The diagram below shows the velocity-time graph of a body. The acceleration of the body in the direction of motion is

The period $T$ of a simple pendulum is given by the formula:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Let $T_1$ be the initial period and $L_1$ be the initial length.
Let $T_2$ be the new period and $L_2$ be the new length.
We have:

$$T_1 = 2\pi \sqrt{\frac{L_1}{g}}$$ $$T_2 = 2\pi \sqrt{\frac{L_2}{g}}$$

Dividing the second equation by the first gives:

$$\frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{L_2}{g}}}{2\pi \sqrt{\frac{L_1}{g}}} = \sqrt{\frac{L_2}{L_1}}$$

Given: $T_1 = 62 \text{ s}$, $L_1 = 40 \text{ cm}$, $T_2 = 72 \text{ s}$.
We need to find $L_2$.

$$\frac{72}{62} = \sqrt{\frac{L_2}{40}}$$

Square both sides:

$$\left(\frac{72}{62}\right)^2 = \frac{L_2}{40}$$ $$L_2 = 40 \times \left(\frac{72}{62}\right)^2$$ $$L_2 = 40 \times (1.16129)^2$$ $$L_2 = 40 \times 1.34859$$ $$L_2 \approx 53.94 \text{ cm}$$

The length is increased by $X$, so $L_2 = L_1 + X$.

$$X = L_2 - L_1$$ $$X = 53.94 \text{ cm} - 40 \text{ cm}$$ $$X = 13.94 \text{ cm}$$

The calculated value $X \approx 13.9 \text{ cm}$ matches option A.

The final answer is $\text{A. } 13.9 \text{ cm}$.