An open tube is in resonance with a string (frequency of vibration of the tube is $n_0$). If the tube is dipped in water so that $75\%$ of the length of the tube is inside the water,then the ratio of the frequency of the tube to the string now will be

The number of possible natural oscillations of an air column in a pipe closed at one end of length $85 \, cm$ whose frequencies lie below $1250 \, Hz$ are (Velocity of sound $= 340 \, m s^{-1}$)

The fundamental frequency of a pipe is $100 \ Hz$ and the next two frequencies are $300 \ Hz$ and $500 \ Hz$. Then:

Two open organ pipes of fundamental frequencies $n_{1}$ and $n_{2}$ are joined in series. The fundamental frequency of the new pipe so obtained will be

$A$ pipe open at both ends produces a note of frequency $f_1$. When the pipe is kept with $3/4$th of its length in water,it produces a note of frequency $f_2$. The ratio $f_1/f_2$ is

$A$ resonance tube completely filled with water has a small hole at the bottom. Length of the tube is $0.8 \ m$. $A$ vibrating tuning fork of frequency $500 \ Hz$ is held near the open end of the tube. Water is slowly removed from the bottom. The maximum number of resonances heard will be (Neglect end correction. Speed of sound in air $= 340 \ m/s$).