JEE Main 12 Jan 2019 Evening Question 19
Question: A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 335m,{s^{-1}} $
B) $ 322m,{s^{-1}} $
C) $ 328m,{s^{-1}} $
D) $ 341m,{s^{-1}} $
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Answer:
Correct Answer: C
Solution:
- Due to jagged end $ \frac{v}{4(11-x)\times {10^{-2}}}=512 $ …(i)
$ \frac{v}{4(27-x)\times {10^{-12}}}=256 $ -(ii)
From eqn. (i) and (ii) $ 2(11-x)=(27-x) $ $ x=-5cm $
From eqn. (i) $ \frac{v}{4\times 16\times {10^{-2}}}=512\Rightarrow v\simeq 328m,{s^{-1}} $
