Engineering Jobs   »   Quiz Electronics Engineering 18 July 2020

Quiz Electronics Engineering 18 July 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 18/07/2020

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

Q1. Which Netware protocol provides link-state routing?
(a) NLSP
(b) RIP
(c) SAP
(d) NCP

Q2. Four 100 W bulbs are connected in parallel across 200 V supply line. If one bulb gets fused.
(a) No bulb will light
(b) All the four bulbs will light
(c) Rest of the three bulbs will light
(d) None of the above

Q3. Process is in a ready state ___________ .
(a) when process is scheduled to run after some execution
(b) when process is unable to run until some task has been completed
(c) when process is using the CPU
(d) None of the above

Q4. A binary with n digits all of which are unity has the value
(a) n^2-1
(b) 2^n
(c) 2^((n-1))
(d) 2^n-1

Q5. How many times the word “ PROCESS” will be printed when executing the following program?
(a) 8
(b) 4
(c) 6
(d) 7

Q6. If a planar graph, having 25 vertices divides the plane into 17 different regions. Then how many edges are used to connect the vertices in this graph.
(a) 20
(b) 30
(c) 40
(d) 50

Q7. Merge sort uses:
(a) Divide and Conquer
(b) Backtracking
(c) Heuristic approach
(d) Greedy approach

Q8. If a random coin is tossed 11 times, then what is the probability that for 7th toss head appears exactly 4 times?
(a) 5/32
(b) 15/128
(c) 35/128
(d) None of the above

Q9. Which of the following is illegal declaration in C language?
(a) char*str = “Raj is a Research Scholar”;
(b) charstr[25] = “Raj is a Research Scholar”;
(c) charstr[40] = “Raj is a Research Scholar”;
(d) char[] str = “Raj is a Research Scholar”;

Q10. The form factor is equal to
(a) (r.m.s value)/(average value)
(b) (average value)/(r.m.s value)
(c) ((r.m.s value)²)/(average value)
(d) None of these


S1. Ans.(a)
Network Link Service Protocol(NLSP) provides link-state routing.
Routing Information Protocol(RIP) is used by routers to exchange routing information on a network.
Netware Core Protocol(NCP) is a Novell client-server protocol for LAN and used to access file, print, directory, clock synchronization, messaging, remote command execution and other network service functions.
Service Advertisement Protocol(SAP) is a Novell NetWare protocol which is used for service providers to advertise their services on the network.

S2. Ans.(c)
Sol. In parallel, If one bulb gets fused then it doesn’t affect the rest of the 3 bulbs. Hence, Rest 3 bulbs will light.

S3. Ans.(a)
Sol. Any process can come into ready state either after Start State or while waiting to have the processor allocated to them by the operating system after execution of any instruction.
The process is in blocked state, when process is unable to run until some task has been completed.

S4. Ans.(d)
Sol. For n-bit binary number its value is 2^n-1.
For example: 11112 = 2^3 x 1 + 2^2 x 1 + 2^1 x 1 + 2^0 x 1
= 8 + 4 + 2 + 1 = 15
And, 11112 = 2^4-1 = 15
So, both have same value.

S5. Ans.(b)
Sol. fflush() function will be necessary to hold the output of the process together or print it when necessary. Each fork() function creates 2 child so two fork() function creates 4 child process hence “PROCESS” will be executed 4 times.

S6. Ans.(c)
Sol. Given:
Number of vertices = 25
Number of regions = 17
According to Euler’s Formula –
If G be a connected planar graph and n, m and f denote the numbers of vertices, edges, and faces (regions) respectively in a plane. Then
⇒ n – m + f = 2
⇒ 25 – m + 17 = 2
⇒ 42 – m = 2
So, m = 42 – 2 = 40

S7. Ans.(a)
Sol. Merge Sort is a recursive algorithm. Input array is divided by this in two halves then calls itself for the two halves and finally merges the two sorted halves in linear time. Merge Sort and Quick Sort are Divide and Conquer algorithm.

S8. Ans.(a)
Sol. As the 7th toss is head so before 7th toss we must get 3 heads to make total number of head exactly 4. 3 heads in previous 6 tosses can appear at any toss so the probability is –
Required probability of getting exactly 4 heads = 6C3 (1/2)^3 × (1/2)^3 × 1/2 = 6!/3!3! × (1/2)^6 × 1/2
= 20/1 × 1/64 × 1/2 = × 5/32

S9. Ans.(d)
Sol. char[] str is a declaration in Java and not in C programming.

S10. Ans. (a)
Sol. Form factor is
F =(r.m.s value)/(average value)

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