AMU Medical AMU Solved Paper-2004

  • question_answer
    A uniform rope of mass 0.1 kg and length 2.5 m hangs from ceiling. The speed of transverse wave in the rope at upper end and at a point 0.5 m distance from lower end will be

    A)  \[\text{5 m}/\text{s},\text{ 2}.\text{24 m}/\text{s}\]

    B)  \[\text{1}0\text{ m}/\text{s},\text{ 3}.\text{23 m}/\text{s}\]

    C)  \[\text{7}.\text{5 m}/\text{s},\text{ 1}.\text{2 m}/\text{s}\]    

    D)  None of the above

    Correct Answer: A

    Solution :

                     Velocity of transverse wave along the string is                 \[{{v}_{T}}=\sqrt{\frac{T}{m}}\] where m is mass per unit length, T the tension.                 Also       \[T=mg\],            \[m=\frac{m}{x}\] \[\therefore \]  \[{{v}_{T}}=\sqrt{\frac{m\,gx}{m}}=\sqrt{gx}\] Given,   \[g=10\,m/{{s}^{2}}\],   \[x=0.5\,m\] \[\therefore \]  \[{{v}_{T}}=\sqrt{10\times 0.5}\]                 \[=\sqrt{5}=2.24\,m/s\] At upper end, the velocity is given by                                 \[v=\sqrt{rg}\]                                 \[=\sqrt{2.5\times 10}=5\,m/s\]


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