To solve this problem, we need to determine when the distance between the two ships becomes shortest.$\\$Let's define the positions of the ships at time $t: \\$• Ship A is moving westwards at $10$ km h$^{-1}. \\$• Ship B is moving northwards at $10$ km h$^{-1}. \\$Initially, ship B is 100 km south of ship A.$\\$At time $t: \\$• The position of ship A is $(-10t, 0). \\$• The position of ship B is $(0, 100 - 10t). \\$The distance $D$ between the ships at time $t$ is given by the distance formula:$\\ D = \sqrt{(-10t - 0)^2 + (0 - (100 - 10t))^2} \\ D = \sqrt{(10t)^2 + (10t - 100)^2} \\ D = \sqrt{100t^2 + (10t - 100)^2} \\ D = \sqrt{100t^2 + (100t^2 - 2000t + 10000)} \\ D = \sqrt{200t^2 - 2000t + 10000} \\$To find the time when the distance is shortest, we need to minimize $D. \\$Since $D$ is a square root function, we can minimize the expression inside the square root:$\\ f(t) = 200t^2 - 2000t + 10000 \\$Take the derivative of $f(t)$ with respect to $t: \\ f'(t) = 400t - 2000 \\$Set the derivative equal to zero to find the critical points:$\\ 400t - 2000 = 0 \\ 400t = 2000 \\ t = \frac{2000}{400} \\ t = 5 \\$Therefore, the time after which the distance between the ships becomes shortest is $5$ h.$\\$This corresponds to Option 1.