For a tube closed at one end, in its lowest frequency mode, the closed end is a node and the open end is a loop with no other nodes in between. Therefore the length L of the tube is 1/4 wavelength.

l = 4L

For the shorter tube: l 1 = 4X0.8 m = 3.2 m and its frequency:

f1 = v/l 1 = 340 m/s/3.2 m = 106 Hz

For the longer tube the fundamental frequency is 106 - 17 = 89 Hz

l 2 = v/f2 = 340/89 = 4L2

Therefore L2 = 0.95 m

Answer is C

Return