**Problem Analysis:**

In this problem, we are given that a girl is riding a bicycle towards the north direction with a speed of 5 m/s. The rain is falling vertically downwards. When the girl increases her speed to 10 m/s, she observes that the rain appears to meet her at an angle of 45 degrees to the vertical.

We are required to find the speed of the rain.

**Solution:**

Let's assume the speed of rain to be V.

When the girl is riding with a speed of 5 m/s, the relative velocity of the rain with respect to her is the vector sum of their individual velocities. Since the rain is falling vertically downwards, its velocity is in the downward direction. Therefore, the magnitude of the relative velocity is given by:

V1 = √(5^2 + V^2) [Using Pythagoras theorem]

When the girl increases her speed to 10 m/s, the relative velocity of the rain with respect to her is also increased. At this speed, the rain appears to meet her at an angle of 45 degrees to the vertical.

We can use trigonometry to find the vertical component of the relative velocity, Vv, and the horizontal component of the relative velocity, Vh.

Vv = V cos(45)

Vh = V sin(45)

Now, the magnitude of the relative velocity when the girl is riding with a speed of 10 m/s is given by:

V2 = √((10 + Vh)^2 + (Vv)^2)

Substituting the values of Vh and Vv, we get:

V2 = √((10 + V sin(45))^2 + (V cos(45))^2)

Simplifying the equation, we get:

V2 = √(100 + 20V + V^2 + V^2)

V2 = √(2V^2 + 20V + 100)

Squaring both sides of the equation, we get:

V2^2 = 2V^2 + 20V + 100

V2^2 - 2V^2 - 20V - 100 = 0

Simplifying further, we get:

V2^2 - 2V^2 - 20V - 100 = 0

(V2 - 10)(V2 + 10) - 2V(V + 10) = 0

(V2 - 10 - 2V)(V2 + 10) = 0

V2 - 10 - 2V = 0

V2 - 2V - 10 = 0

(V - 5)(V + 2) = 0

V = 5 or V = -2

Since the speed of rain cannot be negative, the speed of rain is 5 m/s.

Therefore, the speed of the rain is 5 m/s.

5√2 m/s