Explanation

(d) The various forces acting on the particle, are its weight mg acting vertically downwards, normal reaction N. Equating the vertical forces, we have
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$N\sin{\theta = mg\ \ldots(i)}$
Also, centripetal force,
$\frac{{mv}^{2}}{R} = N\cos\theta\ldots.(ii)$
From Eqs. (i) and (ii), we get
$\tan{\theta = \frac{Rg}{v^{2}}}\ldots.(iii)$
Also, from triangle $OAB,$
$\tan{\theta = \frac{R}{h}\ \ldots..(iv)}$
Equating Eqs. (iii) and (iv), we get
$h = \frac{{v\ }^{2}}{g}$
$Given,\ \ \ v = 0.5{\ ms}^{- 1}\ and\ g = {10\ ms}^{- 2}$
$\therefore\ \ h\frac{{(0.5)}^{2}}{10} = 0.025\ m = 2.5\ cm$