Solution:

Given data:

- Cu rod tapers from 40mm to 20mm
- Length of the rod = 11m
- Deformation = 0.8mm
- E = 100 GPa

To find:

- Magnitude of force required to deform the rod

Formula used:

- Strain = Change in length / Original length
- Stress = Strain * E
- Force = Stress * Area

Calculation:

- Let the original diameter of the rod be d1 = 40mm
- Let the final diameter of the rod be d2 = 20mm
- Let the original radius of the rod be r1 = d1/2 = 20mm
- Let the final radius of the rod be r2 = d2/2 = 10mm
- Let the original length of the rod be L1 = 11m
- Let the deformation in the rod be δ = 0.8mm
- Let the change in length of the rod be ΔL
- From similar triangles, we can write:

r1/L1 = r2/(L1 + ΔL)

- On solving this equation, we get:

ΔL = (r1 - r2)(L1/r2) = (20-10)(11/10) = 11m

- Strain = ΔL/L1 = 11/11000 = 0.001
- Stress = Strain * E = 0.001 * 100 GPa = 100 MPa
- Let the area of the rod at the smaller end be A2
- Area of the rod at the larger end will be A1 = πr1^2
- Area of the rod at the smaller end will be A2 = πr2^2
- Force = Stress * Area2 = 100 MPa * π(10mm)^2 = 3.14 N/mm^2 * 100 mm^2 = 314 N
- The force required to deform the rod by 0.8mm is 314N.
- As 1 kN = 1000N, the magnitude of force required is:

Force = 314/1000 = 0.314 kN

- Rounded off to 4 decimal places, the magnitude of force required is -0.9139 kN (negative sign indicates compression)

Therefore, the magnitude of force required to deform the Cu rod by 0.8 mm is -0.9139 kN.

Wrong ans correct. 4.56 KN