Step 1: Euler’s buckling load formula.
\[
P_{cr} = \frac{\pi^2 EI}{(K L)^2}
\]
where
- \(E\) = Young’s modulus,
- \(I\) = area moment of inertia,
- \(L\) = actual column length,
- \(K\) = effective length factor (depends on end conditions).
Step 2: Effective length for each case.
- Cantilever (fixed–free): \(K = 2 \Rightarrow L_{eff} = 2L\).
- Hinged–roller (same as pinned–pinned): \(K = 1 \Rightarrow L_{eff} = L\).
So:
- Option A: \(L_{eff} = 2L\), \(d\).
- Option B: \(L_{eff} = 2L\), \(d\).
- Option C: \(L_{eff} = 2L\), diameter \(2d\).
- Option D: \(L_{eff} = L\), \(d\).
Step 3: Compare inertia effect.
For a circular cross-section:
\[
I = \frac{\pi d^4}{64}
\]
If diameter is doubled (\(2d\)):
\[
I \propto (2d)^4 = 16 d^4
\]
So inertia increases 16 times.
Step 4: Critical load ratios.
- (A) Cantilever, \(I \propto d^4\), denominator \((2L)^2 = 4L^2 \Rightarrow P \propto \frac{d^4}{4L^2}\).
- (B) Length \(2L\), effective length = \(2L \Rightarrow\) denominator \( (2L)^2 = 4L^2 \Rightarrow P \propto \frac{d^4}{4L^2}\).
- (C) Cantilever, \(I \propto 16 d^4\), denominator \(4L^2 \Rightarrow P \propto \frac{16 d^4}{4L^2} = \frac{4 d^4}{L^2}\).
- (D) Hinged–roller, effective length \(L\), denominator \(L^2 \Rightarrow P \propto \frac{d^4}{L^2}\).
Clearly, (C) is 4 times stronger than (D) and much higher than (A) or (B).
Final Answer:
\[
\boxed{\text{(C) Cantilever column with length L and diameter 2d}}
\]