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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2018 · Paper 2 · Physics · Waves and Sound

In an experiment to measure the speed of sound by a resonating air...

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500Hz500Hz 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 50.7cmand83.9cm50.7cmand83.9cm . Which of the following statements is (are) true?

Options

  1. AThe speed of sound determined from this experiment is 332ms−1332m{s}^{- 1}
  2. BThe end correction in this experiment is 0.9cm0.9cm
  3. CThe wavelength of the sound wave is 66.4cm66.4cm
  4. DThe resonance at 50.7cm50.7cm corresponds to the fundamental harmonic

Correct answer

A. The speed of sound determined from this experiment is 332ms−1332m{s}^{- 1}

Step-by-step solution

Let n1{n}_{1} harmonic is corresponding to 50.7cmandn250.7cmand{n}_{2} harmonic is corresponding 83.9cm83.9cm . since both one consecutive harmonics.
∴∴ There difference =λ2= \dfrac{\lambda}{2}
∴λ2=(83.9−50.7)cm∴ \dfrac{\lambda}{2} = \left(83.9 - 50.7\right)cm
λ2=33.2cm\dfrac{\lambda}{2} = 33.2cm
λ=66.4cm\lambda = 66.4cm
∴λ4=16.6cm∴ \dfrac{\lambda}{4} = 16.6cm
length corresponding to fundamental mode must be close to λ4and50.7cm\dfrac{\lambda}{4}and50.7cm must be an odd multiple of this length 16.6×3=49.8cm16.6 \times 3 = 49.8cm . therefore 50.7is3rd50.7is{3}^{rd} harmonic
If end correction is e, then
e+50.7=3λ4e + 50.7 = \dfrac{3\lambda}{4}
e=49.8−50.7=0.9cme = 49.8 - 50.7 = 0.9cm
Speed of sound, c=fλc = f\lambda
∴v=500×66.4cm/sec=332.000m/sec∴ v = 500 \times 66.4cm / sec = 332.000m / sec
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