Verified answer

I assume you want some indication how to do this rather than just the result – I hope so anyway; it's extra work for me!

Your question is perfectly clear and understandable.

Consider the two points, in the circular cross section, where the water surface meets the tank wall. Draw the radii from the centre O of the tank to these two points, which we name A and B. The angle ABO = arc sin(0.25/0.5) = 30º giving that angle AOB = 120º

The area of the short sector AB = A1 = ½r².θ = ½ x 0.5² x 120 x π/180 = 0.2618 m²

The area of the triangle ABO = A2 = 0.25 x 0.5 x cos30º = 0.1083 m²

The total area of the tank cross section = A3 = 0.5² x π = 0.7854 m²

Hence cross sectional area of water = A3 - A1 + A2 = 0.7854 – 0.2618 + 0.1083 = 0.6319 m²

And finally volume of water = 2 x 0.6319 = 1.2638 m³.

How it is possible to fill water 75 m in horizantal axes with dia 1 metter. It is not possible question is wrong.