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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Physics
Chapter
Waves and Sound

JEE Main 2012 · 26 May · Online · Physics · Waves and Sound

An air column in a pipe, which is closed at one end, will be in...

An air column in a pipe, which is closed at one end, will be in resonance wtih a vibrating tuning fork of frequency 264 Hz264 \mathrm{~Hz} if the length of the column in cm\mathrm{cm} is (velocity of sound =330 m/s=330 \mathrm{~m} / \mathrm{s} )

Options

  1. A125.00125.00
  2. B93.7593.75
  3. C62.5062.50
  4. D187.50187.50

Correct answer

B. 93.7593.75

Step-by-step solution

Given : Frequency of tuning fork, n=264HzLengthn=264 \mathrm{ Hz} Length of columnL=? For closedorganpipeclosed organ pipen=v4l⇒l=v4n=3304×264=0.3125 or, l=0.3125×100=31.25cm\begin{aligned} & n=\frac{v}{4 l} \\ & \Rightarrow l=\frac{v}{4 n}=\frac{330}{4 \times 264}=0.3125 \\ & \text{ or, } l=0.3125 \times 100=31.25 \mathrm{ cm} \end{aligned}Incaseofclosedorganpipeonlyoddharmonicsarepossible.ThereforevalueofIn case of closed organ pipe only odd harmonics are possible. Therefore value oflwillbewill be(2 n-1) lHenceoption(b)i.e.Hence option (b) i.e.3 \times 31.25=93.75 cm\mathrm{ cm} is correct.
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