The correct answer is C. 5.0 mm.
The stress in a wire is given by the formula:
$stress = \frac{load}{area}$
The area of a circle is given by the formula:
$area = \pi r^2$
where $r$ is the radius of the circle.
The load is 1960 N. The maximum stress is 100 N/mm2. The radius of the wire is 5 mm.
Substituting these values into the formula for stress, we get:
$stress = \frac{1960 N}{\pi (5 mm)^2} = 100 N/mm^2$
Therefore, the minimum diameter of the wire so that stress in the wire does not exceed 100 N/mm2 is 5 mm.
The other options are incorrect because they do not result in a stress of 100 N/mm2 or less.
Option A: 4.0 mm. The stress in the wire would be 250 N/mm2, which is greater than the maximum stress of 100 N/mm2.
Option B: 4.5 mm. The stress in the wire would be 200 N/mm2, which is greater than the maximum stress of 100 N/mm2.
Option D: 5.5 mm. The stress in the wire would be 143 N/mm2, which is less than the maximum stress of 100 N/mm2.