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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The disturbance y (x, t) of a wave propagating in the positive x-direction is given by \[y=\frac{1}{1+{{x}^{2}}}\] at time t = 0 and by \[y=\frac{1}{\left[ 1+{{\left( x-1 \right)}^{2}} \right]}\]at t = 2 s, where x and y are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of wave in m/s is.   JEE Main Online Paper (Held On 26-May-2012)  

    A) 2.0                                         

    B) 4.0

    C) 0.5                                         

    D) 1.0

    Correct Answer: C

    Solution :

    The equation of wave at any time is obtained by putting \[X=x-vt\] \[y=\frac{1}{1+{{x}^{2}}}=\frac{1}{1+{{(x-vt)}^{2}}}\]  ..........(i) We know at t = 2 sec, \[y=\frac{1}{1+{{(x-1)}^{2}}}\] .........(ii) On comparing (i) and (ii) we get \[vt=1\] \[V=\frac{1}{t}\] As t = 2 sec \[\therefore \]\[V=\frac{1}{2}=0.5m/s.\]


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