JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    A cylindrical tube, open at both ends, has a fundamental frequncy, f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :   AIEEE  Solved  Paper-2012

    A) \[f\]                                        

    B) \[f/2\]

    C) \[3f/4\]                   

    D) \[2f\]

    Correct Answer: A

    Solution :

                 \[f=\frac{v}{2\ell }\]now, it will act like one end opend and other closed. so, \[{{f}_{0}}=\frac{v}{4\ell '}=\frac{v}{4\frac{\ell }{2}}=\frac{v}{2\ell }=f\]


You need to login to perform this action.
You will be redirected in 3 sec spinner