Subodh
As you should be aware in the wildcard mask a binary 0 is a bit that must match and a binary 1 is a bit that may vary.
So in your mask of 0.0.0.32 the first 3 octets must match exactly (it must be 10.1.1). The mask of the fourth octet has a single binary 1. It may help to write out the 4 octet in binary ( 00100000). So for this mask there are exactly 2 values of the address that will match the mask. These values are 10.1.1.0 and 10.1.1.32.
As your comment indicates this mask is quite unusual in an access list. It is much more common to have the mask be the inverse of common subnet masks (such as the mask .31 which is the inverse of mask 224).
HTH
Rick
HTH
Rick