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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    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]

    A) \[335m\,{{s}^{-1}}\]                           

    B) \[322m\,{{s}^{-1}}\]

    C) \[328m\,{{s}^{-1}}\]               

    D)   \[341m\,{{s}^{-1}}\]

    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) \[\frac{v}{4\times 16\times {{10}^{-2}}}=512\Rightarrow v\simeq 328m\,{{s}^{-1}}\]


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