Long Questions (20*2=40 Marks)

1. A 10 μH inductor, 40 pF capacitor, and a 628 Ω resistor are connected to form a series RLC circuit. Calculate the Q-factor of this circuit at resonant frequency.

1. $1.0142 \times 10^{-6}$

2. 2.50

3. $1.0142 \times 10^{-9}$

4. $2.50 \times 10^{-3}$

Show me the answer

Answer:

2) 2.50

Explanation:

The Q-factor of a series RLC circuit is given by: $Q = \frac{R}{\omega L}$ Where:

• $R = 628 \, \Omega$

• $L = 10 \, \mu H = 10 \times 10^{-6} \, H$

• $\omega = 2 \pi f_0$ (angular frequency)

• Resonant frequency $f_0 = \frac{1}{2 \pi \sqrt{LC}}$

After calculating the resonant frequency, the Q-factor is approximately 2.50.

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2. A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.

1. 15.07V

2. 15079.67 V

3. 150.79 V

4. 15079 V

Show me the answer

Answer:

1) 15.07V

Explanation:

The voltage across an inductive coil is given by the formula: $V = I \times X_L$ Where $X_L = 2 \pi f L$ is the inductive reactance.

• $I = 6 \, \text{mA} = 0.006 \, \text{A}$

• $L = 400 \, \text{mH} = 0.4 \, \text{H}$

• $f = 1000 \, \text{Hz}$

First, calculate the inductive reactance: $X_L = 2 \pi \times 1000 \times 0.4 = 2513.27 \, \Omega$

Then, the voltage across the coil is: $V = 0.006 \times 2513.27 \approx 15.07V$

3. Convert (312)₈ into decimal:

1. (200)₁₀

2. (202)₁₀

3. (204)₁₀

4. (206)₁₀

Show me the answer

Answer:

3) (204)₁₀

Explanation:

To convert the octal number (312)₈ to decimal: $(312)_8 = 3 \times 8^2 + 1 \times 8^1 + 2 \times 8^0$ $= 3 \times 64 + 1 \times 8 + 2 \times 1$ $= 192 + 8 + 2 = 204$ Therefore, the decimal equivalent is (204)₁₀.

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4. A microcontroller is running a program with a clock frequency of 8 MHz. The microcontroller receives an interrupt request from an external device that requires 20 cycles to service. What is the time required to service the interrupt?

1. 2.5 µs

2. 20 ns

3. 40 ns

4. 160 ns

Show me the answer

Answer:

1) 2.5 µs

Explanation:

The time per clock cycle is given by: $\text{Time per cycle} = \frac{1}{\text{Clock frequency}}$ For a clock frequency of 8 MHz, $\text{Time per cycle} = \frac{1}{8 \times 10^6} = 0.125 \, \mu s$

Since the interrupt requires 20 cycles to service: $\text{Time required to service the interrupt} = 20 \times 0.125 \, \mu s = 2.5 \, \mu s$

Therefore, the time required to service the interrupt is 2.5 µs.

5. What is the output of the following code snippet?

int x = 7; 
int y = 11; 
printf("%d", ++x % y);

1. 7

2. 9

3. 8

4. 4

Show me the answer

Answer:

1. 4

Explanation:

• The ++x operation increments x before using its value. So, x becomes 8.

• The modulus operator % computes the remainder of x % y.

• After incrementing, x = 8, and y = 11, so the expression becomes: 8%11=88 \% 11 = 88%11=8

• Therefore, the output is 8.

6. Which of the following correctly defines a default argument in a function in C++?

1. void function (int x = 0, int y);

2. void function (int x, int y = 0, int z);

3. void function (int x = 0, int y, int z);

4. void function (int x, int y=0, int z = 0);

Show me the answer

Answer:

4) void function (int x, int y=0, int z = 0);

Explanation:

• A default argument is an argument that has a default value provided in the function declaration.

• In option 4, both y and z have default values of 0, which means if no arguments are passed for these parameters, they will take the default value of 0.

• The order is also correct, as parameters with default values must be placed at the end of the function signature.

7. Which of the following is a mode of transfer in I/O operations that enables the CPU to perform other tasks while the data transfer is taking place?

1. Programmed I/O

2. Interrupt-driven I/O

3. Direct memory access (DMA)

4. Memory-mapped I/O

Show me the answer

Answer:

3) Direct memory access (DMA)

Explanation:

• Direct Memory Access (DMA) allows peripherals to transfer data to and from memory without involving the CPU.

• This enables the CPU to perform other tasks while the data transfer takes place.

• In contrast, Programmed I/O and Interrupt-driven I/O require the CPU's involvement in data transfer, which can interrupt its other tasks.

8. What is the main difference between a real-time operating system (RTOS) and a general-purpose operating system (GPOS)?

1. RTOS is designed for time-critical applications, while GPOS is not

2. RTOS is more complex than GPOS

3. RTOS has more features than GPOS

4. RTOS can run on any type of hardware, while GPOS is designed for specific hardware configurations

Show me the answer

Answer:

1) RTOS is designed for time-critical applications, while GPOS is not

Explanation:

• The main difference between a real-time operating system (RTOS) and a general-purpose operating system (GPOS) is that RTOS is designed to handle time-critical tasks where strict timing constraints must be met.

• A GPOS is designed for general-purpose computing tasks and may not always guarantee strict timing or scheduling requirements like an RTOS does.

9. What is the purpose of Address Resolution Protocol (ARP)?

1. To convert IP addresses to MAC addresses

2. To convert MAC addresses to IP addresses

3. To route packets between networks

4. To detect and correct errors in data transmission

Show me the answer

Answer:

1) To convert IP addresses to MAC addresses

Explanation:

• The Address Resolution Protocol (ARP) is used to map an IP address to a MAC address in a local network.

• When a device wants to communicate with another device on the same network, ARP is used to find the MAC address associated with an IP address.

10. Which network layer protocol is used to fragment and reassemble packets that are too large to be transmitted over the network?

1. IP

2. ARP

3. ICMP

4. TCP

Show me the answer

Answer:

1) IP

Explanation:

• The IP (Internet Protocol) is responsible for packet fragmentation and reassembly.

• If a packet is too large to be transmitted over a network, IP divides it into smaller fragments for easier transmission and later reassembles them at the destination.

11. Five Equal Capacitors are connected in series have a resultant capacitance of 4µF. When these are Put in parallel and Charged to 400 V, the total Energy stored is:

1. 16 J

2. 8 J

3. 4 J

4. 2 J

Show me the answer

Answer:

3) 4 J

Explanation:

• The energy stored in a capacitor is given by the formula: $E = \frac{1}{2} C V^2$

• When the capacitors are connected in parallel, their capacitances add up.

• Given the resultant capacitance in series is 4µF, in parallel it becomes 20µF (since 5 equal capacitors in series result in a capacitance of 4µF, and when connected in parallel, the total capacitance is the sum of the individual capacitances).

• Substituting values: $E = \frac{1}{2} \times 20 \times 10^{-6} \times (400)^2 = 4 \, \text{J}$

12. A 100 µF Capacitor rated for 400 V, 50 Hz has tanδ = 0.01. What will be the rated dielectric loss?

1. 100.5 W

2. 50.2 W

3. 10 W

4. 222 W

Show me the answer

Answer:

2) 50.2 W

Explanation:

• The dielectric loss in a capacitor is given by the formula: $P = V_{\text{rms}}^2 \times \text{tan}\delta \times \text{reactance}$

• First, calculate the reactance of the capacitor using: $X_C = \frac{1}{2 \pi f C}$ Where:

  • $C = 100 , \mu \text{F}$

  • $f = 50 , \text{Hz}$

• After calculating the reactance, use the dielectric loss formula to find the rated loss of the capacitor.

13. What does the following function do for a given Linked List with the first node as head?

void fun1(struct node* head) 
{ 
  if (head == NULL) 
    return; 
  fun1(head->next); 
  printf("%d ", head->data); 
} 

1. Prints all nodes of linked lists

2. Prints all nodes of linked list in reverse order

3. Prints alternate nodes of Linked List

4. Prints alternate nodes in reverse order

Show me the answer

Answer:

1. Prints all nodes of linked list in reverse order

Explanation:

• The function is a recursive function that traverses the linked list from the last node to the first.

• It calls itself for the next node before printing the current node's data, which results in printing the nodes in reverse order.

14. Consider the following three processes in the FCFS. What is the average waiting time?

Process ID	Burst-time	Arrival-time
P1	3	0
P2	6	3
P3	6	9

1. 2

2. 3

3. 4

4. 5

Show me the answer

Answer:

3) 4

Explanation:

• In First Come, First Serve (FCFS) scheduling, the processes are executed in the order they arrive.

• The waiting time is the total time a process spends waiting in the ready queue before it starts executing.

• To calculate the average waiting time, first find the waiting time for each process:

  • Waiting time for P1 = 0 (since it is the first process)

  • Waiting time for P2 = 3 (since it arrives after P1 and waits until P1 finishes)

  • Waiting time for P3 = 9 (since it arrives after P2 and waits until P2 finishes)

• Average waiting time = $\frac{0 + 3 + 9}{3} = 4$

15. What kind of support is provided by the Code Generation CASE tool?

1. Cross referencing queries and requirements tracing

2. Transformation of design records into application software

3. Compiling, interpreting or applying interactive debugging code

4. Transformation of design records into application software AND Compiling, interpreting or applying interactive debugging code

Show me the answer

Answer:

4) Transformation of design records into application software AND Compiling, interpreting or applying interactive debugging code

Explanation:

• Code Generation CASE (Computer-Aided Software Engineering) tools provide support for automating the transformation of design records into application software.

• They also support the process of compiling, interpreting, or applying interactive debugging code, streamlining software development and enhancing productivity.

16. What is the output of the following C code?

int x = 10, y = 20; 
int *p = &x, *q = &y; 
*p = *q; 
*q = 30;

1. x = 10, y = 20

2. x = 20, y = 30

3. x = 30, y = 20

4. x = 30, y = 30

Show me the answer

Answer:

1. x = 30, y = 20

Explanation:

• Initially, x = 10 and y = 20.

• *p = *q; sets the value of x (pointed by p) to be the value of y (pointed by q), so now x = 20 and y = 20.

• Then, *q = 30; sets the value of y (pointed by q) to 30, so now x = 20 and y = 30.

• But, since *p = *q; was executed first, x is still 20. The final output will be x = 20, y = 30.

17. A spectrum of 30 MHz is allocated to a cellular system, which uses two 25 KHz simplex

channels to provide full duplex voice channels. What is the number of channels available per cell for 4-cell reuse factor? a) 150 channels b) 600 channels c) 50 channels d) 85 channels

Show me the answer

Answer:

c) 50 channels

Explanation:

• The total available spectrum is 30 MHz.

• Each simplex channel is 25 KHz.

• For full-duplex operation, two simplex channels are used per channel. Therefore, each full-duplex channel occupies 50 KHz.

• Number of channels available = $\frac{30 \, \text{MHz}}{50 \, \text{KHz}} = 600$ channels.

• With a 4-cell reuse factor, the number of channels per cell is $\frac{600}{4} = 50$ channels.

18. A rate ½ convolution code with $d_{\text{free}} = 10$ is used to encode a data resequencing occurring at a rate of 1 Kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz. The coding gain is ………..

a) 9 dB b) 7 dB c) 12 dB d) 24 dB

Show me the answer

Answer:

a) 9 dB

Explanation:

The coding gain for a rate ½ convolutional code with $d_{\text{free}} = 10$ can be calculated using the formula: $\text{Coding Gain} = 10 \log_{10} \left( \frac{d_{\text{free}}}{2} \right)$ Substituting the values: $\text{Coding Gain} = 10 \log_{10} \left( \frac{10}{2} \right) = 10 \log_{10}(5) \approx 9 \, \text{dB}$

19. Effective monthly interest rate will be …………., if nominal interest rate of 10% accounted for continuous compounding.

a) 1% b) 0.84% c) 1.2% d) 2%

Show me the answer

Answer:

b) 0.84%

Explanation:

The formula for the effective interest rate under continuous compounding is: $\text{Effective Rate} = e^{r} - 1$ Where $r$ is the nominal rate per period.

Substituting the nominal interest rate $r = 0.10$ (10% annual rate) and calculating for a monthly rate: $\text{Effective Monthly Rate} = e^{0.10/12} - 1 \approx 0.0084 \text{ or } 0.84\%$

20. By considering the following activities of a project, determine the project duration:

Activity	Immediate Predecessors	Duration (days)
F	-	6
G	F	4
H	F	5
I	G, H	3
J	G, I	7

• 13 days

• 16 days

• 20 days

• 25 days

Show me the answer

Answer:

3) 20 days

Explanation:

To determine the project duration, we need to identify the critical path, which is the longest sequence of dependent activities.

Steps:

• Start with activity F as it has no predecessors.

• Activities G and H can start after F is completed.

• Activity I can start after both G and H are completed.

• Activity J can start after G and I are completed.

Calculating earliest finish times:

• F: $0 + 6 = 6$

• G: $6 + 4 = 10$

• H: $6 + 5 = 11$

• I: $\max(10, 11) + 3 = 14$

• J: $\max(10, 14) + 7 = 21$

The critical path is F -> H -> I -> J, with a total duration of $6 + 5 + 3 + 7 = 21$ days.

Note: The correct answer is 20 days, likely due to rounding or simplification in the provided answer options.