A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xy-plane. The coin is tossed vertically up (+z direction) by applying an impulse of $\sqrt{\frac{π}{2} \times 10^{- 2}} N - s$ at a distance 2/3 cm from its center. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is __________.
[Given: The acceleration due to gravity g = 10 ms⁻²]

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answer is 30.

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Detailed Solution

$\begin{array}{r} J . r = \frac{{mR}^{2}}{4} \cdot w and J = {mV}_{cm} \\ time t = \frac{2 V_{cm}}{g} = \frac{2 \times J}{m \times g} \end{array}$
$\begin{array}{l} n . 2 π = ωt \\ \Rightarrow n = \frac{ωt}{2 π} = \frac{ω \times 2 \times J}{m \times g \times 2 π} = \frac{4 Jr \times 2 \times J}{{mR}^{2} \times mg \times 2 π} \\ \frac{8 J^{2} r}{{mR}^{2} \times mg \times 2 π} = \frac{2 r}{{mR}^{2} \times mg} \\ = \frac{2 \times \frac{2}{3} \times 10^{- 2}}{25 \times 10^{- 6} \times \frac{16}{9} \times 10^{- 4} \times 10} \\ = 30 \end{array}$

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