The magnitude of the force (in newtons) acting on a body varies with time $t$ (in micro seconds) as shown in the fig. $AB$, $BC$ and $CD$ are straight line segments. The magnitude of the total impulse of the force on the body from $t = 4\,\mu s$ to $t = 16\,\mu s$ is ............ N s.

Step 1: Identify the coordinates of the key points from the graph.

The straight line segments $AB$, $BC$, and $CD$ connect the following coordinates $(t, F)$ where time $t$ is in $\mu\text{s}$ and force $F$ is in $\text{N}$:

$$A(0, 200)$$ $$B(4, 200)$$ $$C(6, 800)$$ $$D(16, 0)$$

Step 2: Relate impulse to the area under the force-time graph.

The total impulse $J$ from $t = 4\,\mu\text{s}$ to $t = 16\,\mu\text{s}$ is the area under the $F\text{-}t$ graph over this interval.

$$J = \int_{4\,\mu\text{s}}^{16\,\mu\text{s}} F\,dt$$

This interval corresponds to the area under the segments $BC$ and $CD$.

Step 3: Calculate the area under segment $BC$.

The region under $BC$ from $t = 4\,\mu\text{s}$ to $t = 6\,\mu\text{s}$ forms a trapezoid with parallel sides of heights $200\,\text{N}$ and $800\,\text{N}$, and a width of $2\,\mu\text{s}$.

$$A_{BC} = \frac{1}{2}(200 + 800)(6 - 4)$$ $$A_{BC} = 1000\,\text{N}\cdot\mu\text{s}$$

Step 4: Calculate the area under segment $CD$.

The region under $CD$ from $t = 6\,\mu\text{s}$ to $t = 16\,\mu\text{s}$ forms a triangle with a base of $10\,\mu\text{s}$ and a height of $800\,\text{N}$.

$$A_{CD} = \frac{1}{2}(800)(16 - 6)$$ $$A_{CD} = 4000\,\text{N}\cdot\mu\text{s}$$

Step 5: Sum the areas and convert the units to $\text{N}\cdot\text{s}$.

Add the two areas to find the total impulse in $\text{N}\cdot\mu\text{s}$.

$$J = 1000 + 4000 = 5000\,\text{N}\cdot\mu\text{s}$$

Convert microseconds to seconds using $1\,\mu\text{s} = 10^{-6}\,\text{s}$.

$$J = 5000 \times 10^{-6}\,\text{N}\cdot\text{s}$$ $$J = 5 \times 10^{-3}\,\text{N}\cdot\text{s} = 0.005\,\text{N}\cdot\text{s}$$

Answer: $0.005$ (or $5 \times 10^{-3}$)