One Square and one straight line(pipe) Edited:

A farmer has a farm.The farm is a square whose sides have length 1. A single straight water pipe passes somewhere under the farm with depth one meter. He wants to dig furrows to find the pipe. The objective is to minimize the total length of furrows that are guaranteed to cross the pipe.
  The furrows that he makes may be a set of line segments.

One solution of the problem is the diagonals of square but isn't minimum. 

 A: Well, I have a candidate for the smallest configuration (but can not prove it is optimal).  It does, however, beat the two diagonals.
Represent the square in the coordinate plane, with vertices at $$A = (-1,-1) \; B = (-1,1) \;C = (1, 1)\; D=  (1, -1)$$  Thus the sides have length 2 (that will simplify later calculations).  Now draw a horizontal segment on the x-axis from $$P = (-L, 0) \; to \; Q = (L,0)$$ so this segment has length 2L.  Now connect A and B to P and C and D to Q. The final figure looks like a capital H with angled sides.
[NOTE:  I tried to draw this figure here, but failed.  Sorry.]
Now, the length of my figure is a function of L, specifically: $$f(L) = 2L + 4\sqrt{1 + (1-L)^2}.$$  We seek L for which this is minimal.  It is easy to differentiate f and in so doing we see that the minimum comes when L = $\frac{1}{\sqrt{3}}$.  The resulting configuration has total length ~ 5.464 whereas the two diagonals have total length ~ 5.657.  So the angled H is ahead, if not by much.
I have no idea at all how to decide if this configuration is minimal.
