Array index should be integer not real numbers.
Expect the option B, All remaining indexes are not real numbers.
Option A , takes enum value as index which is integer number.
Option C, takes character which is having equivalent decimal value.
Option D, has boolean value as index whose value may be 0 or 1

Phong shading is a per-fragment color computation. The vertex shader provides the normal and position data as out variables to the fragment shader. The fragment shader then interpolates these variables and computes the color

Default subnet mask for Class C is 255.255.255.192
(192)₁₀ = (11000000)₂
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits

IP datagram is encapsulated in the payload field of Ethernet frame, so we can say that it is transported in information field of frame.

While the IP address of source and the destination in a datagram is kept same at each hop, the source MAC address is replaced at each station while the frame is in transit. So for the current station, the source address field will contain the MAC address of the previous station.

Wireless medium can’t be used for CSMA operations as they are not suitable for collision detection methods. As radio already have a large number of users, it can’t detect the collisions between the senders and the receivers.

C is useless in S→ABCc so remove it.
S→ABc
Now see each of remaining production
Bb→bb ( i.e. B→b)
Ab→ab (i.e.A→a)
Aa→aa (i.e. A→a)
So
S→ABc→abc

TCP uses a Sliding Window mechanism for the flow control of data in transit on a network. In Sliding Window, the sending device can send all packets within the TCP window size without receiving an ACK.
The receiving device should acknowledge each packet it received and after receiving the ACK from the receiving device, the sending device slides the window to right side

Let f_h is the highest frequency and f_i is the lowest frequency.
Bandwidth = f_h – f_i = 4000 - 40 KHz = 3960 KHz = 3.96 MHz

→ Token ring is standardized with protocol IEEE 802.5. Token passing is the method of medium access, with only one token allowed to exist on the network at a time.
→ Network devices must acquire the token to transmit data, and may only transmit a single frame before releasing the token to the next station on the ring. When a station has data to transmit, it acquires the token at the earliest opportunity, marks it as busy, and attaches the data and control information to the token to create a data frame, which is then transmitted to the next station on the ring.
→ The frame will be relayed around the ring until it reaches the destination station, which reads the data, marks the frame as having been read, and sends it on around the ring. When the sender receives the acknowledged data frame, it generates a new token, marks it as being available for use, and sends it to the next station.
→ IEEE 802.2 is the logical link control (LLC) as the upper portion of the data link layer of the OSI Model.
→ IEEE 802.6 is a standard governed by the ANSI for Metropolitan Area Networks (MAN).
→ IEEE 802.3 is a working group and a collection of Institute of Electrical and Electronics Engineers (IEEE) standards produced by the working group defining the physical layer and data link layer media access control (MAC) of wired Ethernet.

(X+Y)*(X+Z)*(Y+Z)=(X+Y)*(X+Z) Consensus Law
XY+XZ+YZ=XY+XZ Consensus Law

For min-term: AB + A'C +BC = AB + A'C
AND for max-term : ( A + B ).( A' + C ).( B + C ) = ( A + B ).(A' + C )

(X₀X₁+X₂)X₃+X₄)X₅+⋯+X_N
=(X₀X₁X₃+X₂X₃+X₄)X₅+⋯+X_N

=X₀X₁X₃X₅+X₂X₃X₅+X₄X₅+⋯+X_N

=X₀X₁X₃X₅⋯X_N−1+X₂X₃X₅⋯X_N−1+X₄X₅X₇⋯X_N−1+⋯+X_N

N² = (7601)₈
N² = 7*8*8*8 + 6*8*8 + 0 + 1*8
N² = 3969
N = (63)¹⁰
Now, (241)₅ = 2*5*5 + 4*5 + 1 = 71
Option (A) is incorrect.
(143)₆ = 1*6*6 + 4*6 + 3 = 63
Option (B) is correct.

Here, the baud value represents the serial port is capable of transferring a maximum number of bits per second.
Given 2400 baud value,
Sending maximum of 2400 bits per second with serial port .
Total Data To send = 1 bit(start) + 8 bits(char size) + 2 bits(Stop) = 11 bits.
Number of 8-bit characters that can be transmitted per second = 2400/11 = 218.18
The effective number of characters transmitted = 218 So, the correct option is (C).

1. All processes are arrived at time 0.
2. Algorithm used for scheduling is round robin with time quantum of one unit time.
3. The order of execution of the processes A B C D A C A C A,C,C,C,C,C
4. After 8 context switches, process A completes it execution So the completion time is 9

→ The Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles).
→ A single execution of the algorithm will find the lengths (summed weights) of shortest paths between all pairs of vertices
→ The Floyd-Warshall algorithm is an example of dynamic programming.

In 8086 architecture there are 16 bit data lines and 20 bit address lines.
16 bit data lines means that the word size must be 2 bytes and these 2 bytes can be read in a single memory cycle.
20 bit address lines corresponds that the memory size is 2²⁰ Bytes = 1 Megabyte

1. Pipelining is one way of improving the overall processing performance of a processor.
2. This architectural approach allows the simultaneous execution of several instructions.
3. Pipelining is transparent to the programmer; it exploits parallelism at the instruction level by overlapping the execution process of instructions.
4. It is analogous to an assembly line where workers perform a specific task and pass the partially completed product to the next worker.

Given, (12x)₃ = (123)_x
Since LHS has 3 as the base and RHS has ‘x’ base,
1 * 3*3 + 2 * 3 + x * 1 = 1 * x*x + 2 * x + 3
9 + 6 + x = x2 + 2x + 3
x2 + x - 12 = 0
x2 + 4x - 3x - 12 = 0
x( x + 4 ) - 3(x + 4) = 0
(x + 4)(x - 3) = 0
x = 3 and -4
But, both the values are infeasible.

Alternative explanation :
According to the rules of number systems , the numbers present in a number system should not be greater than the base of the number system.
According to LHS , (12x)3 tells us that the value of x should be less than 3.
According to RHS , (123)x tells us that the value of x should be greater than 3 as largest digit in 123 is 3.
Therefore, any combination is not possible.

Explanation:
→ The Mosfet has a much better response in high frequency than the BJT.
→ Normally the switching speed would be in the us range for the BJT while the Mosfet is in the ns range.
→ The advantage of the Mosfet is also in the circuitry needed to operate it. The BJT is a current driven device while the Mosfet is voltage driven. The circuitry for creating voltage pulses is much simpler than the ones for current pulses and the first ones require less power