Waves

JEE Main level, single correct, medium

100 of 142

Question 100 of 142

A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512Hz512Hz produces first resonance when the tube is filled with water to a mark 11cm11cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256Hz256Hz which produces first resonance when water reaches a mark 27cm27cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to

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Answer: (D) 328ms−1328m{s}^{-1}

λ14=11+e⇒v512×4=11+e...(1)\dfrac{{\lambda }_{1}}{4}=11+e\Rightarrow \dfrac{v}{512\times 4}=11+e...(1)

λ24=27+e⇒v256×4=27+e...(2)\dfrac{{\lambda }_{2}}{4}=27+e\Rightarrow \dfrac{v}{256\times 4}=27+e...(2)

Subtracting the equations,

v256×4×12=16cm=0.16m\dfrac{v}{256\times 4}\times \dfrac{1}{2}=16cm=0.16m

⇒v=0.16×2×4×256\Rightarrow v=0.16\times 2\times 4\times 256

=328ms−1=328m{s}^{-1}

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