The correct answer is $\boxed{\text{B) 3 cm}}$.
The central dip of the cable is given by the following formula:
$$d = \frac{T}{2\pi E I}$$
where:
- $T$ is the tension in the cable,
- $E$ is the Young’s modulus of the cable,
- $I$ is the moment of inertia of the cable, and
- $d$ is the central dip of the cable.
The Young’s modulus of steel is $200 \times 10^9 \text{ Pa}$, and the moment of inertia of a uniform beam of length $L$ and mass per unit length $\mu$ is given by:
$$I = \frac{1}{12}mL^2$$
In this case, the length of the cable is $L = 20 \text{ m}$ and the mass per unit length is $\mu = 1 \text{ kg}/\text{m}$. Substituting these values into the formula for the moment of inertia gives:
$$I = \frac{1}{12} \times 1 \times 20^2 = 400 \text{ m}^4$$
Substituting all of these values into the formula for the central dip gives:
$$d = \frac{1000}{2\pi \times 200 \times 10^9 \times 400} = 0.03 \text{ m} = 3 \text{ cm}$$