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 $k'$. Then they are connected in parallel and force constant is $k"$ . Then $k' : k''$ is :

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

Answer 1

Answer: (B)

1 : 11

Answer 2

Spring constant $\propto \frac{1}{\text { length}}$
$k \propto \frac{1}{l}$
i.e, $k_{1}=6 k$
$k_{2}=3 k$
$k_{3}=2 k$
In series
$\frac{1}{k'}=\frac{1}{6 k}+\frac{1}{3 k}+\frac{1}{2 k}$
$\frac{1}{k'}=\frac{6}{6 k}$
$k=k$
$k''=6 k+3 k+2 k$
$k''=11 k$
$\frac{k'}{k''}=\frac{1}{11} \text { i.e } k': k''=1: 11$