Let the mass of the iron ball be $m = 100 \text{ g} = 0.1 \text{ kg}$ and its velocity be $v = 10 \text{ m/s}$. The ball collides with the wall at an angle $\theta = 30^\circ$ with the wall. The component of velocity perpendicular to the wall before the collision is $v \sin 30^\circ$. Since the ball rebounds with the same angle, the component of velocity perpendicular to the wall after the collision is $-v \sin 30^\circ$ (assuming the collision is elastic). The magnitude of the change in momentum of the ball along the normal is: $\Delta p = m(v \sin 30^\circ - (-v \sin 30^\circ)) = 2mv \sin 30^\circ$ Substituting the given values: $\Delta p = 2 \times 0.1 \times 10 \times \sin 30^\circ = 2 \times 1 \times 0.5 = 1 \text{ kg m/s}$ The force experienced by the wall is equal to the rate of change of momentum (according to Newton's second law of motion) : $F = \frac{\Delta p}{\Delta t}$ Given the period of contact $\Delta t = 0.1 \text{ s}$: $F = \frac{1}{0.1} = 10 \text{ N}$ Thus, the force experienced by the wall is $10 \text{ N}$.