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KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    A solid sphere released from rest from the top of an inclined plane of inclination\[{{\theta }_{1}}\], rolls without sliding and reaches the bottom with speed \[{{v}_{1}}\] and its time of descent is\[{{t}_{1}}\]. The same sphere is then released from rest from the top of another inclined plane of inclination\[{{\theta }_{2}}\] but of the same height, rolls without sliding and reaches the bottom with speed \[{{v}_{2}}\] and its time of descent is \[{{t}_{2}}\]. If \[{{\theta }_{2}}>{{\theta }_{1}}\], then

    A) \[{{v}_{2}}>{{v}_{1}};{{t}_{2}}<{{t}_{1}}\]

    B) \[{{v}_{2}}={{v}_{1}};{{t}_{2}}<{{t}_{1}}\]

    C) \[{{v}_{2}}<{{v}_{1}};{{t}_{2}}>{{t}_{1}}\]

    D) \[{{v}_{2}}={{v}_{1}};{{t}_{2}}={{t}_{1}}\]

    Correct Answer: B

    Solution :

    In rolling without slipping, the mechanical energy is conserved. Since both the inclined planes are of the same height\[{{v}_{2}}={{v}_{1}}\]. The acceleration of the sphere rolling down the plane is
    \[a=\frac{g\sin \theta }{1+\frac{I}{M{{R}^{2}}}}\]
    Since\[{{\theta }_{2}}>{{\theta }_{1}}\,\,;\,\,{{a}_{2}}>{{a}_{1}}\]. Hence
    \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{2}}}{\sin {{\theta }_{1}}}\]
    Thus\[{{t}_{1}}>{{t}_{2}}\]. Hence the correct choice is [B].


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