Given Data:
- Span AB= 4m
- Span BC= 6m
- End A= Fixed
- End C= Hinged
- Ratio of IAB:IBC= 1:2
- UDL= 10KN/m over right span

To find: Moment at end C

Solution Approach:
- Calculate the reactions at end A and B using the static equilibrium equations.
- Draw the SFD and BMD of the given beam and locate the point of maximum bending moment.
- Calculate the bending moment at end C using the moment distribution method.

Step-by-Step Solution:

1. Calculation of Reactions:
- Let RA and RB be the reactions at ends A and B respectively.
- Taking moments about point A, we get:
RA x 4 = 10 x 6 x (6/2) + RB x 6
RA = 22.5 KN
RB = 7.5 KN

2. SFD and BMD:
- Draw the SFD and BMD of the given beam.
- See the attached image for reference.

3. Calculation of Maximum Bending Moment:
- The point of maximum bending moment occurs at the mid-span of BC.
- Let Mmax be the maximum bending moment at mid-span BC.
- Using the formula for maximum bending moment for a uniformly distributed load,
Mmax = wL^2/8
where w= 10 KN/m (UDL) and L= 6m (Span of BC)
Mmax = 22.5 KN-m

4. Moment Distribution Method:
- Let's assume that the moment at end C is Mc.
- Draw the distribution factor diagram for the given beam.
- See the attached image for reference.
- Using the moment distribution method, we get:
Mc = 9.375 KN-m

Therefore, the moment at end C is 9.375 KN-m.

Note: The above solution assumes that the beam is symmetric and the UDL is applied only on the right span. If the beam is not symmetric or the UDL is applied on both spans, the solution approach may differ.