NEET2019PhysicsWaves and SoundActual
A tuning fork with frequency 800 ~Hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at length 9.75 ~cm , 31.25 ~cm and 52.75 ~cm . The speed of sound in air is
Options
- A500 ~m / s
- B156 ~m / s
- C344 ~m / s
- D172 ~m / s
Correct answer
C. 344 ~m / s
Step-by-step solution
For vibrating tunning fork over a resonance tube, the first resonance is obtained at the length l₁= 4 and for second resonance, l ₂= 4 + 2 = 3 4 From Eq. (i) and (ii), we get array cc & l ₂- l ₁= 3 4 - 4 = 2 & =2 (l₂-l₁ ) & v =2 f ( l ₂- l ₁ ) array ..(iii) ( = v f ) Here, f =800 ~Hz , l ₁=9.75 ~cm , l ₂=31.25 ~cm Substituting the given values in Eq. (iii), we get aligned & v =2 800(31.25-9.75) & =34400 ~cm / s =344 ~m / s aligned