The two nearest harmonics of a tube closed at one end and open at the other end are 220Hz and 260Hz. What is the fundamental frequency of the system?

Frequency of nth harmonic vibration in closed pipe Vn = (2n-1)v/4L = 220
Frequency of (n+1)th harmonic vibration Vn+1 = [2(n+1)-1]v/4L = [(2n+1)v]/4L = 260Hz
But Vn+1 - Vn = 260 - 220 = 40
[(2n+1) - (2n-1)]v/4L = 40
2v/4L = 40
v/2L = 40
v/L = 80Hz
So the fundamental frequency of the system is 20Hz (since v/4L = fundamental frequency, and v/L = 80Hz, then v/4L = 80/4 = 20Hz).
Hence C is the correct answer.