According to the Law of Conservation of Energy, the potential energy (PE) at the highest point is converted entirely into kinetic energy (KE) at the lowest point, assuming no energy loss. Potential Energy at the highest point ($PE_{top}$) = $mgh$ Kinetic Energy at the lowest point ($KE_{bottom}$) = $\frac{1}{2}mv^2$ By Conservation of Energy: $PE_{top} = KE_{bottom}$ $mgh = \frac{1}{2}mv^2$ We can cancel 'm' from both sides: $gh = \frac{1}{2}v^2$ Rearranging for $v$: $v^2 = 2gh \implies v = \sqrt{2gh}$ Given: Height ($h$) = 0.2 m Acceleration due to gravity ($g$) = $10 \ m/s^2$ $v = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \ m/s$. Therefore, the speed of the pendulum bob at its lowest point is 2 m/s. Options A, C, and D are incorrect based on the conservation of energy principle and calculation.