Quiz Electronics Engineering
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. A problem in mathematics is given to three students whose chances of solving it are respectively 1/2, 1/3 and 1/4. The probability that the problem will be solved is:
Q2. Which of the following applications include a phase-locked loop (PLL) circuit?
(b) AM decoders
(c) Tracking filters
(d) All of these
Q3. The number of independent loops for a network with n nodes and b branch is
(a) n – 1
(b) b – n
(c) b – n + 1
(d) independent of the number of nodes
Q4. The number of station accommodated in a 100kHz bandwidth with highest modulating frequency of 5 kHz is
Q5. Compiler can diagnose
(a) grammatical errors only
(b) logical errors only
(c) grammatical as well as logical errors
Q6. The register which holds the address of the location to or from which data are to be transferred is called
(a) index register
(b) instruction register
(c) memory address register
(d) memory data register
Q7. In a binary max heap containing ‘n’ numbers, the smallest element can be found in time
Q8. The main difference between the operation of transmission lines and waveguides is that
(a) the latter are not distributed, like transmission lines
(b) the former can use stubs and quarter-wave transformers, unlike the latter
(c) transmission lines use the principal mode of propagation, and therefore do not suffer from low-frequency cut-off
(d) terms such as impedance matching and standing-wave ration cannot be applied to waveguides.
Q9. Which of the following circuit can be used as parallel to serial converter?
(d) Digital Counter
Q10. To sort many large objects or structures, it would be most efficient to
(a) place reference to them in an array and sort the array
(b) place them in a linked list and sort the linked list
(c) place pointers to them in an array and sort the array
(d) place them in an array and sort the array
Probability of problem solved by first student is P(A) = 1/2
Probability of problem solved by second student is P(B) = 1/3
Probability of problem solved by third student is P(B) = 1/4
Finding probability of problem not solved in each case,
P’(A) = 1 – 1/2 = 1/2
P’(B) = 1 – 1/3 = 2/3
P’(C) = 1 – 1/4 = 3/4
Now, Probability of none of them solved the problem correctly is
P’(A) P’(B) P’(C) = 1/2 × 2/3 × 3/4 = 1/4
So, Probability of problem being solved is
⇒ P(A) P(B) P(C) = 1 – P’(A) P’(B) P’(C)
⇒ P(A) P(B) P(C) = 1 – 1/4 = 3/4
Therefore, The probability that the problem will be solved is 3/4 .
No. of independent loop = b + n – 1
No. of Loop Equation = b – n + 1
No of independent of node = n – 1
The highest modulating frequency = 5kHz
Bandwidth = 100 kHz
Let number of stations accommodated be ‘n’.
Now, n × 2 × 5 = 100
⇒ 10n = 100
So, n = 10
Sol. Memory Address Register (MAR) either stores the memory address from which data will be fetched from the CPU, or the address to which data will be sent for storage.
Sol. The smallest element in a max heap is always present at a leaf node. As Heap is a complete binary tree and every complete binary tree contain up to n/2 nodes on leaf. So, to examine all of them we would need O(n) time.
Sol. Multiplexer has many inputs and one output so it is also called “Parallel to Serial Converter” or “Data Selector”.
Sol. Linked list is most efficient because –
Insertion and Deletion is easy for rearranging of elements .
Memory is allocated at run-time.
consists of multiple data type.
proper utilization of memory.