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 the force constant is $K^{\prime \prime }$. Then, $\frac{K^{{}^{\prime }}}{K^{\prime \prime }}$ is

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

Answer 1

The correct option is B $1:11$

Let us assume, the length of the spring is $l$. When we cut the spring in the ratio of lengths 1 : 2 : 3, we get three springs of lengths $\frac{l}{6},\frac{2l}{6}$ and $\frac{3l}{6}$ with force constants,
$\therefore K_1=\frac{Kl}{l_1}=\frac{Kl}{\frac{l}{6}}=6K$
$K_2=\frac{Kl}{l_2}=\frac{Kl}{\frac{2l}{6}}=3K$
$K_3=\frac{Kl}{l_3}=\frac{Kl}{\frac{3l}{6}}=2K$
When connected in series,
$\frac{1}{K^{{}^{\prime }}}=\frac{1}{6K}+\frac{1}{3K}+\frac{1}{2K}=\frac{1+2+3}{6K}=\frac{1}{K}$
$\therefore K^{{}^{\prime }}=K$
When connected in parallel,
$K^{\prime \prime }=6K+3K+2K=11K$
$\therefore \frac{K^{{}^{\prime }}}{K^{\prime \prime }}=\frac{K}{11K}=\frac{1}{11}$