A spring of force constant $k$ is cut into lengths of ratio $1:2:3$. They are connected in series and the new

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 $k'$. If they are connected in parallel and force constant is $k''$, then $k':k''$ is:

Accepted Answer

1. **Spring Constant and Length:** The force constant $k$ of a spring is inversely proportional to its length $l$ ($k \propto \frac{1}{l}$) [NCERT Class 11, Physics Part I, Sec 6.9].
2. **Cutting the Spring:** The spring of length $L$ is cut into ratio $1:2:3$. The lengths of the segments are:
   - $l_1 = \frac{1}{6}L \implies k_1 = 6k$
   - $l_2 = \frac{2}{6}L = \frac{1}{3}L \implies k_2 = 3k$
   - $l_3 = \frac{3}{6}L = \frac{1}{2}L \implies k_3 = 2k$
3. **Series Combination ($k'$):** When connected in series, the effective spring constant is given by:
   $$ \frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} $$
   $$ \frac{1}{k'} = \frac{1}{6k} + \frac{1}{3k} + \frac{1}{2k} = \frac{1+2+3}{6k} = \frac{6}{6k} = \frac{1}{k} $$
   $$ k' = k $$ (This is expected as reconnecting parts in series reconstructs the original spring).
4. **Parallel Combination ($k''$):** When connected in parallel, the effective spring constant is the sum of individual constants:
   $$ k'' = k_1 + k_2 + k_3 $$
   $$ k'' = 6k + 3k + 2k = 11k $$
5. **Ratio:**
   $$ \frac{k'}{k''} = \frac{k}{11k} = 1:11 $$

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