A spring force of constant k is cut into length of ratio 1:2:3. They are connected in series and the new force constant is k’. Then they are connected in parallel and the force constant is k”. Then k’: k” is:

# A spring force of constant k is cut into length of ratio 1:2:3. They are connected in series and the new force constant is k’. Then they are connected in parallel and the force constant is k”. Then k’: k” is:

1. A
1:14
2. B
1:6
3. C
1:9
4. D
1:11

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### Solution:

Concept: We must first determine the stiffness of each spring individually before using the appropriate formula to determine the net stiffness of all the springs in order to calculate the net stiffness or force constant of springs. We shall divide the two situations after determining the net stiffness for each to arrive at the solution.
Assume that the spring has a total length of l and a spring constant of k. The length of the corresponding segments when the spring is divided into pieces with lengths 1:2:3 will be l/6, l/3, and l/2, respectively.
Even after the spring has been divided into several pieces, the product of the spring constant and its length is always constant. Kl is the result of the original spring's length and the spring constant. The many components of this product won't change.
Let spring constant of the springs of length l/6,l/3,l/2 be .
${k}_{1}\frac{l}{6}={k}_{2}\frac{l}{3}={k}_{3}\frac{l}{2}=\mathit{kl}$
We may infer from the equation above that
${k}_{1}=6k,{k}_{2}=3k,{k}_{3}=2k$
The net stiffness (spring constant) of n parallel springs with stiffness k1, k2,..., kn.
${k}_{\mathit{net}}={k}_{1}+{k}_{2}+....+{k}_{n}$
The net stiffness of the spring in this instance with stiffness k1, k2, and k3 is k"
$k\text{'}\text{'}=6k+3k+2k=11\mathit{kk″}=6k+3k+2k=11k\mathrm{}k\text{'}\text{'}=11k$
Net stiffness of n springs k1,k2,….kn
$\frac{1}{k\text{'}}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}=\frac{1}{k}$

Hence, the correct option is 2.

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