JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Waves & Sound

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In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm . If the velocity of sound in air is

330 m / s

, the tuning fork frequency is:

(A)1100 Hz
(B)2200 Hz✓ Correct
(C)3300 Hz
(D)550 Hz

Answer: (B)

Solution

In a tube closed by the water surface, consecutive resonances with the same fork differ by exactly half a wavelength of the air column. Raising the water from

17.0

cm to

24.5

cm shortens the air column by

7.5

cm, so

\frac{λ}{2} = 7.5

cm, giving

λ = 15

= 0.15

m. The frequency is then

f = \frac{v}{λ} = \frac{330}{0.15} = 2200

Hz.

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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