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Waves question

2019 · 12 Apr · Shift 2 · Q45
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Waves question

2019 · 12 Apr · Shift 2 · Q45

JEE MainPhysicsWavesMCQ+4 / −1
A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, v is equal to -
  1. A
    338 ms–1
  2. B
    384 ms–1
  3. C
    379 ms–1
  4. D
    332 ms–1
View written solutionFree

Correct answer: B

  1. Use resonance tube relation

For a resonance tube closed at one end, successive resonant lengths differ by

λ2\frac{\lambda}{2}2λ​

where λ\lambdaλ is the wavelength of sound.

So,

l2−l1=λ2l_2-l_1=\frac{\lambda}{2}l2​−l1​=2λ​

Given:

l1=30 cm,l2=70 cml_1=30\text{ cm},\quad l_2=70\text{ cm}l1​=30 cm,l2​=70 cm

Thus,

l2−l1=70−30=40 cm=0.40 ml_2-l_1=70-30=40\text{ cm}=0.40\text{ m}l2​−l1​=70−30=40 cm=0.40 m

Hence,

λ2=0.40⇒λ=0.80 m\frac{\lambda}{2}=0.40 \Rightarrow \lambda=0.80\text{ m}2λ​=0.40⇒λ=0.80 m
  1. Apply wave speed formula

We know,

v=fλv=f\lambdav=fλ

Given frequency,

f=480 Hzf=480\text{ Hz}f=480 Hz

So,

v=480×0.80=384 m s−1v=480\times 0.80=384\text{ m s}^{-1}v=480×0.80=384 m s−1
  1. Match with options
v=384 m s−1v=384\text{ m s}^{-1}v=384 m s−1

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer is B, which matches the derived answer.

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