JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    DIRECTION (Qs. 54): Read the following passage and answer the questions that follows:
    A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at \[x=2m\] is found at its mean position and it is moving in positive y-direction at \[t=1\] sec. The amplitude of the wave, the wavelength and the angular frequency of the wave are \[0.1\] meter, \[\pi /2\] meter and \[2\pi \] rad/sec respectively.
    The instantaneous power transfer through \[x=2m\] and \[t=1.25\text{ }sec,\] is

    A)  \[10\,J/s\]                                 

    B)  \[~4\pi /3\text{ }\,\,J/s\]

    C)  \[2\pi /3\text{ }\,\,J/s\]                    

    D)   \[0\]

    Correct Answer: D

    Solution :

    Time period of oscillation\[T=\frac{2\pi }{\omega }=\frac{2\pi }{2\pi }=1\sec .\]. second, the particle is at rest at extreme position. Hence instantaneous power at \[x=2\] at t = 1.25 sec is zero.


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