A heavy uniform cable of length 100 m is hanging vertically from a...

The elongation $d(\Delta l)$ of a small element of length $dx$ at a distance $x$ from the lower end is given by Hooke's law:
$d(\Delta l) = \dfrac{F(x) dx}{AY}$

The total elongation of the cable is the integral of this expression over its entire length:
$\Delta l = \int_{0}^{L} \dfrac{F(x) dx}{AY} = \dfrac{1}{AY} \int_{0}^{L} F(x) dx$

The integral $\int_{0}^{L} F(x) dx$ represents the area under the given $F-x$ graph. Since the graph is a straight line from $(0, 0)$ to $(100, 4000)$, the area is that of a triangle:
$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 100 \times 4000 = 2 \times 10^5\text{ J}$

Now, substitute the values of $A$ and $Y$:
$AY = (5 \times 10^{-4}) \times (2 \times 10^{11}) = 10 \times 10^7 = 10^8\text{ N}$

Therefore, the total elongation is:
$\Delta l = \dfrac{2 \times 10^5}{10^8} = 2 \times 10^{-3}\text{ m}$

Converting to millimeters:
$\Delta l = 2\text{ mm}$

Answer: $2$