someone please verify. answer answer is C.
First some background on indexing.
File is stored in form of blocks in hard disk. File is made up of records, each record has some key field on which indexes are created(for faster access of file). suppose record is of 64 byte and key field is of 14 byte. hence we create index on this 14 byte key field(not necessary as you can index on non key field also). Apart from 14 bytes of key field it must also include block address in which record with this key value is present . lets say block address is 2 byte.
Now come to question
size of the index file directly proportional to number of blocks used for the index.
if we have 1000 records. there will be 1000(14+2) Bytes for index. If we store all of them in single block(provided it fits in single block) there will not be multilevel indexing, hence no memory needed to store secondary indexes.
I still have doubt in this .
size of block
If size of block is large we can accommodate more indexes per block hence number of levels in indexing is less hence file size is lesser.
siz of address of block
If block size becomes 1 byte instead of 2 byte , we need less number of block hence file size is less
please correct me if my answer is wrong.