ((2n+1))/3
(c) ((n-1))/n
(d) (n – 1)

Q3. Time to Live(TTL) field in the IP datagram is used _________ .
(a) to optimize throughput.
(b) to reduce delays.
(c) to prevent packet looping.
(d) to prioritize packets.

Q4. ______ number of undirected graphs can be constructed using V = (v_1,v_2,…….,v_n).
(a) 2^((n(n-1))/2)
(b) n^3
(c) n – 1/2
(d) 2^((n-1)/2)

Q5. _______ is holding an entry foreach terminal symbol and is acting as permanent database.
(a) Variable Table
(b) Keyword Table
(c) Terminal Table
(d) Identifier Table

Q6. Identify the true statement from the given statements.
Program relocation at run time :
1. requires transfer complete block to some memory locations.
2. requires both base address and relative address.
3. requires only absolute address.
(a) 1 only.
(b) 1 and 2
(c) 1, 2 and 3
(d) 1 and 3

Q7. The hexadecimal representation of 〖(632)〗_8 is:
(a) 19A
(b) 198
(c) 29A
(d) 291

Q8. The process executes the following code and after execution _____ number of child process get created.
(a) 4
(b) 1
(c) 15
(d) 16

Q9. _______ merges the bodies of two loops.
(a) loop rolling
(b) loop folding
(c) loop merge
(d) loop jamming

Q10. Which one of the following statements about Differential Pulse Code Modulation(DPCM) is true?
(a) The sum of message signal sample with its prediction is quantized
(b) The difference of message signal sample with its prediction is quantized.
(c) The difference of message signal sample and a random signal is quantized.
(d) The message signal sample is directly quantized, and its prediction is not used.


S1. Ans.(d)
Sol. Given:
Page fault service time = 10 ms = 10 × 10^(-3)
Average memory access time = 20 ns = 20 × 10^(-9)
Effective memory access time(EMAT) = Page fault for 10^6 times memory access × Page fault service time + (1 – Page fault for 10^6 times memory access) × average memory access time
⇒ EMAT = 1/10^6 × 10 × 10^(-3) + (1 – 1/10^6 )× 20 × 10^(-9)
⇒ EMAT = 10 × 10^(-9) + 1 × 20 × 10^(-9)
So, EMAT = 30 × 10^(-9) = 30 ns

S2. Ans.(b)

S3. Ans.(c)
Sol. TTL value tells about the longevity of the packet in the network and decides whether it should be discarded or not. TTL field is used mainly to avoid a situation in which IP datagram which are not in use to keep away from repeating again and again.

S4. Ans.(a)
Sol. With n vertices, total possible number of edges are (n(n-1))/2. For every edges there are two possibility, either we choose it or ignore it. So, total number of undirected graphs is 2^((n(n-1))/2).

S5. Ans.(c)
Sol. Terminal Table is a permanent table database which lists all key words and special symbols.

S6. Ans.(b)
Sol. Program relocation at run time transfer complete block to some memory locations. Program relocation is done on the basis of base address and relative address because blocks are addressed through this address .
Program relocation involves blocks transfer but Absolute addressing mode and indirect addressing modes are used only for one instruction at one time.

S7. Ans.(a)
Sol. To convert into hexadecimal digit, at first convert it from octal to binary and then from binary to hexadecimal digit.
Step 1: Octal to Binary conversion.
One octal digit is equivalent to binary digit.
〖(632)〗_8 = 〖(110 011 010)〗_2

Step 2: Binary to hexadecimal conversion.
4 binary digit is equivalent to 1 hexadecimal digit.
〖(110 011 010)〗_2 = 〖(0001 1001 1010)〗_2 = 〖(19A)〗_16
1010 = 10_10 = A_16

S8. Ans.(c)
Sol. With n fork statements, there are always 2^n – 1 child processes.
There are 4 fork statements, So number of child process gets created is 15.

S9. Ans.(d)
Sol. Loop Jamming is the combination of two or more loops into a single loop.

S10. Ans.(b)
Sol. In DPCM, instead of transmitting actual samples, difference of message signal samples are quantized and transmitted.

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