The period of a physical pendulum is $T = 2\pi \sqrt{I/(mgh)}$, where $h$ is the distance from the pivot to the center of mass. For a uniform rod of length $L$ pivoted at one end, the moment of inertia is $I = \frac{1}{3}mL^2$ and the center of mass is at $h = L/2$.

[Q1] Substituting $I$ and $h$ into the period equation:

$$T = 2\pi \sqrt{\frac{\frac{1}{3}mL^2}{mg(L/2)}} = 2\pi \sqrt{\frac{2L}{3g}}$$

Solving for the length $L$:

$$L = \frac{3gT^2}{8\pi^2} = \frac{3(9.8 \text{ m/s}^2)(1.5 \text{ s})^2}{8\pi^2} = 0.84 \text{ m}$$

[Q2] By conservation of energy, the maximum kinetic energy $K_{max}$ equals the maximum change in potential energy, which occurs between the lowest point and the point of maximum angular displacement $\theta_m$.

$$K_{max} = \Delta U_g = mgh(1 - \cos\theta_m) = mg\frac{L}{2}(1 - \cos\theta_m)$$ $$K_{max} = (0.50 \text{ kg})(9.8 \text{ m/s}^2)\frac{0.84 \text{ m}}{2}(1 - \cos 10^\circ) = 0.031 \text{ J}$$