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JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2013
- Hard
- MCQ
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 \(30^o\) with the horizontal. The ball strikes the ground at \(B\). What is the value of the distance \(AB\) ? (Moment of inertia of a spherical shell of mass \(m\) and radius \(R\) about its diameter .......... \(m\). \(= \frac {2}{3}\,mR^2\))

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Answer & Solution
Correct Answer
(B) \(2.08\)
Step-by-step Solution
Detailed explanation
Velocity of the tennis ball on the surface of the earth or ground
\(\begin{array}{l}
v = \sqrt {\frac{{2gh}}{{1 + \frac{{{K^2}}}{{{R^2}}}}}} \,\,(where\,k = radius\,of\,gyration\\
of\,spherical\,shell\, = \sqrt {\frac{2}{3}} R)\\
Horizontal\,range\,AB = \frac{{{v^2}\sin 2\theta }}{g}\\
= \frac{{{{\left( {\sqrt {\frac{{2gh}}{{1 + {k^2}/{R^2}}}} } \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)}}{g}
\end{array}\)
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