A tuning fork with a frequency of 800 Hz produces resonance in a resonance column tube with the upper end open and the lower end closed by the water surface. Successive resonances are observed at lengths 9.75 cm, 31.25 cm, and 52.75 cm. The speed of sound in air is,

  1. 344 m/s
  2. 172 m/s
  3. 500 m/s
  4. 156 m/s

Answer (Detailed Solution Below)

Option 1 : 344 m/s
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Detailed Solution

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EXPLANATION:

For a closed organ pipe length is given as:

L = (2n-1)\(\frac{λ }{4}\) i.e odd multiple of \(\frac{λ }{4}\)

∴ L1\(\frac{λ }{4}\)  , L2\(\frac{3λ }{4}\) , L3 \(\frac{5λ }{4}\)    -----(1)

Also, we know that the velocity of sound is given as:

v = nλ    -----(2)

Given:

Frequency, n = 800 Hz, L1 = 9.75 cm, L2 = 31.25 cm, L3 = 52.75 cm.

From equation (1) we have,

L2 - L1 = \(\frac{λ }{2}\) L3 - L2

\(\frac{λ }{2}\) = 31.25 - 9.75 = 21.50

⇒λ = 43 cm.

From equation (2) we have,

v = nλ = 800×43×10-2 

⇒v = 344 m/s.

Hence option 1) is the correct choice.

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