The order of an internal node in a $B+$ tree index is the maximum number of children it can have. Suppose that a child pointer takes $6$ bytes, the search field value takes $14$ bytes, and the block size is $512$ bytes. What is the order of the internal node?
lets assume order of b+ tree is n.
B+ tree internal node has n-1 keys and n node (child) pointer.
The answer to the first question is $2048 ...