A tuning fork of frequency 340 Hz is vibrated just above the tube of 120 cm height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance (speed of sound in the air = 340 m/sec)

Because the tuning fork is in resonance with air column in the pipe closed at one end, the frequency is $\mathrm{n}=\frac{(2\mathrm{N}-1)\mathrm{v}}{4\mathrm{l}}$ where N = 1, 2, 3 .... corresponds to different mode of vibration 
putting n = 340 Hz, v = 340 m/s, the length of air column in the pipe can be
$\mathrm{l}=\frac{(2\mathrm{N}-1)340}{4\times 340}=\frac{(2\mathrm{N}-1)}{4}\mathrm{m}=\frac{(2\mathrm{N}-1)\times 100}{4}\mathrm{cm}$ 
For N = 1, 2, 3, ... we get l = 25 cm, 75 cm, 125 cm ...
As the tube is only 120 cm long, length of air column after water is poured in it may be 25 cm or 75 cm only, 125 cm is not possible, the corresponding length of water column in the tube will be (120 – 25) cm = 95 cm or (120 – 75) cm = 45 cm.
Thus minimum length of water column is 45 cm.