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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    A sound wave of frequency 5Hz, has velocity 360 m/s. If phase difference between two particles is\[60{}^\circ \], then path difference will be:

    A)  1.2m                    

    B)         12m

    C)  2.1 m                   

    D)         0.21m

    Correct Answer: B

    Solution :

    Phase difference\[\frac{2\pi }{\lambda }\] path difference        \[\phi =\frac{2\pi }{\lambda }\Delta x\]         \[\phi =\frac{2\pi n}{\upsilon }\Delta x\]                 \[\Rightarrow \]                       \[\frac{\pi }{3}=\frac{2\pi \times 5}{360}\times \Delta x\]                                        \[\Delta x=12m\]


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