Problems from IIT JEE

Problem (IIT JEE 1985): An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz, if the length of the column is, (Speed of sound = 330 m/s.)

  1. 31.25 cm
  2. 62.50 cm
  3. 93.75 cm
  4. 125 cm

Solution: Natural frequencies of a pipe of length $l$, closed at one end, are given by, $\nu=nv/(4l)$, where $v={330}\;\mathrm{m/s}$ is the speed of sound and $n$ is an odd integer. Thus, the possible length of pipe (closed at one end) that can resonate with a tuning fork of frequency $\nu={264}\;\mathrm{Hz}$ are, \begin{align} l_n=\frac{nv}{4\nu}=\frac{n(330)}{4(264)}=0.3125 n\;m, \end{align} which gives $l_1={31.25}\;\mathrm{cm}$, $l_3={93.75}\;\mathrm{cm}$, $l_5={156.25}\;\mathrm{cm}$ etc. Hence, correct options are A and C.