In resonance column experiment a tuning fork of frequency f = 400 Hz is held above the pipe as shown in figure. The reservoir is raised and lowered to change the level of water and thus the length of

Question

In resonance column experiment a tuning fork of frequency $f = 400 Hz$ is held above the pipe as shown in figure. The reservoir is raised and lowered to change the level of water and thus the length of the column of air in the tube. The area of cross section of the reservoir is 6 times that of the pipe. Initially, the reservoir is kept so that the pipe is full up to the brim. Tuning fork is sounded and the reservoir is lowered. When the reservoir is lowered by 21 cm,first resonance is recorded. When the reservoir is lowered further by 49 cm the second resonance is heard. Find the speed of sound (in $m / s$ ) in air.

TestHub · Accepted Answer

When reservoir is lowered by x , let the level of water fall by y

$x - \frac{y}{6} = y ∴ x = \frac{7 y}{6} \Rightarrow y = \frac{6 x}{7}$

For $x = 21 cm, y_{1} = 18 cm$

For $x = 21 + 49 = 70 cm; y_{2} = 60 cm$ $∴ \frac{λ}{4} + e = 18$

$\frac{3 λ}{4} + e = 60$

(ii) - (i) gives

$\frac{λ}{2} = 42 \Rightarrow λ = 84 cm = 0.84 m$

$∴ V = λ f = 0.84 \times 400 = 336 ms^{- 1}$

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JEE Advanced 2026