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(A) Let the original length of the spring be $L$. When the spring is cut into lengths in the ratio $1:2:3$,the lengths of the three segments are $l_1

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$1:11$

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(A) Let the original length of the spring be $L$. When the spring is cut into lengths in the ratio $1:2:3$,the lengths of the three segments are $l_1 = L/6$,$l_2 = 2L/6$,and $l_3 = 3L/6$.Since the force constant $k$ is inversely proportional to the length $(k \propto 1/l)$,the new force constants are:$k_1 = k(L/l_1) = 6k$$k_2 = k(L/l_2) = 3k$$k_3 = k(L/l_3) = 2k$When connected in series,the equivalent force constant $k'$ is given by:$1/k' = 1/k_1 + 1/k_2 + 1/k_3 = 1/(6k) + 1/(3k) + 1/(2k) = (1+2+3)/(6k) = 6/(6k) = 1/k$Thus,$k' = k$.When connected in parallel,the equivalent force constant $k''$ is given by:$k'' = k_1 + k_2 + k_3 = 6k + 3k + 2k = 11k$.Therefore,the ratio $k':k'' = k : 11k = 1:11$.

$A$ 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 the force constant is $k''$. Then $k':k''$ is

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