A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is

tan45° = 1
so height = distance of pole from O = 20
Let Bird travels distance x horizontally.
tan30° = 20/(20 + x) = 1/√3
20√3 = 20 + x
x = 20(√3 - 1)
Speed = x/t = 20(√3 - 1)/1 = 20(√3 - 1) m/s
However, this solution is not among the given options. Let's reconsider the problem.
Let the initial distance of the bird from O be x1. Then tan(45) = 20/x1 = 1, so x1 = 20 m.
After 1 second, let the distance of the bird from O be x2. Then tan(30) = 20/x2 = 1/√3, so x2 = 20√3 m.
The horizontal distance covered in 1 second is x2 - x1 = 20√3 - 20 = 20(√3 - 1) m.
The speed of the bird is 20(√3 - 1) m/s. This is not among the options provided. There might be an error in the question or the options.