**Solution:**

Given:
- Deflection at the free end of a cantilever of rectangular cross-section = 0.8 cm
- Depth of the section is doubled while keeping the weight same

To find:
- Deflection at the free end of the cantilever due to the same loading when the depth is doubled

**Assumptions:**
- The cantilever is loaded with a point load at the free end.
- The material of the cantilever is homogeneous and isotropic.
- The cantilever is under the elastic range of deformation.

**Analysis:**

When the depth of the section is doubled while keeping the weight same, the moment of inertia of the section increases. This leads to an increase in the stiffness of the cantilever, resulting in a smaller deflection.

Let's denote the original depth of the section as 'd' and the new depth as '2d'.

**Step 1: Calculation of Moment of Inertia (I)**

For a rectangular cross-section, the moment of inertia (I) is given by:

I = (b * d³) / 12

Where:
- b = width of the section
- d = depth of the section

**Step 2: Calculation of Stiffness (k)**

The stiffness (k) of the cantilever is given by:

k = 3EI / L³

Where:
- E = modulus of elasticity of the material
- I = moment of inertia of the section
- L = length of the cantilever

**Step 3: Calculation of Deflection (δ)**

The deflection (δ) at the free end of the cantilever is given by:

δ = (WL³) / (3EI)

Where:
- W = applied load at the free end

**Step 4: Comparison of Original and New Deflections**

Let's assume that the applied load (W) is the same for both cases.

For the original cantilever:
δ₁ = (WL₁³) / (3EI₁)

For the new cantilever with doubled depth:
δ₂ = (WL₂³) / (3EI₂)

Since the weight is the same, we can equate W₁L₁³ to W₂L₂³.

From the equation of deflection, we can write:
δ₂ = δ₁ * (I₂ / I₁) * (E₁ / E₂) * (L₁ / L₂)³

Since the weight is the same, L₁ = L₂.

Therefore, the deflection at the free end of the cantilever due to the same loading will be:

δ₂ = δ₁ * (I₂ / I₁) * (E₁ / E₂)