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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      The equation of a wave on a string of linear mass density\[0.04\text{ }kg\text{ }{{m}^{1}}\]is given by\[y=0.02\] (m) \[\sin \left[ 2\pi \left( \frac{t}{0.04(s)}-\frac{x}{0.50(m)} \right) \right]\].The tension in the string is ?       AIEEE  Solved  Paper-2010

    A) 6.25 N  

    B)        4.0 N     

    C)        12.5 N  

    D)        0.5 N

    Correct Answer: A

    Solution :

    Putting\[\omega =\frac{2\pi }{.04},k=\frac{2\pi }{0.5}\] in equation \[T=\mu {{v}^{2}}=\mu {{\left( \frac{\omega }{k} \right)}^{2}}\] \[=6.25\text{ }N\]


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