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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    A tennis ball (treated as hollow spherical shell) starting from O rolls down a hill. At point A the ball becomes air borne leaving at an angle of \[{{30}^{0}}\] with the horizontal. The ball strikes the ground at B. What is the value of the distance AB?                 (Moment of inertia of spherical shell of mass m and radius R about its diameter \[=\frac{2}{3}{{\operatorname{mR}}^{2}})\]   JEE Main  Online Paper (Held On 22 April 2013)

    A)  1.87 m                  

    B)  2.08  m

    C)  1.57 m                  

    D)  1.77 m

    Correct Answer: B

    Solution :

     Velocity of the tennis ball on the surface of the earth or ground \[v=\sqrt{\frac{2gh}{1+\frac{{{k}^{2}}}{{{R}^{2}}}}}\] (where k = radius of gyration of spherical shell\[=\sqrt{\frac{2}{3}}R\]) Horizontal range \[AB=\frac{{{v}^{2}}\sin 2\theta }{g}\] \[=\frac{{{\left( \sqrt{\frac{2gh}{1+{{k}^{2}}/{{R}^{2}}}} \right)}^{2}}\sin (2\times {{30}^{o}})}{g}\]


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