Question

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 $k^{\prime }$. Then they are connected in parallel and force constant is $k^{\prime \prime }$. Then $k^{\prime }:k^{\prime \prime }$ is

• A. 1 : 9
• B. 1 : 11
• C. 1 : 14
• D. 1 : 16

Answer 1

The correct option is B 1 : 11

Let the original length of the spring be $l$.
Then length of the spring segments will be $\frac{l}{6}$, $\frac{l}{3}$, $\frac{l}{2}$
As we know $k\propto \frac{1}{l}$
So spring constants for spring segments will be $k_1=6k,k_2=3k,k_3=2k$
For series combination,
$\frac{1}{k^{\prime }}=\frac{1}{6k}+\frac{1}{2k}+\frac{1}{3k}\Rightarrow k^{\prime }=k$
For parallel combination,
$k^{\prime \prime }=6k+3k+2k=11k$
$\frac{k^{\prime }}{k^{\prime \prime }}=\frac{k}{11k}$ = $\frac{1}{11}$