Waves

JEE Main level, single correct, medium

108 of 142

Question 108 of 142

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} )

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Answer: (B) 93.7593.75

Given : Frequency of tuning fork, n=264 Hzn=264 \mathrm{~Hz}
Length of column L=L= ?
For closed organ pipe
n=v4l⇒l=v4n=3304×264=0.3125 or, l=0.3125×100=31.25 cm\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}
In case of closed organ pipe only odd harmonics are possible.
Therefore value of ll will be (2n−1)l(2 n-1) l Hence option (b) i.e. 3×31.25=93.75 cm3 \times 31.25=93.75 \mathrm{~cm} is correct.

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