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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 16.6 cm                        

    B) 10.0 cm

    C) 20.0 cm            

    D)   33.3 cm

    Correct Answer: C

    Solution :

    Speed of wave \[=\text{ }\sqrt{\frac{T}{\mu }}=\text{ }\sqrt{\frac{\frac{8}{5\times {{10}^{-3}}}}{1}}\] \[=\,\,\sqrt{\frac{8\times {{10}^{3}}}{5}}\,=\,40\,m/s\] \[v=F\lambda \] \[\lambda \,\,=\,\,\frac{40}{100\,\,}\,metre\] Distance between two consecutive node \[=\frac{\lambda }{2}=\frac{20}{100}\,\,metre\,\,=\,\,20\,cm\]


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