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Quiz Electronics Engineering 25 June 2020

Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 25/06/2020

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

Q1. ______ pairs of traversals are not sufficient to build a binary tree.
(a) Pre-order and In-order
(b) Post-order and In-order
(c) Post-order and Pre-order
(d) None of the above

Q2. The current i(t) flowing through a resistor R = 10 ohm is i(t) = 5sinωt + 4sin3ωt + 3sin5ωt. The average power is
(a) 160 W
(b) 250 W
(c) 500 W
(d) 720 W

Q3. A linear time invariant system has an impulse response e^2t, t > 0. If the initial conditions are zero and the input is e^3t, the output for t > 0 is
(a) e^3t + e^2t
(b) e^5t
(c) e^3t – e^2t
(d) None of the above

Q4. ______ symbol is used to denote derived attributes in ER model.
(a) Dashed ellipse
(b) Squared ellipse
(c) Ellipse with attribute name underlined.
(d) Rectangular Box

Q5. ______ sorting algorithm has the lowest worst case complexity.
(a) Merge Sort
(b) Bubble Sort
(c) Selection Sort
(d) Quick Sort

Q6. Identify the subnet mask for the given direct broadcast address of subnet is


Q7. Decreasing the RAM causes ______ .
(a) fewer page faults
(b) more page faults
(c) virtual memory get decreases
(d) None of the above

Q8. (A+C ̅)( B ̅+C ̅) simplifies to _________ .
(a) AC ̅ + B ̅
(b) C(A ̅ + B ̅)
(c) BC ̅ + A
(d) AB ̅ + C ̅

Q9. _______ is used to convert from recursive to iterative implementation of an algorithm.
(a) Array
(b) Tree
(c) Stack
(d) Queue

Q10. Shift reduce parsers are
(a) Top down parser
(b) Bottom up parser
(c) Both (a) and (b)
(d) None of the above


S1. Ans.(c)
Sol. A binary tree can be generated by two traversals if one of them is in-order. The following combination can uniquely identify a tree.
In-order and Pre-order.
In-order and Post-order.
In-order and Level-order.
But a full binary tree is generated using post-order and pre-order traversals.

S2. Ans.(b)
Sol. Given:
R = 10 ohm
i(t) = 5sinωt + 4sin3ωt + 3sin5ωt
We know that,
P_avg = I_RMS^2R
RMS current is
⇒ I_RMS = √((5/√2)^2+(4/√2)^2+(3/√2)^2 )
⇒ I_RMS = √((25/2)^ +(16/2)^ +(9/2)^ )
⇒ I_RMS = √(50/2)
So, I_RMS = 5 A

Now, P_avg = 5_ ^2 × 10
⇒ P_avg = 25 × 10
So, P_avg = 250 Watt

S3. Ans.(c)
Sol. Given:
h(t) = e^2t
x(t) = e^3t
In Laplace Domain, Output response is
Y(s) = H(s)X(s)
We know that Laplace of e^at = 1/(s – a)
Converting to Laplace domain, we get
⇒ H(s) = 1/(s – 2)
⇒ X(s) = 1/(s – 3)
Now, Y(s) = 1/(s – 2) . 1/(s – 3) =
⇒ Y(s) = 1/(s – 3) – 1/(s – 2)
Inverse of Laplace domain is
⇒ y(t) = e^3t – e^2t

S4. Ans.(a)
Sol. A Derived attribute can be derived from other attributes of the entity type. In ER model, a derived attribute is represented by a dashed Ellipses.

S5. Ans.(a)
Merge and Heap Sort has lowest worst case complexity of O(nlogn).
Bubble, Insertion, Selection and Quick have worst case complexity of O(n2).

S6. Ans.(d)
Sol. The direct broadcast address(DBA) is a “Class C” IP address. By default Subnet mask of Class C is Last Octet is for Host ID and in broadcast address all the host bits are ‘1’ making last octet 255. But here in DBA last octet has 31 so subnetting is done. is equivalent to 11111111.11111111.11111111.00011111.
So, in subnet mask all bits should be 1 except last 5 bits from right. Therefore, The required Subnet Mask is

S7. Ans.(b)
Sol. Decreasing the RAM causes more page fault. If the size of the RAM decreases then program attempts to access a block of memory which is in its address space but not stored in the Physical Memory or RAM. So, there is increase in the page fault due to lack of data stored in physical memory. Operating system forces use of more virtual memory on Hard Disk, due to this CPU take time to move data between RAM and Hard Disk. Therefore, there is less CPU time available to the applications and more time for paging.

S8. Ans.(d)
(A+C ̅)( B ̅+C ̅)
= AB ̅ + AC ̅ + B ̅C ̅ + C ̅C ̅
= AB ̅ + AC ̅ + B ̅C ̅ + C ̅
= AB ̅ + C ̅(A + B ̅ + 1)
= AB ̅ + C ̅

S9. Ans.(c)
Sol. A recursive function calls itself again to repeat the code but an iterative function use loops to repeat some part of the code.
Stack is the data structure used for converting because function calls are executed in Last In First Out(LIFO) order.

S10. Ans.(b)
Sol. In Shift Reduce parser the construction of parse is done in similar manner like Bottom Up parsing in which the parse tree is constructed from leaves i.e. bottom, to the root i.e. top. LR parser is more generalized form of Shift reduce parser.

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