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(A) Let the initial speed of the block be $v$ and the height of the track be $h$. By the law of conservation of energy,the speed $v'$ of the block at

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(A) Let the initial speed of the block be $v$ and the height of the track be $h$. By the law of conservation of energy,the speed $v'$ of the block at the highest point of the track is given by:$\frac{1}{2}mv^2 = \frac{1}{2}m(v')^2 + mgh$$v' = \sqrt{v^2 - 2gh}$Since $v$ and $h$ are the same for all tracks,the speed $v'$ at the highest point is the same for all tracks.At the highest point,the forces acting on the block are the normal reaction $N$ (downwards) and the weight $mg$ (downwards). These provide the necessary centripetal force:$N + mg = \frac{m(v')^2}{r}$$N = \frac{m(v')^2}{r} - mg$where $r$ is the radius of curvature at the highest point.For $N$ to be maximum,the term $\frac{m(v')^2}{r}$ must be maximum. Since $m$ and $v'$ are constant,$N$ is maximum when the radius of curvature $r$ is minimum.Comparing the four tracks,the track in option $A$ has the smallest radius of curvature at its highest point. Therefore,the normal reaction is maximum in track $A$.

$A$ small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track,the normal reaction is maximum in

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