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 $k^{\prime \prime }$. The ratio $k^{\prime }:k^{\prime \prime }$ is

• A. $1:19$
• B. $1:11$
• C. $1:14$
• D. $1:16$

Answer 1

The correct option is B $1:11$

From the stress-strain relation,

Stress = Young's Modulus $\times$ Strain

$\frac{F}{A}=Y\frac{\Delta l}{l}$

$F=\frac{AY}{l}\Delta l$

Comparing with the spring force relation $F=k\Delta l$,
Spring constant $k=\frac{AY}{l}$
$\Rightarrow kl=$ constant

If the spring is cut into three parts with lengths 1 unit, 2 units and 3 units. Total length of orginal spring is 6 units.

Then, $1\left(k_1\right)=2\left(k_2\right)=3\left(k_3\right)=6\left(k\right)$
$\Rightarrow k_1=6k,k_2=3k,k_3=2k$


If the three springs are connected in series,
$\frac{1}{k^{\prime }}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}=\frac{1}{k}$

$\Rightarrow k^{\prime }=k$

If the three springs are connected in parallel,


$k^{\prime \prime }=k_1+k_2+k_3=6k+3k+2k=11k$

Therefore, $k^{\prime }:k^{\prime \prime }=1:11$