The correct answer is B. 12 m.
The maximum deflection of a simply supported beam is given by the following equation:
$$\delta_{max} = \frac{5wL^4}{384EI}$$
where:
- $\delta_{max}$ is the maximum deflection,
- $w$ is the uniformly distributed load,
- $L$ is the length of the beam,
- $E$ is the Young’s modulus of the material, and
- $I$ is the moment of inertia of the beam.
The slope at the ends of a simply supported beam is given by the following equation:
$$\theta_{max} = \frac{wL^3}{384EI}$$
We can solve the equation for $L$ to get:
$$L = \sqrt{\frac{384EI}{\frac{5wL^4}{384EI} + \theta_{max}}} = \sqrt{\frac{384EI}{\theta_{max}}}$$
We are given that $\delta_{max} = 10$ mm, $\theta_{max} = 0.002$ radian, $E = 200 \times 10^9 \text{ Pa}$, and $I = 1000 \text{ mm}^4$. Substituting these values into the equation for $L$, we get:
$$L = \sqrt{\frac{384 \times 200 \times 10^9 \text{ Pa}}{0.002 \text{ rad}}} = 12 \text{ m}$$
Therefore, the length of the joist is 12 m.
Option A is incorrect because the maximum deflection of a simply supported beam is not equal to 10 mm. Option C is incorrect because the slope at the ends of a simply supported beam is not equal to 0.002 radian. Option D is incorrect because the length of the joist is not equal to 16 m.