Solution:
$k \propto 1/L$. The parts have stiffness $6k, 3k, 2k$. In series, $K' = k$. In parallel, $K'' = 6k + 3k + 2k = 11k$. The ratio is $1:11$.
A spring of force constant $k$ is cut into lengths of ratio $1 : 2 : 3$. They are connected in series and the new force constant is $K'$. Then they are connected in parallel and force constant is $K''$. Then $K' : K''$ is: (2017-Delhi)
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