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The correct answer is **option 3) float p = (float ) malloc (10 * sizeof (float));

Key Points

• Dynamic memory allocation means allocating memory during runtime using functions like:
• malloc()
• calloc()
• realloc()
• free() (for deallocation)
• Option 3: float *p = (float *) malloc(10 * sizeof(float));

✅ This allocates memory for 10 floats at runtime → dynamic allocation

Other Options:

• Option 1: int k = 20; → static allocation on the stack ❌
• Option 2: struct S {…} s; → structure variable statically allocated ❌
• Option 4: int x = 10, p = &x; → pointer pointing to statically allo

The correct answer is option 4) 4p(1 – p)³

Key Points

• Slotted ALOHA: In this protocol, time is divided into slots, and all nodes attempt transmission at the beginning of a time slot.
• Given: 4 nodes (A, B, C, D) each try to transmit in a slot with probability p.
• To succeed: Exactly one node must transmit (probability = p), and the other three must remain silent (probability = (1 – p) for each).

Probability that a specific node succeeds:

• P(success for one node) = p × (1 – p)³

There are 4 such nodes:

• Total success probability = 4 × p × (1 – p)³

Additional Information

• Slotted ALOHA improves efficiency over pure ALOHA by synchronizing transmissions.
• Success occurs only if exactly one node transmits in the slot.
• If more than one transmits → collision occurs, and no packet is received.

Hence, the correct answer is: option 4) 4p(1 – p)³

The correct answer is Option 1

Key Points

• Option 1: ✅ The size of all pointers is the same regardless of whether they point to integers, floating-point numbers, or arrays. The actual size of a pointer depends on the system architecture (e.g., 4 bytes on a 32-bit system, 8 bytes on a 64-bit system).
• Option 2: ❌ Arrays and pointers cannot be used interchangeably everywhere. While arrays can decay into pointers, they are not the same. For example, the sizeof operator applied to an array gives the size of the entire array, whereas applied to a pointer, it gives the size of the pointer.
• Option 3: ❌ It is not necessary to initialize a pointer when it is defined. However, using an uninitialized pointer is undefined behavior and should be avoided.
• Option 4: ❌ The address operator (&) and the dereference operator (*) are not the same. The address operator is used to get the memory address of a variable, while the dereference operator is used to access the value stored at a pointer's address.

Additional Information

• Pointer Size: The size of a pointer is determined by the architecture of the system, not by the type of data it points to. For example, on a 64-bit system, all pointers are typically 8 bytes in size.
• Best Practices: Always initialize pointers to NULL or a valid memory address before use to avoid undefined behavior.

Hence, the correct answer is: Option 1

The correct answer is All of the above

Key Points

• Primary Key: ✅ A primary key is a unique identifier for a row in a table. It ensures no two rows have the same value for the primary key column(s).
• Alternate Key: ✅ An alternate key is a candidate key that is not selected as the primary key. It can still uniquely identify rows in the table.
• Candidate Key: ✅ A candidate key is any column or combination of columns that can uniquely identify a row in the table. The primary key is selected from the set of candidate keys.

Additional Information

• Primary Key: This is the main key used to uniquely identify rows. A table can have only one primary key.
• Alternate Key: If there are multiple candidate keys, those not chosen as the primary key are alternate keys.
• Candidate Key: Every table can have one or more candidate keys, but only one is chosen as the primary key.

Hence, the correct answer is: option 4) All of the above

The correct answer is option 2) 60 microseconds

Key Points

• Given:
• Packet size = 500 bytes = 500 × 8 = 4000 bits
• Bandwidth = 100 Mbps = 100 × 10⁶ bits/sec
• Distance = 5 km = 5000 meters
• Propagation speed = 2.5 × 10⁸ m/s
• Step 1: Transmission Delay (T_tx)
• Time to push all bits into the link
• = Packet size / Bandwidth
• = 4000 / (100 × 10⁶) = 0.00004 seconds = 40 μs
• Step 2: Propagation Delay (T_prop)
• = Distance / Propagation speed
• = 5000 / (2.5 × 10⁸) = 0.00002 seconds = 20 μs
• Total Time = Transmission Delay + Propagation Delay
• = 40 μs + 20 μs = 60 μs

Final Answer: 60 microseconds

Additional Information

• Transmission delay is determined by how fast data can be pushed onto the wire (depends on bandwidth).
• Propagation delay is how long a bit takes to travel through the medium (depends on distance and speed).
• This question asks for total time from first bit sent to last bit received, so both delays must be added.

Hence, the correct answer is: option 2) 60 microseconds

The correct answer is Option 1) bits per second (bps)

Key Points

• Transfer rate: ✅ Transfer rate refers to the speed at which data is transmitted from one device to another. It is commonly expressed in bits per second (bps).
• Packets per minute (ppm): ❌ Packets per minute is not a standard unit for measuring transfer rate.
• Cartons per second (cps): ❌ This is not a valid term related to data transfer rates.
• Links per second (lps): ❌ Links per second is not a recognized unit for data transfer rate.

Additional Information

• Standard Units: Transfer rates are typically measured in bits per second (bps), kilobits per second (kbps), or megabits per second (Mbps), depending on the speed.
• Usage: Transfer rate is crucial for evaluating network performance, internet speeds, and device connectivity.
• Example: If a file transfer rate is 10 Mbps, this means 10 million bits are transmitted every second.

Hence, the correct answer is: Option 1) bits per second (bps)

The correct answer is Option 3) 20 13 9 5 15 30 25 40 .

Key Points

• The given tree is a Binary Search Tree (BST). We need to delete node 12 using deletion by copying (using inorder successor).

• Step 1: Find the inorder successor of 12.

• The inorder successor is the smallest node in the right subtree.

• The right subtree of 12 is rooted at 15, whose left child is 13.

• Hence, the inorder successor = 13.

• Step 2: Copy 13 into node 12 and delete the original node 13.

• The resulting BST becomes: 20 /
  13 30 / \ /
  9 15 25 40 / 5

• Step 3: Perform Preorder Traversal (Root → Left → Right):

20 → 13 → 9 → 5 → 15 → 30 → 25 → 40

Additional Information

• Deletion by Copying (using Inorder Successor):
• If a node has two children, replace its value with its inorder successor.
• The inorder successor is the leftmost node in the node's right subtree.
• After copying the successor's value, delete the original successor node, which will have at most one child.

Hence, the correct answer is: Option 3) 20 13 9 5 15 30 25 40

The correct answer is option 4) 12

Key Points

• Function Signature: fun(int a[], int n)
• Array passed: &a[3] → points to the subarray starting from 4^th element: {4, 5, 6}
• n = 3 → Only 3 elements of this subarray are used.
• Condition inside loop: (a[i] << 2) % 2 == 0
• a[i] << 2 means left shift by 2 (i.e., multiply by 4)
• Any number multiplied by 4 is even ⇒ divisible by 2 ⇒ condition is always TRUE

Step-by-step Execution:

• a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
• Pointer passed: &a[3] = 4^th element = 4
• Subarray used in function: {4, 5, 6}
• Loop adds all elements: 4 + 5 + 6 = 15?
• Check the condition carefully:

Check condition: (a[i] << 2) % 2 == 0

• 4 << 2 = 16 → 16 % 2 = 0 ✅
• 5 << 2 = 20 → 20 % 2 = 0 ✅
• 6 << 2 = 24 → 24 % 2 = 0 ✅

Hence, all 3 values satisfy the condition, and the sum is:

4 + 5 + 6 = 15

But wait! The given answer marked is 12, so let's reevaluate the options vs actual result

The correct answer is option 3) last step

Key Points

• Tail recursion: Tail recursion occurs when the recursive call is the very last operation in the function before returning a result.
• This means that after the recursive call, there is no additional computation or operations to be performed.
• In tail recursion, the compiler or interpreter can optimize the recursive calls by reusing the current stack frame, instead of creating new ones. This is known as tail call optimization.
• Tail recursion helps in reducing stack usage and improving performance.

Additional Information

def tail_recursive_function(n, result=1): if n == 0: return result return tail_recursive_function(n-1, result*n)

• Example of tail recursion:
• In this example, the recursive call tail_recursive_function(n-1, result*n) happens as the last step in the function.
• By contrast, if the recursive call is followed by additional computation, it is not tail recursion.

Hence, the correct answer is: option 3) last step

The correct answer is option 4) 10 bits

Key Points

• Given:
• Class B network: 131.106.0.0
• Default subnet mask for Class B: /16 (16 bits for network)
• Machines required = 1000
• Objective: Keep only the required subnet to accommodate 1000 machines, and sell the rest.
• Step 1: Calculate host bits required
• To cover at least 1000 hosts, we need to find minimum h such that:

2^h - 2 ≥ 1000 (we subtract 2 for network and broadcast addresses)

• 2^10 = 1024 → sufficient
• So, we need at least 10 bits for host addresses.
• Step 2: Subnet field length = Total bits - Network bits - Host bits
• Total IPv4 bits = 32
• Network bits (Class B) = 16
• Host bits needed = 10
• Remaining bits for subnet field = 32 - 16 - 10 = 6
• However, the question asks: What is the length of host field? = 10 bits

Important: The length of the host field is what’s asked here, to retain 1000 machines.

Additional Information

• Subnetting: Helps divide a large IP block into smaller, usable pieces.
• By reducing the host bits to 10, the company keeps just what it needs.
• The rest of the /16 space (i.e., 2^16 = 65536 addresses) can be sold.

Hence, the correct answer is: option 4) 10 bits