Determine the time period of the particle's motion as depicted in...

The time taken by the particle to slide down an incline of angle $\theta$ from a height $h$ is given by $t = \sqrt{\dfrac{2s}{a}}$, where $s = \dfrac{h}{\sin \theta}$ and $a = g \sin \theta$.

Thus, $t = \sqrt{\dfrac{2h}{g \sin^2 \theta}} = \dfrac{1}{\sin \theta} \sqrt{\dfrac{2h}{g}}$.

The total time period $T$ of the motion consists of sliding down the first incline, sliding up the second incline, sliding down the second incline, and sliding up the first incline.

Therefore, $T = 2(t_1 + t_2) = 2 \sqrt{\dfrac{2h}{g}} \left( \dfrac{1}{\sin \theta_1} + \dfrac{1}{\sin \theta_2} \right)$.

Given $h = 10 \text{ cm} = 0.1 \text{ m}$, $\theta_1 = 45^\circ$, $\theta_2 = 60^\circ$, and taking $g = 9.8 \text{ m/s}^2$:

$T = 2 \sqrt{\dfrac{2 \times 0.1}{9.8}} \left( \dfrac{1}{\sin 45^\circ} + \dfrac{1}{\sin 60^\circ} \right)$

$T = 2 \sqrt{\dfrac{1}{49}} \left( \sqrt{2} + \dfrac{2}{\sqrt{3}} \right)$

$T = \dfrac{2}{7} \left( 1.414 + 1.155 \right)$

$T = \dfrac{2}{7} (2.569) \approx 0.734 \text{ s}$.

The closest value is $0.73 \text{ s}$.