UGC NET Dec-2020 and June-2021 Paper-2
December 3, 2023UGC NET Dec-2020 and June-2021 Paper-2
December 3, 2023UGC NET Dec-2020 and June-2021 Paper-2
| Question 9 |
The order of a leaf node in a B+ tree is the maximum number of (value, data record pointer) pairs it can hold. Given that the block size is 1K bytes, data record pointer is 7 bytes long, the value field is 9 bytes long and a block pointer is 6 bytes long, what is the order of the leaf node?
| 63 | |
| 64 | |
| 67 | |
| 68 |
Question 9 Explanation:
Using the formula dp*m+vp*m+bp < =block size we get the order as 63. Here dp is Data
Pointer(7B), vp is value size(9B), bp is Block Pointer and m is the order of leaf.
Pointer(7B), vp is value size(9B), bp is Block Pointer and m is the order of leaf.
Correct Answer: A
Question 9 Explanation:
Using the formula dp*m+vp*m+bp < =block size we get the order as 63. Here dp is Data
Pointer(7B), vp is value size(9B), bp is Block Pointer and m is the order of leaf.
Pointer(7B), vp is value size(9B), bp is Block Pointer and m is the order of leaf.