Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    Mass of 2 kg is moving with a string in horizontal circle with angular velocity 5 cycles/min keeping the radius constant, tension in string is doubled. Now, the angular velocity of the mass will be

    A)  14 cycles/min    

    B)  10 cycles/min

    C)  2.25 cycles/min  

    D)  7 cycles/min

    Correct Answer: D

    Solution :

     Tension in the string should be equal to centripetal force \[T=\frac{m{{v}^{2}}}{r},\] where, \[m=\]mass of body \[r=\]radius of circular path \[v=\]linear velocity of body m and r are constant. Hence,        \[v\propto \sqrt{T}\] \[\therefore \] \[\frac{{{v}_{2}}}{{{v}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{1/2}}={{\left( \frac{2{{T}_{1}}}{{{T}_{1}}} \right)}^{1/2}}\]  \[{{v}_{2}}=\sqrt{2}{{v}_{1}}=\sqrt{2}\times 5\] \[=5\times 1.4=7\]cycles/min


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