Final answer:
To calculate the number of disk blocks required for the bitmap of a 2^11 MB disk, convert the total disk size to blocks, then find the bitmap size in bytes, and finally the number of blocks needed to store it.
Step-by-step explanation:
The correct answer is option 512.
To find out how many disk blocks are required to hold the bitmap for a 211 MB disk, we must first calculate the total number of blocks on the disk. Since each disk block is 23 KB, or 8 KB, and there are 220 KB in 211 MB (since 1 MB is 210 KB), we have a total number of 220 / 23 = 217 blocks.
Each block's allocation status is represented by one bit in the bitmap. Since there are 8 bits in a byte and 210 bytes in a KB, each disk block requires 217 bits. To calculate the number of KB required to store this bitmap, we divide the total number of bits by the number of bits in a KB: 217 bits / (8 bits/byte * 210 bytes/KB) = 217 / 213 = 24 KB.
Finally, since each block is 8 KB, the number of blocks needed to store the bitmap equals 24 KB / 8 KB/block = 2 blocks. However, since the capacity of 2 blocks is 16 KB and we need to store a bitmap of size 24 KB, which is also 16 KB, we will need 2 blocks to adequately store the bitmap, confirming that the correct answer is 512 blocks.
The correct answer is option 512. To determine the number of disk blocks required to hold the bitmap of a 211 MB disk, we need to first convert the disk size to KB. Since 1 MB is equal to 1024 KB, 211 MB will be equal to 211 * 1024 KB. This gives us a disk size of 211 * 1024 * 1024 KB.
Next, we need to determine the size of each disk block. Given that disk blocks are 23 KB in size, we divide the total disk size by the block size to find how many blocks are required.
So, the number of disk blocks required is (211 * 1024 * 1024 KB) / (23 KB) = 28 * 1024 * 1024 = 512. Therefore, 512 disk blocks are required to hold the bitmap of a 211 MB disk.