MHT CET · Physics · Waves and Sound

A pipe closed at one end has length $83 cm$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 Hz$ are (velocity of sound in air $= 332 m s^{- 1}$ )

A $3$
B $4$
C $5$
D $6$

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Answer & Solution

Correct Answer

(C) $5$

Step-by-step Solution

Detailed explanation

Fundamental frequency

n = \frac{V}{4 L} = \frac{332}{4 \times 83 \times 10^{- 2}} = 100

.
Hence possible frequencies are

n = 100

n_{1} = 300

n_{2} = 500

n_{3} = 700.

n_{5} = 900

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (velocity of sound in air =332 m s-1)

Step-by-step Solution

A pipe closed at one end has length $83 cm$ . The number of possible natural oscillations of air column whose frequencies lie below $1000 Hz$ are (velocity of sound in air $= 332 m s^{- 1}$ )