Question

plz answer them asap

27. With load time binding the program cannot be relocated (T/F) 28. Using Dynamic Link Libraries (DLL), makes the program modules smaller (T/F) 29. Interpreted languages are faster than compiled languages (T/F) 30. When we discussed Database software, we discussed a Race condition. Briefly describe that a Race condition is and how it can be solved. 15 pts) 31. The output from the Java compiler is a binary class file that is executed by a A. Java Virtual Machine B. operating system C. compiler D. assembler Extra Credit (5 points) 32. Suppose a disk drive has the following characteristics: • 4 surfaces · 1024 tracks per surface • 128 sectors per track • 512 bytes/sector • Track-to-track seek time of 5 milliseconds • Rotational speed of 5000 RPM. a) What is the capacity of the drive? b) What is the access time

Answer 1

27. The answer is: True

28. The answer is: False

• It will not make the modules smaller.
• It will avoid the duplicate modules by using the module created at one place in another place wherever required.

29. The answer is: False

• Compiled languages are faster than interpreted languages.
  • Because compiled languages are directly converted into machine code that the processor can execute.

30. Race Condition: The condition when several processes are accessing and manipulating the same data concurrently and the outcome of the execution depends on the particular order in which the access takes place.

Solution is:

• If we need to prevent another process from executing while current process is modifying this variable, this is done through Mutual Exclusion

31. The answer is: Java Virtual Machine

32.

a) Capacity = Surfaces * Tracks * Sectors * Bytes

= 4 * 1024 * 128 * 512

= 268435456 bytes

Therefore, Capacity = 268435456 bytes

b) Access time = Seek Time + Latency

We know Seek time (Track to Track Seek Time) = 5 ms.

We need to calculate the Latency which is:

Latency = ½ * [ (60/RPM) * (1000 / 1 sec)]

= ½ * [(60/5000) * (1000)]

= 6 ms

Now, Access time = 6 + 5 = 11 ms