BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    A string of length \[49\,\pi \,m/s\] is fixed at one end and carries a mass \[17.5\,m/s\] at the other end. The string makes \[{{\omega }_{1}}\] rev/s around the vertical axis through the fixed end as shown in the figure, the tension in the string is:                                                    [BHU PMT-2002]                                                  

    A)                  \[{{\omega }_{2}}\]                                       

    B)                  \[{{\omega }_{2}}-{{\omega }_{1}}\]

    C)                  \[{{\omega }_{1}}:{{\omega }_{2}}\]                                     

    D)                  \[\sqrt{{{\omega }_{1}}}:\sqrt{{{\omega }_{2}}}\]

    Correct Answer: A

    Solution :

                     Key Idea: Outward force acting on the mass is known as centrifugal force.                 An object travelling in a circle behaves as if it is experiencing an outward force. This force is known as centrifugal force, which actually does not exist and is given by                 \[v\]                 In this case, \[\lambda \]                 \[\phi \]\[y=7\,\,\sin \left( 7\,\,\pi t-0.4\,\,\pi x+\frac{\pi }{3} \right)\]                 \[\frac{2\pi }{\lambda }v=7\pi \]\[\frac{2\pi }{\lambda }=0.4\pi \]                 \[\Rightarrow \]               \[v=\frac{7\pi }{0.4\,\pi }=17.5\,\,m/s\]                 \[\left( \Delta U \right)\]              \[\left( W \right)\]


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