SATHEE: Wave Motion 2 Question 28

Wave Motion 2 Question 28

28. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency $500 H z$ is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length $50.7 c m$ and $83.9 c m$ . Which of the following statements is (are) true?

(a) The speed of sound determined from this experiment is $332 m s^{- 1}$ .

(b) The end correction in this experiment is $0.9 c m$ .

(d) The resonance at $50.7 c m$ corresponds to the fundamental harmonic.

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Answer:

Correct Answer: 28. $(a, c)$

Solution:

Let $n$ th harmonic is corresponding to $50.7 c m$ and $(n + 1)$ th harmonic is corresponding $83.9 c m$ .

$∴$ Their difference is $\frac{λ}{2}$ .

$\begin{aligned} ∴ \frac{λ}{2} = (83.9 - 50.7) c m \\ or λ = 66.4 c m \\ ∴ \frac{λ}{4} = 16.6 c m \end{aligned}$

Length corresponding to fundamental mode must be close to $\frac{λ}{4}$ and $50.7 c m$ must be an odd multiple of this length. $16.6 \times 3 = 49.8 c m$ . Therefore, 50.7 is 3rd harmonic.

If end correction is $e$ , then

$\begin{aligned} e + 50.7 = & \frac{3 λ}{4} \\ e = 49.8 - 50.7 = - 0.9 c m \end{aligned}$

$∴$ Speed of sound, $v = f λ$

$\Rightarrow v = 500 \times 66.4 c m / s = 332 m / s$

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Wave Motion 2 Question 28

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Wave Motion 2 Question 28

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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