@MiNiPanda
Thanks Bro.... i understood your explanation... i don't know how i said based on repeating keys only, i realize my mistake, another simple way to say it
Comparing Leaf nodes :-
B tree :- (p-1) . (keys + Record pointer) + (p) (Block pointer)
B+ tree :- (p-1) (keys+Record Pointer) + (1) (Block pointer)
∴ B+ tree can accommodate more no.of keys than B tree ===> More levels in B tree
Comparing Internal Nodes :-
B tree takes the size :- (p-1) . (keys + Record pointer) + (p) (Block pointer)
B+ tree takes the size :- (p-1) (keys) + (p) (Block pointer)
Note that one key is repeated in B+ tree but it much smaller than the size of ( (p-1).Record Pointer ) of B tree
∴ B+ tree can accommodate more no.of keys than B tree ===> More levels in B tree
By these two points we an say B tree has more no.of levels compare to B+ tree to accommodate n keys.