A tuning fork having a frequency of 340 Hz is vibrated just above a cylindrical tube. The height of the tube is 120 cm. Water is slowly poured in it. The minimum height of water required for resonance is 15 (x) cm. Then x is? (Velocity of sound in air = 340 m/s)

In a closed tube, the first resonance takes place, when the length of air column $=\lambda /4$   
The wavelength of sound in air
   $\lambda =\frac{v}{n}=\frac{340}{340}$
 $=1m[Asv=340m/sandn=340]$  
 Here length of first resonance can be given as
      $\begin{array}{l}\frac{\lambda }{4}=0.25m\\ =25cm\end{array}$
The length of second resonance
   $\begin{array}{l}\frac{3\lambda }{4}=\frac{3}{4}m\\ =75cm\end{array}$
The length of third resonance
   $\begin{array}{l}\frac{5\lambda }{4}=\frac{5}{4}m\\ =125cm\end{array}$
    As the length of the tube is only 120 cm, hence third resonance is not possible.
    Therefore, the minimum height of water required for resonance = 120-75=45 cm