To prevent the plank from tipping, the torque caused by the person's weight must be balanced by the torque due to the plank's weight. The torque exerted by the person is the weight of the person multiplied by the distance from the support. Let's denote the distance from the support where the person stands as \( x \) meters.
The torque exerted by the person is \( 800 \, \text{N} \times x \, \text{m} \). The torque exerted by the plank is the weight of the plank multiplied by half of its length, since the supports are placed 1.0 m from each end.
Given that the plank's length is 10 m and its weight is 100 N, the torque exerted by the plank is \( 100 \, \text{N} \times \frac{10}{2} = 500 \, \text{Nm} \).
To prevent tipping, the torques must balance:
\[ 800 \, \text{N} \times x = 500 \, \text{Nm} \]
Solving for \( x \):
\[ x = \frac{500 \, \text{Nm}}{800 \, \text{N}} = 0.625 \, \text{m} \]
So, the person can stand up to 0.625 m from one end of the plank without causing it to tip.
Among the given options, the closest one to 0.625 m is \( \textbf{(c) 0.5 m} \).
