A thin circular coin of mass 5gm and radius (4)/(3)cm is initially...

By impulse $-$ momentum theorem:

$J = M{V}_{CM}$

$⇒ {V}_{CM} = \dfrac{J}{M} = \dfrac{\sqrt{\dfrac{\pi}{2}}}{100 \times \dfrac{5}{1000}} = \sqrt{2\pi}m{s}^{- 1}$

$⇒$ Total time of journey $= \dfrac{2{V}_{CM}}{g} = \dfrac{2}{g} \times \sqrt{2\pi}$

$⇒ ∆ t = \dfrac{\sqrt{2\pi}}{5}s$

Also, by angular impulse$-$ momentum theorem:

$J \times \dfrac{R}{2} = I\omega ⇒ J \times \dfrac{R}{2} = \left[\dfrac{M{R}^{2}}{4}\right]\omega$

$⇒ \omega = \dfrac{J \times \dfrac{R}{2}}{\dfrac{M{R}^{2}}{4}} = \dfrac{J}{MR} \times 2$

$= \dfrac{\left(\dfrac{\sqrt{\dfrac{\pi}{2}}}{100}\right) \times 2}{\left(\dfrac{5}{1000}\right) \times \left(\dfrac{4}{3} \times \dfrac{1}{100}\right)}$

$= 150\sqrt{2\pi}rad{s}^{- 1}$

$⇒$ Number of rotations $= \dfrac{\omega ∆ t}{2\pi}$

$= \dfrac{\left(150\sqrt{2\pi}\right) \times \left(\dfrac{\sqrt{2\pi}}{5}\right)}{2\pi} = 30$

Therefore, $n = 30$.