SSC JE Civil Engineering 2013 Solved Paper

Change given option in ans put (c) instead of (a).
Eulers Buckling theory
$\mathit{P}=\frac{{\mathit{\pi }}^2\mathit{E}\mathit{I}}{{\mathit{l}}^2}$
$\mathit{l}$ = effective length of column
Value of effective length = $\mathit{l}$
Both end fix $=\frac{\mathit{l}}{2}$
Both end hinged $=\mathit{l}$
One end fix one is free $=2\mathit{l}$
One-end fix other hinge $=\frac{\mathit{l}}{\sqrt{2}}$
Therefore, for both end fixed eulers buckling load
$\mathit{P}=\frac{{\mathit{\pi }}^2\mathit{E}\mathit{I}}{\left(\mathit{L}∕2\right)}$ $=\frac{4{\mathit{\pi }}^2\mathit{E}\mathit{I}}{{\mathit{l}}^2}$