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. $1:14$
• B. $1:6$
• C. $1:9$
• D. $1:11$

Answer 1

The correct option is D $1:11$

spring in series
$\frac{1}{k^{\prime }}=\frac{1}{k}+\frac{1}{2k}+\frac{1}{3k}\to k^{\prime }=\frac{6k}{11}$

$k"=k+2k+5k=6k$
$\frac{k^{\prime }}{k^{\prime \prime }}=\frac{\frac{6k}{11}}{6k}$ = $\frac{1}{11}$